\input{"preamble.tex"}

\addbibresource{Schemes.bib}

\let\Begin\begin
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\title{
\rule{\linewidth}{1pt} \\
\textbf{
    Schemes
  }
    \\ {\normalsize Lectures by Phil Engel. University of Georgia, Fall
2021} \\
  \rule{\linewidth}{2pt}
}
\titlehead{
    \begin{center}
  \includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png}
  \end{center}
       \begin{minipage}{.35\linewidth}
    \begin{flushleft}
      \vspace{2em}
      {\fontsize{6pt}{2pt} \textit{Notes: These are notes live-tex'd
from a graduate course in Schemes taught by Phil Engel at the University
of Georgia in Fall 2021, with material based on Hartshorne. Any errors
or inaccuracies are almost certainly my own. } } \\
    \end{flushleft}
    \end{minipage}
    \hfill
    \begin{minipage}{.65\linewidth}
    \end{minipage}
  }







\begin{document}

\date{}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
  \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2022-11-12 }
\end{flushleft}


\newpage

% Note: addsec only in KomaScript
\addsec{Table of Contents}
\tableofcontents
\newpage

\hypertarget{presheaves-wednesday-august-18}{%
\section{Presheaves (Wednesday, August
18)}\label{presheaves-wednesday-august-18}}

\hypertarget{definitions-and-examples}{%
\subsection{Definitions and Examples}\label{definitions-and-examples}}

\begin{remark}

We'll be covering Hartshorne, chapter 2:

\begin{itemize}
\item
  Sections 1-5: Fundamental, sheaves, schemes, morphisms, constant
  sheaves.
\item
  Sections 6-9: Divisors, linear systems of differentials, nonsingular
  varieties.
\end{itemize}

Note that most of the important material of this book is contained in
the exercises!

\end{remark}

\begin{remark}

Recall that a \textbf{topological space} \(X\) is collection of
\emph{open} sets \({\mathcal{U}}= \left\{{U_i \subseteq X}\right\}\)
which is closed under arbitrary unions and finite intersections, where
\(X, \emptyset\in {\mathcal{U}}\).

\end{remark}

\begin{definition}[Presheaf]

A \textbf{presheaf of abelian groups} \({\mathcal{F}}\) on \(X\) a
topological space is an assignment to every open \(U \subseteq X\) an
abelian group \({\mathcal{F}}(U)\) and restriction morphisms
\(\rho_{UV}: {\mathcal{F}}(U) \to {\mathcal{F}}(V)\) for every inclusion
\(V \subseteq U\) satisfying

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \({\mathcal{F}}(\emptyset) = 0\)
\item
  \(\rho_{UU}: {\mathcal{F}}(U) \to {\mathcal{F}}(U)\) is
  \(\operatorname{id}_{{\mathcal{F}}(U)}\).
\item
  If \(W \subseteq V \subseteq U\) are opens, then
  \begin{align*}
  \rho_{UW} = \rho_{VW} \circ \rho_{UV}
  .\end{align*}
\end{enumerate}

We'll refer to \({\mathcal{F}}(U)\) as the \textbf{sections of
\({\mathcal{F}}\) over \(U\)}, also denoted
\({{\Gamma}\qty{U; {\mathcal{F}}} }\) and write the restrictions as
\({ \left.{{s}} \right|_{{v}} } = \rho_{UV}(s)\) for \(V \subseteq U\).

\end{definition}

\begin{example}[Presheaf of continuous functions]

Let \(X \coloneqq{\mathbf{R}}^1\) with the standard topology and take
\({\mathcal{F}}= C^0({-}; {\mathbf{R}}^1)\) (continuous real-valued
functions) as the associated presheaf. For any open
\(U \subset {\mathbf{R}}\), the group of sections is
\begin{align*}
{\mathcal{F}}(U) \coloneqq\left\{{f: U\to {\mathbf{R}}^1 {~\mathrel{\Big\vert}~}f \text{ is continuous}}\right\}
.\end{align*}

For restriction maps, given \(U \subseteq V\) take the actual
restriction of functions
\begin{align*}
C^0(V; {\mathbf{R}}^1) &\to C^0(U; {\mathbf{R}}^1) \\
f &\mapsto { \left.{{f}} \right|_{{U}} }
.\end{align*}
We can declare
\(C^0(\emptyset; {\mathbf{R}}^1) = \left\{{0}\right\} = 0\in {\mathsf{Grp}}\),
and the remaining conditions in the definition above follow immediately.

\end{example}

\hypertarget{constant-presheaves}{%
\subsection{Constant Presheaves}\label{constant-presheaves}}

\begin{definition}[Constant presheaves]

The \textbf{constant presheaf} associated to \(A\in {\mathsf{Ab}}\) on
\(X\in {\mathsf{Top}}\) is denoted \(\underline{A}\), where
\begin{align*}
\underline{A}(U) \coloneqq
\begin{cases}
A &  U \neq \emptyset
\\
0 & U = \emptyset.
\end{cases}
\end{align*}
and
\begin{align*}
\rho_{UV} \coloneqq
\begin{cases}
\operatorname{id}_A & V \neq \emptyset 
\\
0 & V=\emptyset .
\end{cases}
.\end{align*}

\end{definition}

\begin{warnings}

The constant sheaf is not the sheaf of constant functions! Instead these
are \emph{locally} constant functions.

\end{warnings}

\begin{remark}

Let \({\mathsf{Open}}_{/ {X}}\) denote the category of open sets of
\(X\), defined by

\begin{itemize}
\tightlist
\item
  Objects:
  \({\operatorname{Ob}}({\mathsf{Open}}_{/ {X}} ) \coloneqq\left\{{U_i}\right\}\),
  so each object is an open set.
\item
  Morphisms:
  \begin{align*}
  {\mathsf{Open}}_{/ {X}} (U, V) \coloneqq
  \begin{cases}
  \emptyset & V \not\subset U 
  \\
  \text{The singleton } \left\{{U \xhookrightarrow{\iota} V}\right\} & \text{otherwise}.
  \end{cases}
  .\end{align*}
\end{itemize}

\end{remark}

\begin{example}[Of $\Open\slice{X}$ ]

Take \(X\coloneqq\left\{{p, q}\right\}\) with the discrete topology to
obtain a category with 4 objects:

\begin{center}
\begin{tikzcd}
    & {\left\{{p, q}\right\}} \\
    {\left\{{p}\right\}} && {\left\{{q}\right\}} \\
    & \emptyset
    \arrow[from=3-2, to=1-2]
    \arrow[from=3-2, to=2-1]
    \arrow[from=3-2, to=2-3]
    \arrow[from=2-1, to=1-2]
    \arrow[from=2-3, to=1-2]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMSwwLCJcXHRze3AsIHF9Il0sWzAsMSwiXFx0c3twfSJdLFsyLDEsIlxcdHN7cX0iXSxbMSwyLCJcXGVtcHR5c2V0Il0sWzMsMF0sWzMsMV0sWzMsMl0sWzEsMF0sWzIsMF1d}{Link
to Diagram}
\end{quote}

Similarly, the indiscrete topology yields
\(\emptyset \to \left\{{p, q}\right\}\), a category with two objects.

\end{example}

\begin{remark}

A presheaf is a contravariant functor
\({\mathcal{F}}: {\mathsf{Open}}_{/ {X}} \to {\mathsf{Ab}}\) which sends
the cofinal/initial object
\(\mathsf{\emptyset} \in {\mathsf{Open}}_{/ {X}}\) to the final/terminal
object \(\left\{{{\operatorname{pt}}}\right\}\in {\mathsf{Ab}}\). More
generally, we can replace \({\mathsf{Ab}}\) with any category
\(\mathsf{C}\) admitting a final object:

\begin{itemize}
\tightlist
\item
  \(\mathsf{C} \coloneqq\mathsf{CRing}\) the category of commutative
  rings, which we'll use to define schemes.
\item
  \(\mathsf{C} = {\mathsf{Grp}}\), the category of (potentially
  nonabelian) groups.
\item
  \(\mathsf{C} \coloneqq{\mathsf{Top}}\), the category of (arbitrary)
  topological spaces.
\end{itemize}

\end{remark}

\begin{example}[of presheaves]

Let \(X\in {\mathsf{Var}}_{/ {k}}\) a variety over
\(k\in \mathsf{Field}\) equipped with the Zariski topology, so the opens
are complements of vanishing loci. Given \(U \subseteq X\), define a
presheaf of regular functions \({\mathcal{F}}\coloneqq{\mathcal{O}}\)
where

\begin{itemize}
\item
  \({\mathcal{O}}(U)\) are the regular functions \(f:U\to k\),
  i.e.~functions on \(U\) which are locally expressible as a ratio
  \(f = g/h\) with \(g, h\in k[x_1, \cdots, x_{n}]\).
\item
  Restrictions are restrictions of functions.
\end{itemize}

Taking \(X = {\mathbf{A}}^1_{/ {k}}\), the Zariski topology is the
cofinite topology, so every open \(U\) is the complement of a finite set
and \(U = \left\{{t_1, \cdots, t_m}\right\}^c\). Then
\({\mathcal{O}}(U) = \left\{{\phi: U\to k}\right\}\) which is locally a
fraction, and it turns out that these are all globally fractions and
thus
\begin{align*}
{\mathcal{O}}(U) 
&= \left\{{ {f(t) \over g(t)} {~\mathrel{\Big\vert}~}f,g\in k[t], \quad g(t) \neq 0 \,\,\, \forall t\in U}\right\} \\
&= \left\{{{ f(t) \over  \prod_{i=1}^m (t-t_i)^{m_i}} {~\mathrel{\Big\vert}~}f\in k[t] }\right\} \\
&= k[t] \left[ { \scriptstyle { {S}^{-1}}  } \right] 
,\end{align*}
where \(S = \left\langle{\prod_{i=1}^m t-t_i}\right\rangle\) is the
multiplicative set generated by the factors. This forms an abelian group
since we can take least common denominators, and we have restrictions.

\end{example}

\begin{warnings}

Note that there are two similar notations for localization which mean
different things! For a multiplicative set \(S\), the ring
\(R \left[ { \scriptstyle { {S}^{-1}} } \right]\) literally means
localizing at that set. For \({\mathfrak{p}}\in \operatorname{Spec}R\),
the ring \(R \left[ { \scriptstyle { {{\mathfrak{p}}}^{-1}} } \right]\)
means localizing at the multiplicative set
\(S \coloneqq{\mathfrak{p}}^c\).

\end{warnings}

\hypertarget{sheaves-stalks-local-rings-friday-august-20}{%
\section{Sheaves, Stalks, Local Rings (Friday, August
20)}\label{sheaves-stalks-local-rings-friday-august-20}}

\hypertarget{sheaves}{%
\subsection{Sheaves}\label{sheaves}}

\begin{definition}[Sheaf]

Recall the definition of a presheaf, and the main 3 properties:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \({\mathcal{F}}( \mathsf{\emptyset} ) = \left\{{{\operatorname{pt}}}\right\}\)
  where \(\left\{{{\operatorname{pt}}}\right\}= 0 \in {\mathsf{Ab}}\),
\item
  \(\rho_{UU} = \operatorname{id}_{{\mathcal{F}}(U)}\)
\item
  For all \(W \subseteq V \subseteq U\), a cocycle condition:
  \begin{align*}
  \rho_{UW} = \rho_{VW} \circ \rho_{UV}
  .\end{align*}
  Write \(s_i \in {\mathcal{F}}(U_i)\) to be a section.

  A presheaf is a \textbf{sheaf} if it additionally satisfies
\item
  When restrictions are compatible on overlaps, so
  \begin{align*}
  { \left.{{s_i}} \right|_{{U_i \cap U_j}} } = { \left.{{s_j}} \right|_{{U_i \cap U_j}} }
  ,\end{align*}
  there exists a uniquely glued section \({\mathcal{F}}(\cup U_i)\) such
  that \({ \left.{{s}} \right|_{{U_i}} } = s_i\) for all \(i\).
\end{enumerate}

\end{definition}

\begin{example}[?]

Take \(C^0({-}; {\mathbf{R}})\) the sheaf of continuous real-valued
functions on a topological space. For \(f_i: U_i \to {\mathbf{R}}\)
agreeing on overlaps, there is a continuous function
\(f: \cup U_i\to {\mathbf{R}}\) restricting to \(f_i\) on each \(U_i\)
by just defining \(f(x) \coloneqq f_i(x)\) for \(x\in U_i\) and
assembling these into a piecewise function, which is well-defined by
agreement of the \(f_i\) on overlaps.

\end{example}

\begin{example}[A presheaf which is not a sheaf]

Let \(X\) be a topological space and \(A\in \mathsf{CRing}\), then take
the constant sheaf
\begin{align*}
\underline{A}(U) \coloneqq
\begin{cases}
A & U\neq \emptyset 
\\
0 & \text{else}.
\end{cases}
.\end{align*}
This is not a sheaf -- let \(X = {\mathbf{R}}\) and
\(A = {\mathbf{Z}}/2\), let \(U_1 = (0, 1)\) and \(U_2 = (2, 3)\), and
take \(s_1 = 0\) on \(U_1\) and \(s_2 = 1\) on \(U_2\). Since
\(U_1 \cap U_2 = \emptyset\), the sections trivially agree on overlaps,
but there is no constant function on \(U_1 \cup U_2\) restricting to 1
on \(U_2\) and 0 on \(U_1\)

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{26pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2022-11-11_22-25.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{example}

\begin{definition}[Locally constant sheaves]

The \textbf{(locally) constant sheaf} \(\underline{A}\) on any
\(X\in {\mathsf{Top}}\) is defined as
\begin{align*}
\underline{A}(U) \coloneqq\left\{{ f: U\to A {~\mathrel{\Big\vert}~}f \text{ is locally constant} }\right\}
.\end{align*}

\end{definition}

\begin{remark}

As a general principle, this is a sheaf since the defining property can
be verified locally.

\end{remark}

\begin{example}[?]

Let \(C^0_{\mathrm{bd}}\) be the presheaf of bounded continuous
functions on \(S^1\). This is not a sheaf, but one needs to go to
infinitely many sets: take the image of \([{1\over n}, {1\over n+1}]\)
with (say) \(f_n(x) = n\) for each \(n\). Then each \(f_n\) is bounded
(it's just constant), but the full collection is unbounded, so these can
not glue to a bounded function.

\end{example}

\hypertarget{stalks-and-local-rings}{%
\subsection{Stalks and Local Rings}\label{stalks-and-local-rings}}

\begin{definition}[Stalks]

Let \({\mathcal{F}}\in \underset{ \mathsf{pre} } {\mathsf{Sh} }(X)\) and
\(p\in X\), then the \textbf{stalk} of \({\mathcal{F}}\) at \(p\) is
defined as
\begin{align*}
{\mathcal{F}}_p(U) \coloneqq\lim_{U\ni p} \coloneqq\left\{{(s, U) {~\mathrel{\Big\vert}~}U\ni p \text{ open}, \, s\in {\mathcal{F}}(U)}\right\}/{_{\scriptstyle / \sim} }
,\end{align*}
where \((s, U) \sim (t ,V)\) iff there exists a \(W \ni p\) with
\(W \subset U \cap V\) with
\({ \left.{{s}} \right|_{{W}} } = { \left.{{t}} \right|_{{W}} }\). An
equivalence class \([(s, U)] \in {\mathcal{F}}_p\) is referred to as a
\textbf{germ}.

\end{definition}

\begin{example}[Stalks of sheaves of analytic functions]

Let \(C^\omega({-}; {\mathbf{R}})\) be the sheaf of analytic functions,
i.e.~those locally expressible as convergent power series. This is a
sheaf because this condition can be checked locally. What is the stalk
\(C_0^\omega\) at zero? An example of a function in this germ is
\([(f(x) = {1\over 1-x}, (-1, 1))\). A first guess is
\({\mathbf{R}} {\llbracket t \rrbracket }\), but the claim is that this
won't work.

Note that there is an injective map
\(C_0^\omega \hookrightarrow{\mathbf{R}} {\llbracket t \rrbracket }\)
because \(f, g\) have analytic power series expansions at zero, and if
these expressions are equal then
\({ \left.{{f}} \right|_{{I}} } = { \left.{{g}} \right|_{{I}} }\) for
some \(I\) containing zero. This map won't be surjective because there
are power series with a non-positive radius of convergence, for example
taking \(f(t) \coloneqq\sum_{k=0}^\infty {kt}^k\) which only converges
at \(t=0\). So the answer is that
\(C_0^\omega \leq {\mathbf{R}} {\llbracket t \rrbracket }\) is the
subring of power series with positive radius of convergence.

\end{example}

\begin{definition}[Local ring of the structure sheaf, $\OO_p$]

Let \(X \in \mathsf{Alg} {\mathsf{Var}}\) and \({\mathcal{O}}\) its
sheaf of regular functions. For \(p\in X\), the stalk
\({\mathcal{O}}_p\) is the \textbf{local ring} of \(X\) at \(p\).

\end{definition}

\begin{example}[Local rings of affine space]

For \(X \coloneqq{\mathbf{A}}^1_{/ {k}}\) for
\(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\), the
opens are cofinite sets and
\({\mathcal{O}}(U) = \left\{{f/g {~\mathrel{\Big\vert}~}f, g\in k[t]}\right\}\).
Consider the stalk \({\mathcal{O}}_p\) for some fixed
\(p\in {\mathbf{A}}^1_{/ {k}}\). Applying the definition, we have
\begin{align*}
{\mathcal{O}}_p \coloneqq\left\{{(f/g, U) {~\mathrel{\Big\vert}~}p\in U,\, g\neq 0 \text { on } U}\right\} / {_{\scriptstyle / \sim} }
.\end{align*}
Given any \(g\in k[t]\) with \(g(p) \neq 0\), there is a Zariski open
set \(U = V(g)^c = D_g\), the distinguished open associated to \(g\),
where \(g\neq 0\) on \(U\) by definition. Thus \(p\in U\), and so any
\(f/g\in \operatorname{ff}(k[t]) = k { \left( {t} \right) }\) with
\(p\neq 0\) defines an element \((f/g, D_g) \in {\mathcal{O}}_p\).
Concretely:
\begin{align*}
{ \left.{{f/g}} \right|_{{W}} } = { \left.{{f/g}} \right|_{{W'}} } \implies f/g = f'/g' \in k(t)
,\end{align*}
and \(fg' = f'g\) on the cofinite set \(W\), making them equal as
polynomials. We can thus write
\begin{align*}
{\mathcal{O}}_p = \left\{{f/g \in k(t) {~\mathrel{\Big\vert}~}g(p) \neq 0}\right\} = k[t] \left[ { \scriptstyle { {\left\langle{t-p}\right\rangle}^{-1}}  } \right] , \quad \left\langle{t-p}\right\rangle\in \operatorname{mSpec}k[t]
,\end{align*}
recalling that
\(k[t] \left[ { \scriptstyle { {{\mathfrak{p}}}^{-1}} } \right] \coloneqq\left\{{f/g {~\mathrel{\Big\vert}~}f,g\in k[t],\,\, g\not\in {\mathfrak{p}}}\right\}\).

\end{example}

\begin{remark}

Note that for \(X\in {\mathsf{Aff}}{\mathsf{Var}}\), writing
\(X = V(f_i) = V(I)\) for a radical ideal \(I\), we have the coordinate
ring
\begin{align*}
k[X] \coloneqq k[x_1, \cdots, x_{n}]/I = R
\implies 
{\mathcal{O}}_p = R \left[ { \scriptstyle { {{\mathfrak{m}}_p}^{-1}}  } \right] ,\quad {\mathfrak{m}}_p \coloneqq\left\{{f\in R {~\mathrel{\Big\vert}~}f(p) = 0}\right\}
.\end{align*}
We thus have the following:

\end{remark}

\begin{slogan}

The local ring at \(p\) is the localization at the maximal ideal of all
functions in the coordinate ring vanishing at \(p\).

\end{slogan}

\begin{warnings}

This doesn't quite hold for non-algebraically closed fields:
\begin{align*}
f(x) \coloneqq x^p-x \in { \mathbf{F} }_p[x] \implies f(x) = 0 \quad \forall x\in { \mathbf{F} }_p \implies f \equiv 0 \in { \mathbf{F} }_p[x]
.\end{align*}

\begin{quote}
DZG: I missed something here, so I'm not sure \textbf{what} isn't
supposed to hold!
\end{quote}

\end{warnings}

\begin{remark}

Next time: morphisms of sheaves/presheaves, and isomorphisms of sheaves
can be checked on stalks.

\end{remark}

\hypertarget{more-sheaves-monday-august-23}{%
\section{More Sheaves (Monday, August
23)}\label{more-sheaves-monday-august-23}}

\hypertarget{morphisms-of-presheaves}{%
\subsection{Morphisms of Presheaves}\label{morphisms-of-presheaves}}

\begin{remark}

Recall that the \textbf{stalk} of a presheaf \({\mathcal{F}}\) at \(p\)
is defined as
\begin{align*}
{\mathcal{F}}_p \coloneqq\colim_{U\ni p} {\mathcal{F}}(U) = \left\{{ (s, U) {~\mathrel{\Big\vert}~}s\in {\mathcal{F}}(U) }\right\}_{/ {\sim}} 
.\end{align*}

\end{remark}

\begin{definition}[Morphisms of presheaves]

Let
\({\mathcal{F}}, {\mathcal{G}}\in \underset{ \mathsf{pre} } {\mathsf{Sh} }(X)\),
then a \textbf{morphism} \(\phi: {\mathcal{F}}\to {\mathcal{G}}\) is a
collection
\(\left\{{\phi(U): {\mathcal{F}}(U) \to {\mathcal{G}}(U)}\right\}\) of
morphisms of abelian groups for all \(U\in {\mathsf{Open}}(X)\) such
that for all \(V \subset U\), the following diagram commutes:

\begin{center}
\begin{tikzcd}
    {{\mathcal{F}}(U)} && {{\mathcal{G}}(U)} \\
    \\
    {{\mathcal{F}}(V)} && {{\mathcal{G}}(V)}
    \arrow["{\phi(U)}", from=1-1, to=1-3]
    \arrow["{\phi(V)}", from=3-1, to=3-3]
    \arrow["{\operatorname{res}(UV)}"{description}, from=1-1, to=3-1]
    \arrow["{\operatorname{res}'(UV)}"{description}, from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXG1jZihVKSJdLFswLDIsIlxcbWNmKFYpIl0sWzIsMCwiXFxtY2coVSkiXSxbMiwyLCJcXG1jZyhWKSJdLFswLDIsIlxccGhpKFUpIl0sWzEsMywiXFxwaGkoVikiXSxbMCwxLCJcXHJlcyhVVikiLDFdLFsyLDMsIlxccmVzJyhVVikiLDFdXQ==}{Link
to Diagram}
\end{quote}

An \textbf{isomorphism} is a morphism with a two-sided inverse.

\end{definition}

\begin{remark}

Note that if we regard a sheaf as a contravariant functor, a morphism is
then just a natural transformation.

\end{remark}

\begin{remark}

A morphism \(\phi: {\mathcal{F}}\to {\mathcal{G}}\) defines a morphisms
on stalks \(\phi_p: {\mathcal{F}}_p \to {\mathcal{G}}_p\).

\end{remark}

\begin{example}[of a nontrivial morphism of sheaves]

Let \(X \coloneqq{\mathbf{C}}^{\times}\) with the classical topology,
making it into a real manifold, and take
\(C^0({-}; {\mathbf{C}}) \in {\mathsf{Sh}}(X, {\mathsf{Ab}})\) be the
sheaf of continuous functions and let
\(C^0({-}; {\mathbf{C}})^{\times}\) the sheaf of of nowhere zero
continuous continuous functions. Note that this is a sheaf of abelian
groups since the operations are defined pointwise. There is then a
morphism
\begin{align*}
\exp({-}): C^0({-}; {\mathbf{C}}) &\to C^0({-}; {\mathbf{C}})^{\times}\\
f &\mapsto e^f && \text{ on open sets } U\subseteq X
.\end{align*}

Since exponentiating and restricting are operations done pointwise, the
required square commutes, yielding a morphism of sheaves.

\end{example}

\hypertarget{kernel-and-cokernel-sheaves}{%
\subsection{Kernel and cokernel
sheaves}\label{kernel-and-cokernel-sheaves}}

\begin{definition}[(co)kernel and image sheaves]

Let \(\phi: {\mathcal{F}}\to {\mathcal{G}}\) be morphisms of presheaves,
then define the presheaves
\begin{align*}
\ker(\phi)(U) &\coloneqq\ker(\phi(U)) \\
\operatorname{coker}^{{\mathsf{pre}}}(\phi)(U) &\coloneqq{\mathcal{G}}(U) / \phi({\mathcal{F}}(U))\\ 
\operatorname{im}(\phi)(U) &\coloneqq\operatorname{im}(\phi(U)) \\
.\end{align*}

\end{definition}

\begin{warnings}

If \({\mathcal{F}}, {\mathcal{G}}\in {\mathsf{Sh}}(X)\), then for a
morphism \(\phi: {\mathcal{F}}\to {\mathcal{G}}\), the image and
cokernel presheaves need not be sheaves!

\end{warnings}

\begin{example}[of why the cokernel presheaf is not a sheaf]

Consider \(\ker \exp\) where
\begin{align*}
\exp: C^0({-}; {\mathbf{C}})\to C^0({-}; {\mathbf{C}})^{\times}\qquad \in {\mathsf{Sh}}({\mathbf{C}}^{\times})
.\end{align*}
One can check that \(\ker \exp = 2\pi i \underline{{\mathbf{Z}}}(U)\),
and so the kernel is actually a sheaf. We also have
\begin{align*}
\operatorname{coker}^{{\mathsf{pre}}} \exp(U) \coloneqq{ C^0(U; {\mathbf{C}})\over  \exp(C^0(U;{\mathbf{C}})^{\times}) }
.\end{align*}
On opens,
\(\operatorname{coker}^{{\mathsf{pre}}} \exp(U) = \left\{{1}\right\} \iff\)
every nonvanishing continuous function \(g\) on \(U\) has a continuous
logarithm, i.e.~\(g = e^f\) for some \(f\). Examples of opens with this
property include any contractible (or even just simply connected) open
set in \({\mathbf{C}}^{\times}\). Consider
\(U\coloneqq{\mathbf{C}}^{\times}\) and
\(z\in C^0({\mathbf{C}}^{\times}; {\mathbf{C}})^{\times}\), which is a
nonvanishing function. Then the equivalence class
\([z] \in \operatorname{coker}^{{\mathsf{pre}}} \exp({\mathbf{C}}^{\times})\)
is nontrivial -- note that \(z\neq e^f\) for any
\(f\in C^0({\mathbf{C}}^{\times}; {\mathbf{C}})\), since any attempted
definition of \(\log(z)\) will have monodromy.

On the other hand, we can cover \({\mathbf{C}}^{\times}\) by
contractible opens \(\left\{{U_i}\right\}_{i\in I}\) where
\({ \left.{{[z]}} \right|_{{U_i}} } = 1 \in \operatorname{coker}^{{\mathsf{pre}}} \exp (U_i)\)
and similarly
\({ \left.{{1}} \right|_{{\operatorname{id}}} } = 1 \in \operatorname{coker}^{{\mathsf{pre}}} \exp(U_i)\),
showing that the cokernel fails the unique gluing axiom and is not a
sheaf.

\end{example}

\hypertarget{sheafification}{%
\subsection{Sheafification}\label{sheafification}}

\begin{definition}[Sheafification]

Given any
\({\mathcal{F}}\in \underset{ \mathsf{pre} } {\mathsf{Sh} }(X)\) there
exists an \({\mathcal{F}}^+ \in {\mathsf{Sh}}(X)\) and a morphism of
presheaves \(\theta: {\mathcal{F}}\to {\mathcal{F}}^+\) such that for
any \({\mathcal{G}}\in {\mathsf{Sh}}(X)\) with a morphism
\(\phi: {\mathcal{F}}\to {\mathcal{G}}\) there exists a unique
\(\psi: {\mathcal{F}}^+ \to {\mathcal{G}}\) making the following diagram
commute:

\begin{center}
\begin{tikzcd}
    {\mathcal{F}}&& {\mathcal{G}}\\
    \\
    && {{\mathcal{F}}^+}
    \arrow["\theta"', from=1-1, to=3-3]
    \arrow["\phi", from=1-1, to=1-3]
    \arrow["{\exists! \psi}"', from=3-3, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXG1jZiJdLFsyLDIsIlxcbWNmXisiXSxbMiwwLCJcXG1jZyJdLFswLDEsIlxcdGhldGEiLDJdLFswLDIsIlxccGhpIl0sWzEsMiwiXFxleGlzdHMhIFxccHNpIiwyXV0=}{Link
to Diagram}
\end{quote}

The sheaf \({\mathcal{F}}^+ \in {\mathsf{Sh}}(X)\) is called the
\textbf{sheafification} of \({\mathcal{F}}\). This is an example of an
adjunction of functors:
\begin{align*}
\mathop{\mathrm{Hom}}_{ \underset{ \mathsf{pre} } {\mathsf{Sh} }(X)}({\mathcal{F}}, {\mathcal{G}}^{\mathsf{pre}}) \cong \mathop{\mathrm{Hom}}_{{\mathsf{Sh}}(X)}({\mathcal{F}}^+, {\mathcal{G}})
,\end{align*}
where we use the forgetful functor
\({\mathcal{G}}\to {\mathcal{G}}^{\mathsf{pre}}\). This can be expressed
as the adjoint pair
\begin{align*}
\adjunction{({-})^+}{({-})^{\mathsf{pre}}}{ \underset{ \mathsf{pre} } {\mathsf{Sh} }(X)}{{\mathsf{Sh}}(X)}
.\end{align*}

\end{definition}

\begin{proof}[of existence of sheafification]

We construct it directly as
\({\mathcal{F}}^+ \coloneqq\left\{{s:U \to {\textstyle\coprod}_{p\in U} {\mathcal{F}}_p }\right\}\)
such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(s(p) \in {\mathcal{F}}_p\),
\item
  The germs are compatible locally, so for all \(p\in U\) there is a
  \(V\supseteq p\) such that for some \(t\in {\mathcal{F}}(V)\),
  \(s(p) = t_p\) for all \(p\) in \(V\).
\end{enumerate}

So about any point, there should be an actual function specializing to
all germs in an open set.

\end{proof}

\begin{slogan}

The sheafification is constructed from collections of germs which are
locally compatible.

\end{slogan}

\begin{remark}

This process will make \(\operatorname{coker}\exp\) zero as a sheaf,
since it will be zero on a sufficiently small set.

\end{remark}

\hypertarget{exactness-for-sheaves-wednesday-august-25}{%
\section{Exactness for Sheaves (Wednesday, August
25)}\label{exactness-for-sheaves-wednesday-august-25}}

\hypertarget{some-examples}{%
\subsection{Some examples}\label{some-examples}}

\begin{remark}

Recall the definition of sheafification: let
\({\mathcal{F}}\in \underset{ \mathsf{pre} } {\mathsf{Sh} }(X; {\mathsf{Ab}}{\mathsf{Grp}})\).
Construct a sheaf
\({\mathcal{F}}^+\in {\mathsf{Sh}}(X, {\mathsf{Ab}}{\mathsf{Grp}})\) and
a morphism \(\theta: {\mathcal{F}}\to {\mathcal{F}}^+\) of presheaves
satisfying the appropriate universal property:

\begin{center}
\begin{tikzcd}
    {{\mathcal{F}}^+} \\
    \\
    {\mathcal{F}}&& {\mathcal{G}}\\
    \\
    {}
    \arrow["\psi", from=3-1, to=3-3]
    \arrow["\theta", from=3-1, to=1-1]
    \arrow["{\exists \tilde \psi}", dashed, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJcXG1jZiJdLFswLDRdLFswLDAsIlxcbWNmXisiXSxbMiwyLCJcXG1jZyJdLFswLDMsIlxccHNpIl0sWzAsMiwiXFx0aGV0YSJdLFsyLDMsIlxcZXhpc3RzIFxcdGlsZGUgXFxwc2kiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=}{Link
to Diagram}
\end{quote}

So any presheaf morphism to a sheaf factors through the sheafification
uniquely (via \(\theta\)). Note that this is a instance of a general
free/forgetful adjunction.

We can construct it as
\begin{align*}
{\mathcal{F}}^+(U) \coloneqq\left\{{s:U\to {\textstyle\coprod}_{p\in U} {\mathcal{F}}_p,\quad s(p) \in {\mathcal{F}}_p, \cdots}\right\}
.\end{align*}
where the addition condition is that for all \(q\in U\) there exists a
\(V\nu q\) and \(t\in {\mathcal{F}}(V)\) such that \(t_p = s(p)\) for
all \(p\in V\). Note that \(\theta\) is defined by
\(\theta(U)(s) = \left\{{s:p\to s_p}\right\}\), the function assigning
points to germs with respect to the section \(s\). Idea: this is like
replacing an analytic function on an interval with the function sending
a point \(p\) to its power series expansion at \(p\).

\end{remark}

\begin{example}[?]

Recall \(\exp: C^0 \to (C^0)^{\times}\) on \({\mathbf{C}}^{\times}\),
then
\(\operatorname{coker}^{\mathsf{pre}}(\exp)(U) = \left\{{1}\right\}\) on
contractible \(U\), using that one can choose a logarithm on such a set.
However
\(\operatorname{coker}^{\mathsf{pre}}(\exp)({\mathbf{C}}^{\times}) \neq \left\{{1}\right\}\)
since
\([z]\in (C^0)^{\times}({\mathbf{C}}^{\times})/\exp(C^0({\mathbf{C}}^{\times}))\).

\end{example}

\begin{remark}

Letting \(\phi: {\mathcal{F}}\to {\mathcal{G}}\) be a morphisms of
sheaves, then we defined
\(\operatorname{coker}(\phi) \coloneqq(\operatorname{coker}^{\mathsf{pre}}(\phi))^+\)
and
\(\operatorname{im}(\phi) \coloneqq(\operatorname{im}^{\mathsf{pre}}(\phi))^+\).
Then
\begin{align*}
\operatorname{coker}^{\mathsf{pre}}(\exp) &\to \operatorname{coker}(\exp) \\
s\in {\mathcal{F}}(U) &\mapsto s(p) = s_p
.\end{align*}
The claim is that \([z]_p = 1\) for all \(p\in {\mathbf{C}}^{\times}\),
since we can replace \([([z], {\mathbf{C}}^{\times})]\) with
\(([z]_U, U)\) for \(U\) contractible.

\end{remark}

\begin{example}[?]

A useful example to think about: \(X = \left\{{p, q}\right\}\) with

\begin{itemize}
\tightlist
\item
  \({\mathcal{F}}(p) = A\)
\item
  \({\mathcal{F}}(q) = B\)
\item
  \({\mathcal{F}}(X) = 0\)
\end{itemize}

Then local sections don't glue to a global section, so this isn't a
sheaf, but it is a presheaf. The sheafification satisfies
\({\mathcal{F}}^+(X) = A\times B\).

\end{example}

\hypertarget{subsheaves}{%
\subsection{Subsheaves}\label{subsheaves}}

\begin{definition}[Subsheaves, injectivity, surjectivity]

\({\mathcal{F}}'\) is a \textbf{subsheaf} of \({\mathcal{F}}\) if

\begin{itemize}
\tightlist
\item
  \({\mathcal{F}}'(U) \leq {\mathcal{F}}(U)\) for all \(U\),
\item
  \(\mathop{\mathrm{Res}}'(U, V) = { \left.{{ \mathop{\mathrm{Res}}(U, V) }} \right|_{{{\mathcal{F}}'(U)}} }\).
\end{itemize}

\(\phi: {\mathcal{F}}\to {\mathcal{G}}\) is \textbf{injective} iff
\(\ker \phi = 0\), \textbf{surjective} if
\(\operatorname{im}(\phi) = {\mathcal{G}}\) or
\(\operatorname{coker}\phi = 0\).

\end{definition}

\begin{exercise}[?]

Check that \(\ker \phi\) already satisfies the sheaf property.

\end{exercise}

\hypertarget{exact-sequences-of-sheaves}{%
\subsection{Exact Sequences of
sheaves}\label{exact-sequences-of-sheaves}}

\begin{definition}[Exact sequences of sheaves]

Let
\(\cdots \to {\mathcal{F}}^{i-1} \xrightarrow{\phi^{i-1}} {\mathcal{F}}^i \xrightarrow{\phi^i} {\mathcal{F}}^{i+1}\to \cdots\)
be a sequence of morphisms in \({\mathsf{Sh}}(X)\), this is
\textbf{exact} iff \(\ker \phi^i = \operatorname{im}\phi^{i-1}\).

\end{definition}

\begin{lemma}[?]

\(\ker \phi\) is a sheaf.

\end{lemma}

\begin{proof}[?]

By definition,
\(\ker(\phi)(U) \coloneqq\ker \qty{ \phi(U): {\mathcal{F}}(U) \to {\mathcal{G}}(U) }\),
satisfying part (a) in the definition of presheaves. We can define
restrictions
\({ \left.{{\mathop{\mathrm{Res}}(U, V)}} \right|_{{\ker(\phi)(U)}} } \subseteq \ker(\phi)(V)\).
Use the commutative diagram for the morphism
\(\phi: {\mathcal{F}}\to {\mathcal{G}}\).

Now checking gluing: Let \(s_i \in \ker(\phi)(U_i)\) such that
\(\mathop{\mathrm{Res}}(s_i, U_i \cap U_j) = \mathop{\mathrm{Res}}(s_j, U_i \cap U_j)\)
for all \(i, j\). This holds by viewing \(s_i \in {\mathcal{F}}(U_i)\),
so \(\exists ! s\in {\mathcal{F}}(\bigcup_i U_i)\) such that
\(\mathop{\mathrm{Res}}(s, U_i) = s_i\). We want to show
\(s\in \ker(\phi)\qty{\bigcup U_i}\), so consider
\begin{align*}
t\coloneqq\phi\qty{ \bigcup_i U_i}(s) \in {\mathcal{G}}\qty{\bigcup U_i}
,\end{align*}
which is zero. Now
\begin{align*}
\mathop{\mathrm{Res}}(t, U_i) = \phi(U_i)(\mathop{\mathrm{Res}}(s, U_i)) = \phi(U_i)(s_i) = 0
\end{align*}
by assumption, using the commutative diagram. By unique gluing for
\({\mathcal{G}}\), we have \(t=0\), since \(0\) is also a section
restricting to \(0\) everywhere.

\end{proof}

\begin{definition}[Quotients]

For \({\mathcal{F}}' \leq {\mathcal{F}}\) a subsheaf, define the
\textbf{quotient}
\({\mathcal{F}}/{\mathcal{F}}' \coloneqq(({\mathcal{F}}/{\mathcal{F}}')^{\mathsf{pre}})^+\)
where
\begin{align*}
({\mathcal{F}}/{\mathcal{F}}')^{\mathsf{pre}}(U) \coloneqq{\mathcal{F}}(U)/ {\mathcal{F}}'(U)
.\end{align*}

\end{definition}

\hypertarget{passing-to-stalks-pushforwardinverse-image-friday-august-27}{%
\section{Passing to stalks, pushforward/inverse image (Friday, August
27)}\label{passing-to-stalks-pushforwardinverse-image-friday-august-27}}

\hypertarget{isomorphism-iff-isomorphism-on-stalks}{%
\subsection{\texorpdfstring{Isomorphism \(\iff\) isomorphism on
stalks}{Isomorphism \textbackslash iff isomorphism on stalks}}\label{isomorphism-iff-isomorphism-on-stalks}}

\begin{theorem}[Sheaf isomorphism $\iff$ isomorphism on stalks]

Let \(\phi:{\mathcal{F}}\to{\mathcal{G}}\) be a morphism in
\({\mathsf{Sh}}(X)\), then \(\phi\) is an isomorphism \(\iff\)
\(\phi_p: {\mathcal{F}}_p \to{\mathcal{G}}_p\) is an isomorphism for all
\(p\in X\).

\end{theorem}

\begin{proof}[$\implies$]

Suppose \(\phi\) is an isomorphism, so there exists a
\(\psi: {\mathcal{G}}\to {\mathcal{F}}\) which is a two-sided inverse
for \(\phi\). Then \(\psi_p\) is a two-sided inverse to \(\phi_p\),
making it an isomorphism. This follows directly from the formula:
\begin{align*}
\phi_p: {\mathcal{F}}_p &\to {\mathcal{G}}_p \\
(s, U) & \mapsto (\phi(U)(s), U)
.\end{align*}

\end{proof}

\begin{proof}[$\impliedby$]

It suffices to show \(\phi(U): {\mathcal{F}}(U) \to {\mathcal{G}}(U)\)
is an isomorphism for all \(U\). This is because we could define
\(\psi(U):{\mathcal{G}}(U) \to {\mathcal{F}}(U)\) and set
\(\phi^{-1}(U) \coloneqq\psi(U)\), then reversing the arrows in the
diagram for a sheaf morphism again yields a commutative diagram.

\begin{claim}

\(\phi(U)\) is injective.

\end{claim}

For \(s\in {\mathcal{F}}(U)\), we want to show \(\phi(U)(s) = 0\)
implies \(s=0\). Consider the germs \((s, U) \in {\mathcal{F}}_p\) for
\(p\in U\), we have \(\phi_p(s, U) = (0, U) = 0\in {\mathcal{F}}_p\). So
\(S_p = 0\) for all \(p\in U\). Since we have a germ, there exists
\(V_p \ni p\) open such that \({ \left.{{s}} \right|_{{V_p}} } = 0\).
Noting that
\(\left\{{V_p {~\mathrel{\Big\vert}~}p\in U}\right\}\rightrightarrows U\),
by unique gluing we get an \(s\) where
\({ \left.{{s}} \right|_{{V_p}} } = 0\) for all \(V_p\), so
\(s\equiv 0\) on \(U\).

\begin{claim}

\(\phi(U)\) is surjective.

\end{claim}

Let \(t\in {\mathcal{G}}(U)\), and consider germs
\(t_p\in {\mathcal{G}}_p\). There exists a unique
\(s_p\in {\mathcal{F}}_p\) with \(\phi_p(s_p) = t_p\), since \(\phi_p\)
is an isomorphism of stalks by assumption. Use that \(s_p\) is a germ to
get an equivalence class \((s_p, V)\) where \(V \subseteq U\). We have
\(\phi(V)(s(p), V) \sim (t, U)\), noting that \(s\) depends on \(p\).
Having equivalent germs means there exists a \(W(p) \subseteq V\) with
\(p\in W\) with
\(\phi(W(p)) \qty{{ \left.{{s(p)}} \right|_{{W}} }} = { \left.{{t}} \right|_{{W(p)}} }\).
We want to glue these
\(\left\{{ { \left.{{s(p)}} \right|_{{W(p)}} } {~\mathrel{\Big\vert}~}p\in U }\right\}\)
together. It suffices to show they agree on intersections. Taking
\(p, q\in U\), both \({ \left.{{s(p)}} \right|_{{W(p) \cap W(q)}} }\)
and \({ \left.{{s(q)}} \right|_{{W(p) \cap W(q)}} }\) map to
\({ \left.{{t}} \right|_{{W(p) \cap W(q)}} }\) under
\(\phi(W(p) \cap W(q) )\). Injectivity will force these to be equal, so
\(\exists ! s \in {\mathcal{F}}(U)\) with
\({ \left.{{s}} \right|_{{W(p)}} } = s(p)\). We want to now show that
\(\phi(U)(s) = t\). Using commutativity of the square, we have
\(\phi(U)(s) { \left.{{}} \right|_{{W(p)}} } = \phi(W(p)) \qty{{ \left.{{s}} \right|_{{W(p)}} } }\).
This equals \(\phi(W(p))(s(p)) = { \left.{{t}} \right|_{{W(p)}} }\).
Therefore \(\phi(U)(s)\) and \(t\) restrict to sections
\(\left\{{w(p) {~\mathrel{\Big\vert}~}p\in U}\right\}\). Using unique
gluing for \({\mathcal{G}}\) we get \(\phi(U)(s) = t\).

\end{proof}

\begin{remark}

Note: we only needed to check overlaps because of exactness of the
following sequence:
\begin{align*}
0 \to {\mathcal{F}}(U) \to \prod_{i\in I} {\mathcal{F}}(U_i) \to \prod_{i < j} {\mathcal{F}}(U_{ij}) \to \cdots
.\end{align*}

\end{remark}

\hypertarget{inverse-image-and-pushforward}{%
\subsection{Inverse image and
pushforward}\label{inverse-image-and-pushforward}}

\begin{definition}[Pushforward and inverse image sheaves]

Let \(f\in {\mathsf{Top}}(X, Y)\), let
\({\mathcal{F}}\in {\mathsf{Sh}}(X)\) and define the \textbf{pushforward
sheaf} \(f_* {\mathcal{F}}\in {\mathsf{Sh}}(Y)\) by
\begin{align*}
f_*{\mathcal{F}}(V) \coloneqq{\mathcal{F}}( f^{-1}(V))
.\end{align*}

The \textbf{inverse image} sheaf is define as
\begin{align*}
(f^{-1}{\mathcal{F}})(U) \coloneqq\lim_{\text{open } V\supseteq f(U)} F(V)
.\end{align*}

\end{definition}

\begin{remark}

The inverse image sheaf generalizes stalks, recovering
\({\mathcal{F}}_p\) when \(f(U) = p\). Note that \(f(U)\) need not be
open unless \(f\) is an open map, and checking that \(f(U)\) is
(co?)final in the system
\(\left\{{\text{open } V\supseteq f(U)}\right\}\) yields
\begin{align*}
(f ^{-1}{\mathcal{F}})(U) = {\mathcal{F}}(f(U))
.\end{align*}

\end{remark}

\begin{warnings}

We will have a notion of \(f^*\), but this will not generally be the
pullback!

\end{warnings}

\begin{exercise}[?]

Show that \(f_* {\mathcal{F}}\) makes sense precisely because \(f\) is
continuous. Check that \(f_* {\mathcal{F}}\) satisfies the sheaf axioms.
Use that the set of opens of the form \(f^{-1}(V)\) are e.g.~closed
under intersections, and thus inherit all of the sheaf axioms from
\({\mathcal{F}}\).

\end{exercise}

\hypertarget{operatornamespeca-as-a-space-monday-august-30}{%
\section{\texorpdfstring{\(\operatorname{Spec}A\) as a space (Monday,
August
30)}{\textbackslash operatorname\{Spec\}A as a space (Monday, August 30)}}\label{operatornamespeca-as-a-space-monday-august-30}}

\hypertarget{the-prime-spectrum-operatornamespeca}{%
\subsection{\texorpdfstring{The prime spectrum
\(\operatorname{Spec}A\)}{The prime spectrum \textbackslash operatorname\{Spec\}A}}\label{the-prime-spectrum-operatornamespeca}}

\begin{remark}

Let \(R\in\mathsf{CRing}\) be a commutative unital ring in which
\(0\neq 1\) unless \(R=0\). The goal is to define a space \(X\) such
that \(R\) is the ring of functions on \(X\), imitating the
correspondence between \(X\in {\mathsf{Mfd}}\) and
\(R \coloneqq C^0(X; {\mathbf{R}})\). Recall that an ideal
\({\mathfrak{p}}\in \operatorname{Id}(R)\) is \textbf{prime} iff
\({\mathfrak{p}}\subset A\) is a proper subset and
\(fg\in {\mathfrak{p}}\implies f\in {\mathfrak{p}}\) or
\(g\in {\mathfrak{p}}\), or equivalently \(R/{\mathfrak{p}}\) is a
field.

\end{remark}

\begin{slogan}

Ideals are ``contagious'' under multiplication, and \emph{prime} ideals
have ``reverse contagion''.

\end{slogan}

\begin{definition}[Spectrum of a ring]

For \(A\in\mathsf{CRing}\) as above,
\begin{align*}
\operatorname{Spec}A \coloneqq\left\{{{\mathfrak{p}}{~\trianglelefteq~}A {~\mathrel{\Big\vert}~}{\mathfrak{p}}\text{ is a prime ideal}}\right\} \qquad \in {\mathsf{Set}}
.\end{align*}
We topologize \(\operatorname{Spec}A\) by defining a topology of closed
sets as follows:
\begin{align*}
\tau(A) \coloneqq\left\{{V(I) {~\mathrel{\Big\vert}~}I{~\trianglelefteq~}A}\right\},\qquad V(I) \coloneqq\left\{{ {\mathfrak{p}}\in \operatorname{Spec}(A) {~\mathrel{\Big\vert}~}{\mathfrak{p}}\supseteq I }\right\}
.\end{align*}

\end{definition}

\begin{exercise}[The topology is really a topology]

Prove that \((\operatorname{Spec}A, \tau(A))\) yields a well-defined
topological space.

\end{exercise}

\begin{example}[Spec of a field]

For \(A\) a field,
\(\operatorname{Spec}(A) = \left\{{\left\langle{0}\right\rangle}\right\}\)
is a point -- any other nonzero element
\({\mathfrak{p}}\in \operatorname{Spec}A\) would contain a unit \(u\),
in which case
\(u^{-1}u = 1\in {\mathfrak{p}}\implies {\mathfrak{p}}= A\).

\end{example}

\begin{example}[Spec of a polynomial ring]

For \(k\) an algebraically closed field,
\begin{align*}
\operatorname{Spec}k[t] = \left\{{ \left\langle{0}\right\rangle, \left\langle{t-a}\right\rangle {~\mathrel{\Big\vert}~}a\in k}\right\}
.\end{align*}
This is a PID, so every ideal is of the form
\(I = \left\langle{f}\right\rangle\), and one can check that
\begin{align*}
V(\left\langle{f}\right\rangle)
= 
\begin{cases}
\operatorname{Spec}k[t] & f=0 
\\
\left\langle{x-a_1, \cdots, a-a_k}\right\rangle & f(x) = \displaystyle\prod_{i=1}^k (x-a_i)
\end{cases}
.\end{align*}

Note that this is \textbf{not} the cofinite topology on
\(\operatorname{Spec}A\), since \(f=0\) defines a generic point
\(\eta \coloneqq\left\langle{0}\right\rangle\).

\end{example}

\hypertarget{the-structure-sheaf-mathcalo-wednesday-september-01}{%
\section{\texorpdfstring{The structure sheaf \({\mathcal{O}}\)
(Wednesday, September
01)}{The structure sheaf \{\textbackslash mathcal\{O\}\} (Wednesday, September 01)}}\label{the-structure-sheaf-mathcalo-wednesday-september-01}}

\hypertarget{ringed-spaces-are-finer-than-topological-spaces}{%
\subsection{Ringed spaces are finer than topological
spaces}\label{ringed-spaces-are-finer-than-topological-spaces}}

\begin{example}[Polynomial rings]

Let \(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
be algebraically closed, then
\begin{align*}
\operatorname{Spec}k[x] = \left\{{ \left\langle{x-a}\right\rangle {~\mathrel{\Big\vert}~}a\in k}\right\}\cup\left\langle{0}\right\rangle
.\end{align*}
Similarly,
\begin{align*}
\operatorname{Spec}k[x, y] = \left\{{ \left\langle{x-a, y-b}\right\rangle {~\mathrel{\Big\vert}~}a,b\in k}\right\} \cup\left\{{\left\langle{f}\right\rangle {~\mathrel{\Big\vert}~}f\in k[x,y] \text{ irreducible}}\right\} \cup\left\langle{0}\right\rangle
.\end{align*}
Note that both have non-closed, generic points
\(\eta = \left\langle{0}\right\rangle\).

\end{example}

\begin{example}[Distinct ringed spaces which are homeomorphic]

Consider
\(X \coloneqq\operatorname{Spec}{ {\mathbf{Z}}_{\widehat{p}} }\) and
\(Y\coloneqq\operatorname{Spec}{\mathbf{C}} {\llbracket t \rrbracket }\),
then
\begin{align*}
X = \left\{{\left\langle{p}\right\rangle, \left\langle{0}\right\rangle}\right\},\qquad
Y = \left\{{ \left\langle{t}\right\rangle, \left\langle{0}\right\rangle }\right\}
.\end{align*}
Both are two point spaces, with exactly one open/generic point
\(\left\langle{0}\right\rangle\) and one closed point
(\(\left\langle{p}\right\rangle\) and \(\left\langle{t}\right\rangle\)
respectively). These spaces are isomorphic as topological spaces
(i.e.~there is a homeomorphism between them), but later we'll see that
they can be distinguished as ringed spaces.

\end{example}

\hypertarget{properties-of-v}{%
\subsection{\texorpdfstring{Properties of
\(V\)}{Properties of V}}\label{properties-of-v}}

\begin{remark}

Recall that for \(A\in\mathsf{CRing}\), we defined
\(\operatorname{Spec}A\) to have closed sets of the form
\begin{align*}
V(I) = \left\{{p \in \operatorname{Spec}(A) {~\mathrel{\Big\vert}~}p\supseteq I}\right\} \qquad \forall I{~\trianglelefteq~}A
.\end{align*}

\end{remark}

\begin{lemma}[$V$ sends finite products to unions]

\begin{align*}
V(IJ) = V(I) \cup V(J)
,\end{align*}

\end{lemma}

\begin{corollary}[?]

If a prime ideal \(p\) contains \(IJ\) then \(p\supseteq I\) or
\(p\supseteq J\).

\end{corollary}

\begin{proof}[of lemma]

\(\impliedby\): If \(I \subseteq P\) or \(J \subseteq P\), then
\(IJ \subseteq I\) and \(IJ \subseteq J\), so \(IJ \subset p\).

\(\implies\): Suppose \(IJ \subset p\) but \(J \not\subset p\), so pick
\(j\in J \setminus p\). Then for all \(i\in I\), we have
\(ij\in IJ \subseteq p\), forcing \(i\in p\).

\end{proof}

\begin{lemma}[$V$ sends arbitrary sums to intersections]

An arbitrary intersection satisfies
\begin{align*}
V\qty{ \sum_{i\in J} I_i} = 
\bigcap_{i\in J} V(I_i) 
.\end{align*}

\end{lemma}

\begin{proof}[of lemma]

\(\implies\): For \(p\in \operatorname{Spec}(A)\), we want to show that
\(p \supseteq\sum I_i\) iff \(p \supseteq I_i\) for all \(i\), so
\(I_i \subseteq \sum I_i \subset P\).

\(\impliedby\): Ideals are additive groups, regardless of whether or not
they're prime!

\end{proof}

\begin{proof}[of proposition]

\envlist

\begin{itemize}
\tightlist
\item
  \(\emptyset\) is closed, since \(\emptyset = V(A)\)
\item
  \(X\) is closed, since \(X = V(0)\) and \(O\) is contained in every
  prime ideal.
\item
  Closure under finite unions: by induction, it's enough to show that
  \(V(I) \cup V(J)\) is closed. This follows from the 1st lemma above.
\item
  Closure under arbitrary unions: this follows from the 2nd lemma.
\end{itemize}

\end{proof}

\begin{proposition}[$V(I) = V(\sqrt I)$]

\begin{align*}
V(I) = V(\sqrt I)
.\end{align*}

\end{proposition}

\begin{proof}[?]

The proof is simple: prime ideals are radical.

\end{proof}

\begin{example}[?]

Note that
\begin{align*}
\operatorname{Spec}{\mathbf{Z}}= \left\{{\left\langle{0}\right\rangle}\right\} \cup\left\{{ \left\langle{p}\right\rangle {~\mathrel{\Big\vert}~}p{~\trianglelefteq~}{\mathbf{Z}}\text{ is prime}}\right\}
.\end{align*}
In general, maximal ideals are always closed points, and
\(\left\langle{0}\right\rangle\) is not a closed point. This is
homeomorphic to e.g.~
\begin{align*}
\operatorname{Spec}{ \mkern 1.5mu\overline{\mkern-1.5mu \mathbf{Q} \mkern-1.5mu}\mkern 1.5mu }[t] = \left\{{\left\langle{0}\right\rangle}\right\} \cup\left\{{\left\langle{t-a}\right\rangle {~\mathrel{\Big\vert}~}a\in { \mkern 1.5mu\overline{\mkern-1.5mu \mathbf{Q} \mkern-1.5mu}\mkern 1.5mu }}\right\}
,\end{align*}
since both are comprised of countably many closed points and a single
open point.

\end{example}

\hypertarget{localization-and-the-structure-sheaf}{%
\subsection{Localization and the structure
sheaf}\label{localization-and-the-structure-sheaf}}

\begin{definition}[Localization]

Suppose \(p \subseteq A\) is a prime ideal, then the
\textbf{localization} of \(A\) at \(p\), is defined as
\begin{align*}
A_p \coloneqq A \left[ { \scriptstyle { { ({p}^c) }^{-1}}  } \right] \coloneqq\left\{{ {a\over f} {~\mathrel{\Big\vert}~}a, f\in A,\, f\not\in p}\right\}{_{\scriptstyle / \sim} }\\ \\ \quad
{a\over f} \sim {b\over g}\iff \exists \, h\in A \text{ s.t. } h(ag-bf)=0
.\end{align*}
This makes the elements of \(p^c\) invertible, and is a local ring with
residue field \(\kappa = \operatorname{ff}(A/p)\) and maximal ideal
\(pA_p\). Ideals of \(A_p\) biject with ideals of \(A\) contained in
\(p\).

\end{definition}

\begin{remark}

Idea: \(A_p\) should look like germs of functions at the point \(p\).
Note that localizing at the ideal \(p\) is like deleting
\({ \operatorname{cl}} _X(V(p))\), which is also useful. We now want to
construct a sheaf
\({\mathcal{O}}= {\mathcal{O}}_{\operatorname{Spec}A}\) which has stalks
\(A_p\). We'll construct something that's obviously a sheaf, at the cost
of needing to work hard to prove things about it!

\end{remark}

\begin{definition}[Structure sheaf]

For \(U\in \operatorname{Spec}(A)\) open, so \(U = V(I)^c\), define the
\textbf{structure sheaf of \(X\)} as the sheaf given
\begin{align*}
{\mathcal{O}}(U) \coloneqq\left\{{ s:U \to \coprod{p\in U} A_p {~\mathrel{\Big\vert}~}s(p) \in A_p, \text{ and } s \text{ is locally a fraction}}\right\}
.\end{align*}
Here \emph{locally a fraction} means that for all \(p\in U\) there is an
open \(p\in V \subseteq U\) and elements \(a, f\in A\) such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(f\not\in Q\) for any \(Q\in V\) and
\item
  \(s(Q) = a/f\) for all \(Q \in V\).
\end{enumerate}

Restriction is defined for \(V \subseteq U\) as honest function
restriction on \({\mathcal{O}}(U) \to {\mathcal{O}}(V)\).

\end{definition}

\begin{remark}

Note that this is sheafifying the presheaf that sends \(U = D_f\) for
\(f\in A\) to the ring \(A_f\).

\end{remark}

\begin{example}[Structure sheaf of a field]

Let \(k \in \mathsf{Field}\) and
\(X\coloneqq\operatorname{Spec}(k) = \left\{{\left\langle{0}\right\rangle}\right\}\).
Then \({\mathcal{O}}_X\) is determined by
\begin{align*}
{{\Gamma}\qty{X; {\mathcal{O}}_X} } = \left\{{s: \operatorname{Spec}k \to k {~\mathrel{\Big\vert}~}\text{ conditions above} }\right\} = k
,\end{align*}
since the conditions are vacuous here.

\end{example}

\begin{example}[Structure sheaf of formal power series rings]

Let
\(X = \operatorname{Spec}{\mathbf{C}} {\llbracket t \rrbracket } = \left\{{ \left\langle{0}\right\rangle, \left\langle{t}\right\rangle}\right\}\).
Then
\begin{align*}
{\mathcal{O}}_X(X) = {\mathbf{C}} {\llbracket t \rrbracket  } \qquad
{\mathcal{O}}_X(\left\langle{0}\right\rangle) = {\mathbf{C}} { \left( {t} \right) } 
.\end{align*}

\end{example}

\hypertarget{sections-of-the-structure-sheaf-friday-september-03}{%
\section{Sections of the structure sheaf (Friday, September
03)}\label{sections-of-the-structure-sheaf-friday-september-03}}

\hypertarget{sections-of-puncturing-at-zero}{%
\subsection{Sections of puncturing at
zero}\label{sections-of-puncturing-at-zero}}

\begin{remark}

Last time: we defined \(\operatorname{Spec}A\) as a topological space
and \({\mathcal{O}}_{\operatorname{Spec}A}\), a sheaf of rings on
\(\operatorname{Spec}A\) which evidently satisfied the gluing condition:
\begin{align*}
{\mathcal{O}}_{\operatorname{Spec}A}(U) \coloneqq\left\{{s: U\to \coprod_{p\in U} A_p {~\mathrel{\Big\vert}~}s(p) \in A_{p} \, \forall p \text{ and } s \text{ is locally a fraction}}\right\}
.\end{align*}

\end{remark}

\begin{example}[?]

Set
\(X\coloneqq{\mathbf{A}}^1_{/ {k}} \coloneqq\operatorname{Spec}k[t]\)
for \(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\).
Take a point
\(\left\langle{t}\right\rangle = \left\langle{t-0}\right\rangle \in \operatorname{Spec}k[t]\)
corresponding to \(0\in X\), then
\begin{align*}
{\mathcal{O}}_{X}(X\setminus\left\{{ 0 }\right\}) = k[t, t^{-1}]
= \left\{{{f(t) \over t^\ell } {~\mathrel{\Big\vert}~}f\in k[t], \ell \geq 0 }\right\}
.\end{align*}
Generally for \(p = \left\langle{t-a_1, \cdots, t-a_m}\right\rangle\) we
get
\(s_p \in k[t] \left[ { \scriptstyle { {\left\{{t-a_i}\right\}_{1\leq i\leq m} }^{-1}} } \right]\).
Note that for \(p = \left\langle{0}\right\rangle\), we get
\(s_p \in k { \left( {t} \right) }\).

\end{example}

\begin{claim}

A section \(s\) is determined by \(s_{p}\) for
\(p = \left\langle{0}\right\rangle\), so there is an injective map
\begin{align*}
{\mathcal{O}}_{\operatorname{Spec}k[t]}(\operatorname{Spec}k[t]\setminus\left\{{ 0 }\right\}) &\to k(t) \\
s &\mapsto s_{\left\langle{0}\right\rangle}
.\end{align*}

\end{claim}

\begin{proof}[?]

Note that \(\left\langle{0}\right\rangle\) is in every open set, so
locally near \(p\) there exists a \(P\in V\) and \(a,f\) with
\(f\not\in Q\) for all \(Q\) and \(s_Q = a/f\) for all \(Q\in V\). Since
\(\left\langle{0}\right\rangle \in V\), we have
\(s_{\left\langle{0}\right\rangle} = a/f \in k(t)\) and
\(s_p = a/f\in A_p\). Since \(A_p \subseteq k(t)\), we get
\(s_p = s_{\left\langle{0}\right\rangle}\) under this inclusion.

\end{proof}

\begin{claim}

\begin{align*}
{\mathcal{O}}_{\operatorname{Spec}k[t]}(\operatorname{Spec}k[t]\setminus\left\{{ 0 }\right\}) = k[t, t^{-1}] = k[t] \left[ { \scriptstyle { {\left\{{t^\ell}\right\}_{\ell\geq 0}}^{-1}}  } \right] 
.\end{align*}

\end{claim}

\begin{proof}[?]

We showed that the LHS is a subset of \(k(t)\), so which subsets can be
written as things that are locally fractions on the complement of zero.

\(\supseteq\): This can clearly be done in \(k[t, t^{-1}]\) since every
element is locally the fraction \(f/t^k\).

\(\subseteq\): Suppose \(f/g\) with \(f,g\) coprime (this is a PID!)
with a pole away from zero, so \(g\in Q\) for some
\(Q\neq \left\langle{0}\right\rangle\). But then \(f/g\) isn't in
\(A_Q\).

\end{proof}

\begin{remark}

Note that
\(X \coloneqq\operatorname{mSpec}k[t] \subseteq X' \coloneqq\operatorname{Spec}k[t]\)
as the set of closed points, and restricting \({\mathcal{O}}_{X'}\) to
\(X\) yields the sheaf of regular functions. Having the extra generic
point was useful!

\end{remark}

\begin{exercise}[?]

Show that the maximal ideals \(m{~\trianglelefteq~}A\) correspond
precisely to closed points of \(X=\operatorname{Spec}A\).

\end{exercise}

\begin{example}[of a function that is locally but not globally a fraction]

Take \(A \coloneqq k[x,y,z,w]/\left\langle{xy-zw}\right\rangle\), which
is the cone over a smooth quadric surface and
\(X\coloneqq\operatorname{Spec}A\). Then take
\(U = \operatorname{Spec}(A) \setminus V(y, w) = V(y)^c \cap V(w)^c\)
and consider the section
\begin{align*}
s(p) \coloneqq
\begin{cases}
x/w & p\in V(w)^c 
\\
z/y & p\in V(y)^c.
\end{cases}
\end{align*}

For \(p\in U\), it makes sense to consider \(x/w\) and \(z/y\). Are they
equal? The answer is yes because \(xy-zw = 0\). Check that this can't be
a global fraction, which is a consequence of this random open set not
being the complement of localizing at a prime ideal.

\end{example}

\hypertarget{distinguished-opens}{%
\subsection{Distinguished opens}\label{distinguished-opens}}

\begin{definition}[Distinguished open sets]

Given \(f\in A\), the \textbf{distinguished open} \(D(f)\) corresponding
to \(f\) is defined as
\begin{align*}
D(f) = V(\left\langle{f}\right\rangle)^c \coloneqq\left\{{p\in \operatorname{Spec}(A) {~\mathrel{\Big\vert}~}f\in p}\right\}^c = \left\{{p\in \operatorname{Spec}A {~\mathrel{\Big\vert}~}f\not\in p}\right\}
,\end{align*}

i.e.~the points of \(\operatorname{Spec}(A)\) where \(f\) doesn't
vanish.

\end{definition}

\begin{remark}

The sets \(\left\{{D(f) {~\mathrel{\Big\vert}~}f\in A}\right\}\) form a
basis for the topology on \(\operatorname{Spec}(A)\). This follows from
writing \(V(I)^c = \bigcup_{f\in I} D(f)\).

\end{remark}

\hypertarget{the-fundamental-theorem-of-mathcalo_operatornamespeca-hartshorne-proposition-2.2}{%
\subsection{\texorpdfstring{The fundamental theorem of
\({\mathcal{O}}_{\operatorname{Spec}A}\) (Hartshorne Proposition
2.2)}{The fundamental theorem of \{\textbackslash mathcal\{O\}\}\_\{\textbackslash operatorname\{Spec\}A\} (Hartshorne Proposition 2.2)}}\label{the-fundamental-theorem-of-mathcalo_operatornamespeca-hartshorne-proposition-2.2}}

\begin{theorem}[Hartshorne Prop 2.2]

Let \(A\in \mathsf{CRing}\) be unital with \(1\neq 0\) unless \(A=0\)
and consider \((\operatorname{Spec}A, {\mathcal{O}})\). Then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  For any \(p\in \operatorname{Spec}A\), the stalk
  \({\mathcal{O}}_p \cong A_{p}\).
\item
  For any \(f\in A\), \({\mathcal{O}}(D(f)) = A_{f}\).
\item
  Taking \(f=1\), \(\Gamma(\operatorname{Spec}A, {\mathcal{O}}) = A\).
\end{enumerate}

\end{theorem}

\begin{remark}

Note that (b) gives the values of \({\mathcal{O}}\) on a basis of opens,
which determines the sheaf.

\end{remark}

\begin{proof}[of a]

Define a map
\begin{align*}
f_p: {\mathcal{O}}_p &\to A_p \\
(U, s) &\mapsto s(p)
.\end{align*}
This is well-defined since \(p\in W\) for any \(W \subseteq U \cap V\)
for equivalent germs \((U, s) \sim (V, t)\).

Surjectivity: given \(x=a/g \in A_p\), we want
\((U, s)\in {\mathcal{O}}_p\) such that \(f_p(U, s) = a/g\), so just
take \(U = D(g)\) and \(s=a/g\) (which makes sense!) and evidently maps
to \(a/g\).

Injectivity: supposing \(f_p(U, s) = 0\) in \(A_p\), we want
\((U, s) = 0\). If \(s(p) = 0\), then there exists some \(h\in P\) with
\(h\cdot s(p) = 0\). Since \(s(p)\) is locally a fraction, we can find
\(p\in V \subseteq U\) with \(s=a/g\) on \(V\) with \(g\neq 0\) on
\(V\), so consider \(V \cap D(h)\). The claim is that \(s\) is 0 here,
which follows from \(h\cdot (a/g) = 0\).

\end{proof}

\hypertarget{the-fundamental-theorem-of-mathcalo_operatornamespeca-wednesday-september-08}{%
\section{\texorpdfstring{The fundamental theorem of
\({\mathcal{O}}_{\operatorname{Spec}A}\) (Wednesday, September
08)}{The fundamental theorem of \{\textbackslash mathcal\{O\}\}\_\{\textbackslash operatorname\{Spec\}A\} (Wednesday, September 08)}}\label{the-fundamental-theorem-of-mathcalo_operatornamespeca-wednesday-september-08}}

\begin{remark}

Recall that we defined a first version of \emph{affine schemes}, namely
pairs \((\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\)
where for \(U \subseteq \operatorname{Spec}A\) open we have
\(s\in {\mathcal{O}}_{\operatorname{Spec}A}(U)\) locally represented by
\({ \left.{{s}} \right|_{{V}} } = a/f\) for \(V \subseteq U\) where
\(a, f\in A\) and \(V(f) \cap V = \emptyset\), so \(f\) doesn't vanish
on \(V\). Note that the \(D(f)\) form a topological basis for
\(\operatorname{Spec}A\), and the gluing condition is difficult,
i.e.~\({\mathcal{O}}_{\operatorname{Spec}A}(U)\) may be hard to compute,
even given an open cover \({\mathcal{V}}\rightrightarrows U\). We proved
that
\({\mathcal{O}}_{\operatorname{Spec}A, {\mathfrak{p}}} = A_{\mathfrak{p}}\)
last time, and today we're showing

\begin{itemize}
\tightlist
\item
  \({\mathcal{O}}_{\operatorname{Spec}A}(D(f)) = A_{f}\),
\item
  \({{\Gamma}\qty{\operatorname{Spec}A; {\mathcal{O}}_{\operatorname{Spec}A}} } \cong A\).
\end{itemize}

\end{remark}

\hypertarget{proof-of-the-fundamental-theorem}{%
\subsection{Proof of the fundamental
theorem}\label{proof-of-the-fundamental-theorem}}

\begin{proof}[of b and c]

\(b\implies c\): Take \(f=1\in A\), then
\({\mathcal{O}}(\operatorname{Spec}A) = {\mathcal{O}}(D(1)) = A\) using
(b), so the only hard part is showing (b).

To prove (b), by definition of \({\mathcal{O}}\) there is a ring
morphism
\begin{align*}
\psi: A_{f} &\to {\mathcal{O}}(D(f)) \\
{a\over f^n} &\mapsto {a\over f^n}
.\end{align*}

Note that this is just a careful statement, since the morphisms on
stalks \(\psi_{{\mathfrak{p}}}: A_{f} \to A_{{\mathfrak{p}}}\) by not be
injective in general. The proof will follow if \(\psi\) is both
injective and surjective.

\end{proof}

\begin{claim}

\(\psi\) is bijective.

\end{claim}

\begin{proof}[of injectivity]

Suppose \(\psi(s) = 0\), we then want to show \(s=0\). Write
\(s = a/f^n\), then for all \({\mathfrak{p}}\in D(f)\) we know
\(a/f^n = 0 \in A_{{\mathfrak{p}}}\). So for each \({\mathfrak{p}}\)
there is some \(h_{{\mathfrak{p}}} \not\in{\mathfrak{p}}\) where
\begin{align*}
h_{{\mathfrak{p}}}(a\cdot 1 - f^n\cdot 0) = 0 && \text{in } A
\end{align*}
in \(A\). Consider the ideal
\({\mathfrak{a}}\coloneqq\operatorname{Ann}(a) \coloneqq\left\{{b\in A {~\mathrel{\Big\vert}~}ab=0 \in A}\right\} \ni h_{{\mathfrak{p}}}\).
So take the closed subset \(V({\mathfrak{a}})\), which does not contain
\({\mathfrak{p}}\) since \({\mathfrak{a}}\not\subseteq{\mathfrak{p}}\).
Now iterating over all \({\mathfrak{p}}\in D(f)\), we get
\(V({\mathfrak{a}}) \cap D(f) = \emptyset\). So
\(V({\mathfrak{a}}) \subseteq V(f) = D(f)^c\), thus
\(f\in \sqrt{{\mathfrak{a}}}\) and \(f^m a = 0\) for some \(m\). Then
\(f^m(a\cdot 1 - f^n\cdot 0) = 0\) in \(A\), so \(a/f^n = 0\) in
\(A_{f}\).

\end{proof}

\begin{proof}[of surjectivity]

\textbf{Step 1}: Expressing \(s\in {\mathcal{O}}(D(f))\) nicely locally.

By definition of \({\mathcal{O}}_{D(f)}\), there exist \(V_i\) with
\({ \left.{{s}} \right|_{{V_i}} } = a_i/g_i\) for \(a_i, g_i\in A\).
We'd like \(g_i = h_i^{m_i}\) for some \(m_i\), so \(g\) is a power of
\(h_i\), but this may not be true a priori. Fix \(V_i = D(h_i)\), then
\(a_i / g_i\in {\mathcal{O}}(D(h_i))\) implies that
\(g_i\not\in {\mathfrak{p}}\) for any \({\mathfrak{p}}\in D(h_i)\). This
implies that \(D(h_i) \subseteq D(g_i)\), and taking complements yields
\(V(h_i) \supseteq V(g_i)\), and
\(h_i \in \sqrt{\left\langle{g_i}\right\rangle}\) and \(h_i^{n} = g_i\).
Writing \(g_i = h_i^n/c\) we have \(a_i/g_i = ca_i / h_i^n\). Note that
\(D(h_i) = D(h_i^n)\). Now replace \(a_i\) with \(ca_i\) and \(g_i\)
with \(h_i\) to get
\begin{align*}
{ \left.{{s}} \right|_{{D(h_i)}} } = a_i / h_i
.\end{align*}

\textbf{Step 2}: Quasicompactness of \(D(f)\).

Note that \(\left\{{D(h_i)}\right\}_{i\in I} \rightrightarrows D(f)\),
so take a finite subcover \(\left\{{D(h_i)}\right\}_{i\leq m}\).

Proof of quasicompactness: since
\(D(f) \supseteq\bigcup_{i\in I} D(h_i)\), we get
\begin{align*}
V(f) \subseteq \bigcap_{i\in I} V(h_i) = V\qty{ \sum h_i}
.\end{align*}
So \(f^u \in \sum h_i\), and up to reordering we can conclude
\(f^u = \sum_{i\leq m} b_i h_i\) for some \(b_i \in A\). Then
\(D(f) \subseteq \bigcup_{i\leq m} D(h_i)\).

\begin{remark}

Since we can write \(\operatorname{Spec}A = D(1)\), it is quasicompact.

\end{remark}

\textbf{Step 3}: Showing surjectivity.

Next time.

\end{proof}

\hypertarget{friday-september-10}{%
\section{Friday, September 10}\label{friday-september-10}}

\hypertarget{sections-and-stalks-of-mathcalo_operatornamespeca-as-localizations}{%
\subsection{\texorpdfstring{Sections and Stalks of
\({\mathcal{O}}_{\operatorname{Spec}A}\) as
Localizations}{Sections and Stalks of \{\textbackslash mathcal\{O\}\}\_\{\textbackslash operatorname\{Spec\}A\} as Localizations}}\label{sections-and-stalks-of-mathcalo_operatornamespeca-as-localizations}}

\begin{remark}

Last time: any affine scheme is quasicompact, so for each ring \(A\) and
an open cover \({\mathcal{U}}\rightrightarrows D(f)\) then there is a
finite subcover \(\left\{{D(h_i)}\right\}\rightrightarrows D(f)\). We
were looking at proposition: for the ringed space
\((\operatorname{Spec}A, {\mathcal{O}}_A)\),

\begin{itemize}
\tightlist
\item
  \({\mathcal{O}}_{\mathfrak{p}}\cong A \left[ { \scriptstyle { {{\mathfrak{p}}}^{-1}} } \right]\)
  for all \({\mathfrak{p}}\in \operatorname{Spec}A\),
\item
  \({\mathcal{O}}(d(f))\cong A \left[ { \scriptstyle { {f}^{-1}} } \right]\)
  for all \(f\in A\),
\item
  \({{\Gamma}\qty{\operatorname{Spec}A; {\mathcal{O}}_A} } \cong A\).
\end{itemize}

Note that \({\mathcal{O}}_A\) is uniquely characterized by these
properties.

\end{remark}

\begin{remark}

We can write \(D(1) = \operatorname{Spec}A\) and write
\(\left\{{ D(h_i) }\right\} \rightrightarrows\operatorname{Spec}A\) to
obtain \(1^n = \sum b_i h_i\). This is analogous to a partition of
unity, where \(b_i h_i\) vanishes on \(D(h_i)^c = V(h_i)\)

\end{remark}

\begin{proof}[of 2.2b]

Let
\(\psi:A \left[ { \scriptstyle { {f}^{-1}} } \right] \hookrightarrow{\mathcal{O}}(D(f))\)
where we just take restrictions of functions. We know this is injective,
and we want to show surjectivity.

\textbf{Step 1}: Let \(s\in {\mathcal{O}}(D(f))\). For each open
\(D(h_i)\), write \({ \left.{{s}} \right|_{{D(h_i)}} } = a_i /h_i\) for
some \(a_i \in A\).

\textbf{Step 2}: By quasicompactness, write
\(f^n = \sum_{1 \leq i\leq m} b_i h_i\).

\textbf{Step 3}: Glue the \(a_i/h_i\) into an element \(a/f\) of
\(A \left[ { \scriptstyle { {f}^{-1}} } \right]\).

\emph{Part 1}: For any \(1\leq i\neq j\leq m\),
\(D(h_i h_j) = D(h_i) \cap D(h_j)\). Note that \(a_i/h_i = a_j/h_j\) in
\({\mathcal{O}}(D(h_i h_j))\), and these are elements of
\(A \left[ { \scriptstyle { {h_i h_j}^{-1}} } \right]\) since
\(a_i /h_i = a_ih_j/h_i h_j\). Using injectivity of \(\psi\) for
\(h_i h_j\), we get an equality \(a_i/h_i = a_j/h_j\) in
\(A_{h_i h_j}\). Then for \(\ell\) large enough,
\((h_i h_j)^\ell( a_i h_j - a_j h_i) = 0 \in A\). Regrouping yields
\(h_j^{n+1}(h_i^n a_i) - h_i^{n+1}(h_j a_j) = 0\), so
\begin{align*}
{a_i h_i^n \over h_i ^{n+1}} = {a_j h_j^r \over h_j^{n+1}} \coloneqq{\tilde a_i \over \tilde h_i} = {\tilde a_j \over \tilde h_j}
,\end{align*}
noting that \(D(\tilde h_i) = D(\tilde h_i)\) since \(\tilde h_i\) is a
power of \(h_i\).

Now use POU gluing to write \(f^n = \sum_i \tilde b_i \tilde h_i\) and
\(a \coloneqq\sum \tilde a_i \tilde h_i\in A\) be a global function on
\(D(f)\). Then (claim) \(a_j/f^n = \tilde a_j/\tilde h_j\) on
\(D(\tilde h_j)\). We can rewrite
\begin{align*}
\tilde h_j a = \sum_i \tilde b_i \tilde a_i \tilde h_j = \sum_i \tilde b_i \tilde a_j \tilde h_i
.\end{align*}
But then \(a/f^n = { \left.{{s}} \right|_{{D(\tilde h_i)}} }\), so
\(s= a/f^n\).

\end{proof}

\begin{example}[?]

Consider \({\mathbf{P}}^1_{/ {k}}\) as a scheme -- we know the space,
and the claim is that we can glue sheaves of affines to obtain a
structure sheaf for it. Cover \({\mathbf{P}}^1\) by
\(U_0 = {\mathbf{P}}^1\setminus\left\{{\infty}\right\} \cong {\mathbf{A}}^1\)
and
\(U_1 = {\mathbf{P}}^1\setminus\left\{{0}\right\} \cong {\mathbf{A}}^1\).
The gluing data is the following:

\begin{center}
\begin{tikzcd}
    && {{\mathbf{P}}^1_{/ {k}} } \\
    \\
    {{\mathbf{A}}^1} & {U_0} && {U_1\cong {\mathbf{A}}^1} & {{\mathbf{A}}^1} \\
    \\
    & {{\mathbf{A}}^1\setminus\left\{{0}\right\}} & {U_0 \cap U_1 \cong D(t) \subseteq {\mathbf{A}}^1} & {{\mathbf{A}}^1\setminus\left\{{0}\right\}}
    \arrow["{\phi_0}", hook, from=3-2, to=1-3]
    \arrow["{\phi_1}"', hook', from=3-4, to=1-3]
    \arrow[hook', from=5-3, to=3-2]
    \arrow[hook, from=5-3, to=3-4]
    \arrow[hook, two heads, from=5-3, to=5-4]
    \arrow[hook, two heads, from=5-3, to=5-2]
    \arrow[hook, from=5-2, to=3-1]
    \arrow[hook', two heads, from=3-1, to=3-2]
    \arrow[hook, from=5-4, to=3-5]
    \arrow[hook', two heads, from=3-5, to=3-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

Here the transition maps are
\begin{align*}
\phi_1 \circ \phi_0^{-1}: \phi_0( U_0 \cap U_1) &\to \phi_1(U_0 \cap U_1) \\
t &\mapsto t^{-1}
.\end{align*}
What is the map on sheaves? We need a map
\({ \left.{{{\mathcal{O}}}} \right|_{{U_0\setminus\left\{{0}\right\}}} } { \, \xrightarrow{\sim}\, }{ \left.{{{\mathcal{O}}}} \right|_{{U_1 \setminus\left\{{ \infty }\right\}}} }\).

\end{example}

\begin{definition}[Ringed Space]

A \textbf{ringed space}
\((X, {\mathcal{O}}_X) \in {\mathsf{Top}}\times {\mathsf{Sh}}(X, \mathsf{Ring})\)
as a topological space with a sheaf of rings. A morphism is a pair
\((f, f^\#) \in C^0(X, Y) \times \in \mathop{\mathrm{Mor}}_{\mathsf{Sh}}({\mathcal{O}}_Y, f_* {\mathcal{O}}_X)\).

\end{definition}

\begin{example}[?]

The scheme
\((\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\) is a
ringed space.

\end{example}

\begin{example}[?]

Consider \({\mathbf{R}}\) in the Euclidean topology, then
\(({\mathbf{R}}, C^0({-}, {\mathbf{R}}))\) with the sheaf of continuous
functions is a ringed space. Then consider the morphism given by
projection onto the first coordinate of \({\mathbf{R}}^2\):
\begin{align*}
(\pi, \pi^\#): ({\mathbf{R}}^2, C^0({-}, {\mathbf{R}})) &\to ({\mathbf{R}}, C^{\infty }({-}, {\mathbf{R}})) \\
(x, y) &\mapsto x
.\end{align*}
For \(\pi^\#\), we can consider
\(\pi_* C^0({-}, {\mathbf{R}})(U) \coloneqq C^0(\pi^{-1}(U)) = C^0(U\times {\mathbf{R}})\),
so
\begin{align*}
\pi^\#: C^\infty(U, {\mathbf{R}}) &\to C^0(U\times {\mathbf{R}}) \\
f &\mapsto f\circ \pi
,\end{align*}
which is a well-defined map of rings since smooth functions are
continuous.

\end{example}

\begin{warnings}

Not every scheme is built out of affine opens!

\end{warnings}

\hypertarget{monday-september-13}{%
\section{Monday, September 13}\label{monday-september-13}}

\hypertarget{affine-schemes}{%
\subsection{Affine Schemes}\label{affine-schemes}}

\begin{definition}[Restricted sheaves]

Let \((X, {\mathcal{O}}_X) \in \mathsf{RingSp}\) and \(U \subseteq X\)
be open, then for \(V \subseteq U\) open, define the restricted sheaf
\({\mathcal{O}}_{X}{ \left.{{}} \right|_{{V}} }(V) \coloneqq{\mathcal{O}}_X(V)\).

\end{definition}

\begin{warnings}

\begin{align*}
{\mathsf{Sh}}_{/ {X}}  \ni{ \left.{{{\mathcal{O}}_X}} \right|_{{U}} }\neq {\mathcal{O}}_X(U) \in \mathsf{Ring}
.\end{align*}

\end{warnings}

\begin{remark}

Recall the definition of a ringed space \((X, {\mathcal{O}}_X)\). The
quintessential example: \(X\) a smooth manifold and
\({\mathcal{O}}_X \coloneqq C^{\infty}({-}; {\mathbf{R}})\) the sheaf of
smooth functions. For defining morphisms, consider a map \(f:X\to Y\),
then an alternative way of defining \(f\) to be smooth is that there is
a pullback
\begin{align*}
f^*: C^0(V, {\mathbf{R}}) &\to C^0(U, {\mathbf{R}}) \\
g &\mapsto g \circ f
\end{align*}
for \(U \subseteq X, V \subseteq Y\), and that \(f^*\) in fact restricts
to \(f^*: C^\infty(V; {\mathbf{R}}) \to C^\infty(U; {\mathbf{R}})\),
i.e.~preserving smooth functions.

\end{remark}

\begin{definition}[Morphisms of ringed spaces]

A morphism of ringed spaces is a pair
\begin{align*}
(M, {\mathcal{O}}_M) \xrightarrow{(\varphi, \varphi^\#)} (N, {\mathcal{O}}_N)
.\end{align*}
where \(\varphi \in C^0(M, N)\) and
\(\varphi^\# \in \mathop{\mathrm{Mor}}_{{\mathsf{Sh}}_{/ {N}} }({\mathcal{O}}_N, \varphi_* {\mathcal{O}}_M)\).

This is an \textbf{isomorphism} of ringed spaces if

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\varphi\) is a homeomorphism, and
\item
  \(\varphi^\#\) is an isomorphism of sheaves.
\end{enumerate}

\end{definition}

\begin{remark}

In the running example,
\begin{align*}
\varphi^\#(U): {\mathcal{O}}_N(U) \to \varphi_* {\mathcal{O}}_M(M) = {\mathcal{O}}_M(\varphi^{-1}(U))
.\end{align*}
This implies that maps of ringed spaced here induce smooth maps, and so
there is an embedding
\({\mathsf{sm}}{\mathsf{Mfd}}_{/ {{\mathbf{R}}}} \hookrightarrow\mathsf{RingSp}\).

\end{remark}

\begin{remark}

We'll try to set up schemes the same way one sets up smooth manifolds. A
topological manifold is a space locally homeomorphic to
\({\mathbf{R}}^n\), and a smooth manifold is one in which it's locally
isomorphic as a ringed space to
\(({\mathbf{R}}^n, C^\infty({-}; {\mathbf{R}}))\) with its sheaf of
smooth functions.

\end{remark}

\begin{definition}[Smooth manifolds, alternative definition]

A \textbf{smooth manifold} is a ringed space \((M, {\mathcal{O}}_M)\)
that is locally isomorphic to
\(({\mathbf{R}}^d, C^\infty({-}; {\mathbf{R}}))\), i.e.~there is an open
cover \({\mathcal{U}}\rightrightarrows M\) such that
\begin{align*}
(U_i, { \left.{{{\mathcal{O}}_M}} \right|_{{U_i}} }) \cong ({\mathbf{R}}^n, C^{\infty}({-}; {\mathbf{R}}))
.\end{align*}

\end{definition}

\begin{example}[?]

An example of a morphism of ringed spaces that is not an isomorphism:
take \(({\mathbf{R}}, C^0) \to ({\mathbf{R}}, C^\infty)\) given by
\((\operatorname{id}, \operatorname{id}^\#)\) where
\(\operatorname{id}^\#: C^\infty \to \operatorname{id}_* C^0\) is given
by \(\operatorname{id}^\#(U): C^\infty(U) \to C^0(U)\) is the inclusion
of continuous functions into smooth functions.

\end{example}

\begin{remark}

We'll define schemes similarly: build from simpler pieces, namely an
open cover with isomorphisms to affine schemes. A major difference is
that there may not exist a \emph{unique} isomorphism type among all of
the local charts, i.e.~the affine scheme can vary across the cover.

\end{remark}

\begin{remark}

Recall that for \(A\) a ring we defined
\((\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\), where
\(\operatorname{Spec}A \coloneqq\left\{{\text{Prime ideals } {\mathfrak{p}}{~\trianglelefteq~}A}\right\}\),
equipped with the Zariski topology generated by closed sets
\(V(I) \coloneqq\left\{{{\mathfrak{p}}{~\trianglelefteq~}A {~\mathrel{\Big\vert}~}I\supseteq{\mathfrak{p}}}\right\}\).
We then defined
\begin{align*}
{\mathcal{O}}_{\operatorname{Spec}A}(U) \coloneqq\left\{{s: U\to \coprod_{{\mathfrak{p}}\in U} A \left[ { \scriptstyle { {{\mathfrak{p}}}^{-1}}  } \right] 
{~\mathrel{\Big\vert}~}
s({\mathfrak{p}}) \in A \left[ { \scriptstyle { {{\mathfrak{p}}}^{-1}}  } \right] , \, s\text{ locally a fraction}
}\right\}
.\end{align*}
We saw that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  We can identify stalks:
  \({\mathcal{O}}_{\operatorname{Spec}A, {\mathfrak{p}}} = A \left[ { \scriptstyle { {{\mathfrak{p}}}^{-1}} } \right]\)
\item
  We can identify sections on distinguished opens:
  \begin{align*}
  {\mathcal{O}}_{\operatorname{Spec}A}(D_f) = A \left[ { \scriptstyle { {f}^{-1}}  } \right]  = \left\{{a/f^k {~\mathrel{\Big\vert}~}a\in A, k\in {\mathbf{Z}}_{\geq 0}}\right\}
  ,\end{align*}
  where
  \(D_f \coloneqq V(f)^c = \left\{{{\mathfrak{p}}\in \operatorname{Spec}A {~\mathrel{\Big\vert}~}f\not\in {\mathfrak{p}}}\right\}\).
\end{enumerate}

As a corollary, we get
\({\mathcal{O}}_{\operatorname{Spec}A}(\operatorname{Spec}A) = A\),
noting \(\operatorname{Spec}A = d_1\) and
\(A \left[ { \scriptstyle { {1}^{-1}} } \right] = A\). Thus we can
recover the ring \(A\) from the ringed space
\((X, {\mathcal{O}}_X) \coloneqq(\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\)
by taking global sections,
i.e.~\({{\Gamma}\qty{\operatorname{Spec}A; {\mathcal{O}}_{\operatorname{Spec}A}} } = A\).

\end{remark}

\hypertarget{affine-varieties}{%
\subsection{Affine Varieties}\label{affine-varieties}}

\begin{remark}

Let \(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
and set \({\mathbf{A}}^n_{/ {k}} = k^n\) whose regular functions are
given by \(k[x_1, \cdots, x_{n}]\), regarded as maps to \(k\).

\end{remark}

\begin{definition}[Affine variety]

An affine variety is any set of the form
\begin{align*}
X \coloneqq V(f_1,\cdots, f_n) = \left\{{p\in {\mathbf{A}}^n_{/ {k}}  {~\mathrel{\Big\vert}~}f_1(p) = \cdots = f_m(p) = 0}\right\}
\end{align*}
for \(f_i \in k[x_1, \cdots, x_{n}]\),

\end{definition}

\begin{remark}

Writing \(I = \left\langle{f_1,\cdots, f_m}\right\rangle\), we have
\(X = V(\sqrt I)\). Letting
\(I(S) = \left\{{f\in k[x_1, \cdots, x_{n}]{~\mathrel{\Big\vert}~}{ \left.{{f}} \right|_{{S}} } = 0}\right\}\),
then by the Nullstellensatz, \(IV(I) = \sqrt{I}\). This gives a
bijection between affine varieties in \({\mathbf{A}}^n_{/ {k}}\) and
radical ideals \(I {~\trianglelefteq~}k[x_1, \cdots, x_{n}]\).

\end{remark}

\begin{definition}[Coordinate rings of affine varieties]

The \textbf{coordinate ring} of an affine variety \(X\) is
\(k[X] \coloneqq k[x_1, \cdots, x_{n}]/I(X)\), regarded as polynomial
functions on \(X\).

\end{definition}

\begin{remark}

We quotient here because if the difference of functions is in \(I(X)\),
these functions are equal when restricted to \(X\). For example, \(y=x\)
in \(k[x, y]/ \left\langle{x-y}\right\rangle\), which are different
functions where for \(X\coloneqq\Delta\), we have
\({ \left.{{x}} \right|_{{\Delta}} } = { \left.{{y}} \right|_{{\Delta}} }\).

\end{remark}

\begin{remark}

As an application of the Nullstellensatz, there is a correspondence

\begin{align*}
\adjunction{I({-})}{V({-})}{\left\{{\text{Points } p\in X}\right\}}{\operatorname{mSpec}k[X]}
\end{align*}

\end{remark}

\begin{remark}

Why is an affine variety \(X\) an example of an affine scheme
\(\operatorname{Spec}k[X]\)? These don't have the same underlying
topological space:
\begin{align*}
\tau(X) &\coloneqq\left\{{V(I) \coloneqq\left\{{p\in X {~\mathrel{\Big\vert}~}f_i(p) = 0 \,\, \forall f_i \in I}\right\} {~\mathrel{\Big\vert}~}I{~\trianglelefteq~}k[X]}\right\} \\
\tau(\operatorname{mSpec}k[X]) &\coloneqq\left\{{ V(I) \coloneqq\left\{{ {\mathfrak{m}}\in \operatorname{mSpec}k[X] {~\mathrel{\Big\vert}~}{\mathfrak{m}}\supseteq I}\right\} {~\mathrel{\Big\vert}~}I{~\trianglelefteq~}k[X] }\right\}
.\end{align*}

However, they are closely related:
\begin{align*}
{ \left.{{\tau(\operatorname{mSpec}k[X])}} \right|_{{\operatorname{Spec}k[X]}} } = \tau(X_{\mathrm{zar}})
,\end{align*}
i.e.~the space \(\operatorname{Spec}k[X]\) with the restricted topology
from \(\operatorname{mSpec}k[X]\) is homeomorphic to \(X\) with the
Zariski topology. I.e. restricting to \emph{closed points} recovers
regular functions on \(X\).

\end{remark}

\begin{warnings}

Defining things that are locally isomorphic to schemes would work for
objects but not morphisms!

\end{warnings}

\hypertarget{wednesday-september-15}{%
\section{Wednesday, September 15}\label{wednesday-september-15}}

\hypertarget{asdsadas}{%
\subsection{asdsadas}\label{asdsadas}}

\begin{remark}

Last time: for \({\mathsf{Aff}}{\mathsf{Var}}\) we considered
\(X = V(I) \subseteq {\mathbf{A}}^n_{/ {k}}\), and for
\({\mathsf{Aff}}{\mathsf{Sch}}\) we considered
\(\operatorname{Spec}k[X]\) where
\(k[X] \coloneqq k[x_1, \cdots, x_{n}]/ I(X)\). Both had the Zariski
topology, and
\(X = \operatorname{mSpec}k[X] \subseteq \operatorname{Spec}k[X]\). We
had structure sheaves \({\mathcal{O}}_{\operatorname{Spec}k[X]}\), and
for \(X\), we have \(U' \coloneqq U \cap\operatorname{mSpec}k[X]\). On
\(\operatorname{mSpec}k[X]\), we have the notion of a regular function,
and
\({\mathcal{O}}_X(U') = {\mathcal{O}}_{\operatorname{Spec}k[X]}(U')\).
The previous setup only worked for rings finitely generated over a
field, whereas for schemes, we can take things like
\(\operatorname{Spec}{\mathbf{Z}}\), so they're much more versatile
(e.g.~for number theory).

\end{remark}

\hypertarget{adssads}{%
\subsection{adssads}\label{adssads}}

\begin{example}[?]

Compare \({\mathbf{A}}^2_{/ {k}} \in {\mathsf{Aff}}{\mathsf{Var}}\) to
\(\operatorname{Spec}k[x, y]\). Note that
\(\left\langle{0}\right\rangle \in \operatorname{Spec}k[x, y]\), and
taking its closure yields
\begin{align*}
{ \operatorname{cl}} (\left\langle{0}\right\rangle) 
&= \bigcap_{V(I)\supseteq\left\langle{0}\right\rangle } V(I) \\
&= \bigcap_{V(I)\ni 0 } V(I) \\
&= V(0) \\
&= \operatorname{Spec}k[x, y]
,\end{align*}
so \(0\) is a dense point!

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{20pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-09-15_18-17.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

But there are prime ideals of height \(>1\). For example, any
irreducible subvariety of \({\mathbf{A}}^2\) yields a generic point.

\todo[inline]{Krull's dimension theorem?}

\end{example}

\begin{exercise}[?]

Try to draw \(\operatorname{Spec}{\mathbf{Z}}\) and
\(\operatorname{Spec}{\mathbf{Z}}[t]\).

\end{exercise}

\begin{remark}

We'll now try a naive definition of schemes, which we'll find won't
quite work.

\end{remark}

\begin{definition}[A wrong definition of a scheme!]

A \textbf{scheme} is a ringed space \((X, {\mathcal{O}}_X)\) which is
locally an affine scheme, i.e.~there exists an open cover
\({\mathcal{U}}\rightrightarrows X\) such that there is a collection of
rings \(A_i\) with
\begin{align*}
(U_i, { \left.{{{\mathcal{O}}_{X}}} \right|_{{U_i}} } )  { \, \xrightarrow{\sim}\, }(\operatorname{Spec}A_i, {\mathcal{O}}_{\operatorname{Spec}A_i})
.\end{align*}

\end{definition}

\begin{remark}

This produces the right objects, but not the correct morphisms. This is
a subtle issue! With this definition, a morphism of schemes would be a
morphism of ringed spaces \((f, f^\#)\) with
\(f\in {\mathsf{Top}}(X, Y)\) and
\(f^\# \in {\mathsf{Sh}}_{/ {Y}} ({\mathcal{O}}_Y, f_* {\mathcal{O}}_X)\),
where \(f^\#\) is supposed to capture ``pullback of functions''. The
issue: \(f^\#\) may not notice that \(p \xrightarrow{f} f(p)\)! In
particular, it may not preserve the fact that \(f(p) = 0\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{42pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-09-15_11-49.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Hartshorne exercises for how this issue can actually arise.

\end{remark}

\begin{remark}

Let \((f, f^\#)\) be a map of ringed spaces, then there is an induced
map
\begin{align*}
f_p^\#: {\mathcal{O}}_{Y, f(p)} &\to {\mathcal{O}}_{X, p} \\
(U, s) &\mapsto (f^{-1}(U), f^\#(U)(s))
.\end{align*}

\end{remark}

\begin{definition}[Locally ringed space]

A \textbf{locally ringed space} \((X, {\mathcal{O}}_X)\) is a ringed
space for which the stalks
\({\mathcal{O}}_{X, p} \in \mathsf{Loc}\mathsf{Ring}\) are local rings,
i.e.~there exists a unique maximal ideal.

\end{definition}

\begin{example}[of a locally ringed space]

For
\((X, {\mathcal{O}}_X) \coloneqq(\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\),
we saw that
\({\mathcal{O}}_{X, p} = A \left[ { \scriptstyle { {p}^{-1}} } \right]\),
which is local.

\end{example}

\begin{definition}[Morphisms of locally ringed spaces]

A \textbf{morphism of locally ringed spaces} is a morphism of ringed
spaces
\begin{align*}
(f, F^\#): (X, {\mathcal{O}}_X) \to (Y, {\mathcal{O}}_Y)
\end{align*}
such that \(f^\#_p: {\mathcal{O}}_{Y, f(p)} \to {\mathcal{O}}_{X, p}\)
is a homomorphism of local rings,
i.e.~\(f^\#({\mathfrak{m}}_{f(p)}) \subseteq {\mathfrak{m}}_p\).

\begin{quote}
Here we should also require that \(f^\# \neq 0\).
\end{quote}

\end{definition}

\begin{remark}

Morally: this extra condition enforces that pulling back functions
vanishing at \(f(p)\) yields functions which vanish at \(p\).

\end{remark}

\begin{remark}

Alternatively one could require that
\((f^\#)^{-1}({\mathfrak{m}}_p) = {\mathfrak{m}}_{f(p)}\), and (claim)
this is equivalent to the above definition. Use that
\((f^\#)^{-1}({\mathfrak{m}}_p)\) is a prime ideal containing
\({\mathfrak{m}}_p\).

\end{remark}

\begin{example}[of a locally ringed space]

Take
\((X, {\mathcal{O}}_X) \coloneqq({\mathbf{R}}, C^0({-}; {\mathbf{R}}))\).
Why this is in \(\mathsf{Loc}\mathsf{RingSp}\): write a stalk as
\begin{align*}
C_p^0 = \left\{{(f, I) {~\mathrel{\Big\vert}~}I\ni p \text{ an interval}, f\in {\mathsf{Top}}(I, {\mathbf{R}})}\right\}_{/ {\sim}} 
.\end{align*}
Why is this local? Take
\({\mathfrak{m}}_p \coloneqq\left\{{(f, I) {~\mathrel{\Big\vert}~}f(p) = 0}\right\}\),
which is maximal since \(C_p^0/{\mathfrak{m}}\cong {\mathbf{R}}\) is a
field. Then
\({\mathfrak{m}}_p^c = \left\{{(f, I) {~\mathrel{\Big\vert}~}f(p) \neq 0}\right\}\),
and any continuous function that isn't zero at \(p\) is nonzero in some
neighborhood \(J \supseteq I\), so
\({ \left.{{f}} \right|_{{J}} }\neq 0\) anywhere. Then
\((f, I) \sim ({ \left.{{f}} \right|_{{J}} }, J)\), which is invertible
in the ring, so any element in the complement is a unit.

\end{example}

\begin{example}[?]

Consider
\begin{align*}
({\mathbf{R}}, C^0({-}; {\mathbf{R}})) \xrightarrow{(f, f^\#)} ({\mathbf{R}}, C^\infty({-}; {\mathbf{R}}))
.\end{align*}
Take \(f = \operatorname{id}\) and the inclusion
\begin{align*}
f^\# : C^\infty({-}; {\mathbf{R}})\hookrightarrow\operatorname{id}_* C^0({-}; {\mathbf{R}}) = C^0({-}; {\mathbf{R}})
.\end{align*}
Then
\begin{align*}
f_p^\#: C_p^\infty({-}; {\mathbf{R}}) \to C_p^0({-}; {\mathbf{R}})
.\end{align*}
satisfies
\(f_p^\#({\mathfrak{m}}^\infty_{\operatorname{id}(p)}) = {\mathfrak{m}}^0_p\).
We also have
\((f_p^\#)^{-1}({\mathfrak{m}}_p^0) = {\mathfrak{m}}_p^\infty\), since
the germs are just equal.

\end{example}

\begin{definition}[Scheme]

A \textbf{scheme} \((X, {\mathcal{O}}_X)\) is a locally ringed space
which is locally isomorphic to
\((\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\) in
\(\mathsf{Loc}\mathsf{RingSp}\). A \textbf{morphism of schemes} is a
morphism in \(\mathsf{Loc}\mathsf{RingSp}\).

\end{definition}

\begin{remark}

Note that we can restrict to opens, since this doesn't change the
stalks.

\end{remark}

\begin{remark}

As a proof of concept that this is a good notion, we'll see that there's
a fully faithful contravariant functor
\(\operatorname{Spec}: \mathsf{CRing}\to {\mathsf{Sch}}\), so
\begin{align*}
\operatorname{Spec}(\mathop{\mathrm{Mor}}_\mathsf{Ring}(B, A)) = \mathop{\mathrm{Mor}}_{\mathsf{Sch}}( \operatorname{Spec}A, \operatorname{Spec}B)
.\end{align*}

\end{remark}

\hypertarget{friday-september-17}{%
\section{Friday, September 17}\label{friday-september-17}}

\begin{remark}

Recall from last time: a \emph{locally ringed space}
\((X, {\mathcal{O}}_X)\) is a ringed space (so
\(X\in {\mathsf{Top}}, {\mathcal{O}}_X \in {\mathsf{Sh}}(X, \mathsf{Ring}))\)
such that the stalks
\({\mathcal{O}}_{X, p} \in \mathsf{Loc}\mathsf{Ring}\) for all points
\(p\in X\). Some examples:

\begin{itemize}
\tightlist
\item
  \((M, {\mathcal{O}}_M)\) with \(M\) a manifold and \({\mathcal{O}}_M\)
  any sheaf of reasonable functions,
\item
  \((\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\)
\end{itemize}

We defined a scheme as a locally ringed space which is locally
isomorphic in \(\mathsf{Loc}\mathsf{RingSp}\) to
\((\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\). Recall
that \textbf{locally} meant there exists an open cover \({\mathcal{U}}\)
with the property holding for every element in the cover. Note that most
``local'' conditions from commutative algebra (that can be checked at
all localizations) will correspond to properties that hold on \emph{all}
open covers.

\begin{quote}
There are generally two ways to define properties of schemes: either it
holds for every affine open cover, or there exists an affine open cover.
\end{quote}

\end{remark}

\begin{proposition}[?]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  If \(A\in \mathsf{Ring}\), then
  \((\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\in \mathsf{Loc}\mathsf{RingSp}\).
\item
  If \(f\in \mathsf{CRing}(A, B)\) is a ring morphism, it induces a
  morphism
  \((f, f^\#) \in \mathsf{Loc}\mathsf{RingSp}(\operatorname{Spec}B, \operatorname{Spec}A)\).
\item
  Moreover, every
  \((f, f^\#) \in \mathsf{Loc}\mathsf{RingSp}(\operatorname{Spec}B, \operatorname{Spec}A)\)
  is induced by some \(f\in {\mathsf{Top}}(A, B)\).
\end{enumerate}

\end{proposition}

\begin{remark}

Note that morphisms in \(\mathsf{RingSp}\) don't necessarily restrict to
morphisms in \(\mathsf{Loc}\mathsf{RingSp}\), i.e.~this is not a full
subcategory, since morphisms of rings need not be morphisms of local
rings (i.e.~those preserving the maximal ideal).

\end{remark}

\begin{proof}[of (a) and (b)]

Part (a): This follows from the theorem last week that
\({\mathcal{O}}_{\operatorname{Spec}A, p} = A \left[ { \scriptstyle { {p}^{-1}} } \right]\).

Part(b): There's only ever one thing to do! Define the set-theoretic map
\begin{align*}
f: \operatorname{Spec}B \to \operatorname{Spec}A \\
p &\mapsto \phi^{-1}(p)
.\end{align*}
Why this is also continuous: we'll show preimages preserve closed sets.
We can write
\begin{align*}
f^{-1}(V(I)) 
&= f^{-1}\qty{\left\{{Q {~\mathrel{\Big\vert}~}Q \supseteq I}\right\}} \\
&= \left\{{{\mathfrak{p}}{~\mathrel{\Big\vert}~}\phi^{-1}({\mathfrak{p}}) \supseteq I}\right\} \\
&= \left\{{{\mathfrak{p}}{~\mathrel{\Big\vert}~}{\mathfrak{p}}\supseteq I}\right\} \\
&= V(\left\langle{ \phi(I)}\right\rangle )
,\end{align*}
using that
\(f^{-1}(Q) \coloneqq\left\{{{\mathfrak{p}}{~\mathrel{\Big\vert}~}\phi^{-1}({\mathfrak{p}}) = Q}\right\}\).

Now localize to get
\(\phi_p: A \left[ { \scriptstyle { {\phi^{-1}(p)}^{-1}} } \right] \to B \left[ { \scriptstyle { {p}^{-1}} } \right]\).
We now need a sheaf map:
\begin{align*}
f^\#: {\mathcal{O}}_{\operatorname{Spec}A} \to f_* {\mathcal{O}}_{\operatorname{Spec}B}
,\end{align*}
i.e.~an assignment
\(f^\#(V): {\mathcal{O}}_{\operatorname{Spec}A}(V) \to {\mathcal{O}}_{\operatorname{Spec}B}(f^{-1}(V))\)
for all \(V \subseteq \operatorname{Spec}A\) open. We can write
\begin{align*}
{\mathcal{O}}_{\operatorname{Spec}A}(V) \coloneqq\left\{{ s \in {\mathsf{Top}}(V, \prod_{p\in V} A \left[ { \scriptstyle { {p}^{-1}}  } \right]  ) {~\mathrel{\Big\vert}~}s(p)\in A \left[ { \scriptstyle { {p}^{-1}}  } \right] , s \text{ locally a fraction} }\right\}
&\longrightarrow
{\mathcal{O}}_{\operatorname{Spec}B}(f^{-1}V) \coloneqq\left\{{ t \in {\mathsf{Top}}(f^{-1}(V) , \prod_{q\in f^{-1}(V) } B \left[ { \scriptstyle { {q}^{-1}}  } \right]  ) {~\mathrel{\Big\vert}~}t(q)\in B \left[ { \scriptstyle { {q}^{-1}}  } \right] , t \text{ locally a fraction} }\right\} \\
(s_p)_{p\in V} &\mapsto (p_q(s_p))_{q\in f^{-1}(V)}
.\end{align*}
But then \(q\in f^{-1}(p)\) for some \(p\in V\) iff
\(p\in \phi^{-1}(q)\). So using the map on stalks gives a map on
sections, and it preserves the property of locally being a fraction, so
this yields a morphism of sheaves of rings, and it remains to check that
it's a local morphism.

\begin{quote}
Note that you can get this by composing
\(f^{-1}(V) \xrightarrow{f} V \xrightarrow{s} \prod A \left[ { \scriptstyle { {p}^{-1}} } \right] \xrightarrow{\prod \phi_p} \prod B \left[ { \scriptstyle { {\phi(p)}^{-1}} } \right]\).
\end{quote}

We now claim \(f^\#\) is a local homeomorphism. This is clear: we can
write
\(f^\#_p: A \left[ { \scriptstyle { {f(p)}^{-1}} } \right] \to B \left[ { \scriptstyle { {p}^{-1}} } \right]\),
and \(f_p^\# = \phi_p\) by construction, which is a local morphism of
rings. So \(f^\#\) is a morphism in \(\mathsf{Loc}\mathsf{RingSp}\).

\end{proof}

\begin{proof}[of (c)]

Let
\((f, f^\#): (\operatorname{Spec}B, {\mathcal{O}}_{\operatorname{Spec}B} ) \to (\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\)
be a morphism in \(\mathsf{Loc}\mathsf{RingSp}\). Goal: define
\(\phi \in \mathsf{CRing}(A, B)\), inducing \((f, f^\#)\) in the sense
of part (b). Note that by definition,
\(f^\#(\operatorname{Spec}A): {\mathcal{O}}_{\operatorname{Spec}A}(\operatorname{Spec}A) \to {\mathcal{O}}_{\operatorname{Spec}B}(\operatorname{Spec}B)\).
By the previous theorem, global sections recovers rings on affines, so
\(f^\#(\operatorname{Spec}A): A\to B\).

\begin{claim}

\(\phi^{-1}(p) = f(p)\).

\end{claim}

For any \(p\in \operatorname{Spec}B\), we can localize \(f^\#\) to
obtain a local ring morphism
\begin{align*}
f^\#_p: {\mathcal{O}}_{\operatorname{Spec}A, f(p)} \to {\mathcal{O}}_{\operatorname{Spec}B, p}
.\end{align*}
We also have a commutative diagram

\begin{center}
\begin{tikzcd}
    A && B \\
    \\
    {A \left[ { \scriptstyle { {f(p)}^{-1}}  } \right] } && {B \left[ { \scriptstyle { {p}^{-1}}  } \right] }
    \arrow["{f^\#(\operatorname{Spec}A) = \phi}", from=1-1, to=1-3]
    \arrow["{f^\#(\operatorname{Spec}A)_p = \phi_p}"', from=3-1, to=3-3]
    \arrow["{\text{localize}}"{description}, from=1-1, to=3-1]
    \arrow["{\text{localize}}"{description}, from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJBIl0sWzIsMCwiQiJdLFswLDIsIkFcXGxvY2FsaXple2YocCl9Il0sWzIsMiwiQlxcbG9jYWxpemV7cH0iXSxbMCwxLCJmXlxcIyhcXHNwZWMgQSkgPSBcXHBoaSJdLFsyLDMsImZeXFwjKFxcc3BlYyBBKV9wID0gXFxwaGlfcCIsMl0sWzAsMiwiXFx0ZXh0e2xvY2FsaXplfSIsMV0sWzEsMywiXFx0ZXh0e2xvY2FsaXplfSIsMV1d}{Link
to Diagram}
\end{quote}

Now we use locality in a key way to conclude \(\phi^{-1}(p) = f(p)\) by
commutativity: check that
\((f^\#_p)^{-1}({\mathfrak{m}}_p) = {\mathfrak{m}}_{f(p)} \xrightarrow{{\mathsf{loc}}^{-1}} f(p)\),
and
\({\mathsf{loc}}^{-1}({\mathfrak{m}}_p) = p \xrightarrow{\phi} \phi(p)\).

\end{proof}

\hypertarget{monday-september-20}{%
\section{Monday, September 20}\label{monday-september-20}}

\begin{remark}

Last time: we proved the following:

\end{remark}

\begin{proposition}[?]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  If \(A\in \mathsf{Ring}\), then
  \((\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\in \mathsf{Loc}\mathsf{RingSp}\).
\item
  If \(f\in \mathsf{CRing}(A, B)\) is a ring morphism, it induces a
  morphism
  \((f, f^\#) \in \mathsf{Loc}\mathsf{RingSp}(\operatorname{Spec}B, \operatorname{Spec}A)\).
\item
  Moreover, every
  \((f, f^\#) \in \mathsf{Loc}\mathsf{RingSp}(\operatorname{Spec}B, \operatorname{Spec}A)\)
  is induced by some \(f\in {\mathsf{Top}}(A, B)\).
\end{enumerate}

\end{proposition}

\begin{remark}

Recall that a \emph{scheme} \((X, {\mathcal{O}}_X)\) is a LRS which is
locally isomorphic to some affine scheme
\((\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-09-20_11-35.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{remark}

\begin{definition}[Complete Ring]

A ring \(R\) is \textbf{complete} with respect to
\(I{~\trianglelefteq~}R\) if \(R = \invlim_{\leftarrow} R/I^k\).
Elements can be written as sequence \((a_k)_{k\geq 0}\) such that
\(a_k \equiv a_{k-1} \operatorname{mod}I^{k-1}\).

\end{definition}

\begin{example}[?]

A non-example: let \(R = {\mathbf{C}}\dagger{t}\) and
\(I = \left\langle{t}\right\rangle\), then set

\begin{itemize}
\tightlist
\item
  \(a_0 = 1\),
\item
  \(a_1 = 1 + t\),
\item
  \(a_2 = 1 + t + t^2\),
\item
  \(a_3 = 1 + t + t^2 + t^3\)
\end{itemize}

This is an element in \(\colim_n R/I^n\), but is not realized by any
polynomial, since any polynomial is annihilated by quotienting by a
large enough power of \(t\). Note that
\({\mathbf{C}} {\llbracket t \rrbracket } = {\mathbf{C}} { \left[ \scriptstyle {[} \right] } t] { {}_{ \widehat{\left\langle{t}\right\rangle} } }\).

\end{example}

\begin{example}[?]

Part (c) of the proposition would be false if we considered all ringed
spaces. Let \(R \in \mathsf{DVR}\), so \(R\) is local with a principal
maximal idea, or equivalently equipped with a valuation
\(v: R\setminus\left\{{0}\right\}\to {\mathbf{Z}}_{\geq 0}\) satisfying

\begin{itemize}
\tightlist
\item
  \(v(x+y) = v(x) + v(y)\)
\item
  \(v(x+y) \geq \min{v(x), v(y)}\)
\end{itemize}

Then \(v^{-1}({\mathbf{Z}}_{\geq 0})\) is the maximal ideal. Here we'll
additionally require that \(R\) is \textbf{complete} with respect to its
maximal ideal \({\mathfrak{m}}\).

An example is \(R = {\mathbf{C}} {\llbracket t \rrbracket }\), with
\(v\) the \(t{\hbox{-}}\)adic valuation. Another is
\(R = { {\mathbf{Z}}_{\widehat{p}} }= {\mathbf{Z}}{ {}_{ \widehat{p} } }\),
the completion of \({\mathbf{Z}}\) at the prime \(p\), given by elements
\(a_n\cdots a_0\) with \(a_i \in \left\{{0, \cdots, p-1}\right\}\). This
has maximal ideal \({\mathfrak{m}}= p{ {\mathbf{Z}}_{\widehat{p}} }\).

\end{example}

\begin{example}[A morphisms of ringed spaces that isn't a morphism of locally ringed spaces]

Let \(K = \operatorname{ff}(R)\) be the fraction field of \(R\), then
\begin{align*}
\operatorname{ff}({\mathbf{C}} {\llbracket t \rrbracket  } ) = {\mathbf{C}} { \left( {t} \right) }  = \left\{{\sum_{k\geq -N} a_k t^k {~\mathrel{\Big\vert}~}N\in {\mathbf{Z}}_{\geq 0} }\right\}
.\end{align*}
Also
\(\operatorname{ff}({ {\mathbf{Z}}_{\widehat{p}} }) = { {\mathbf{Q}}_p }\),
and these are both examples of complete DVRs.

Consider
\((\operatorname{Spec}R, {\mathcal{O}}_{\operatorname{Spec}R})\) and
\((\operatorname{Spec}K, {\mathcal{O}}_{\operatorname{Spec}K})\). Then
\(X_1\coloneqq\operatorname{Spec}R = \left\{{ \left\langle{0}\right\rangle, {\mathfrak{m}}}\right\}\),
and the closed sets are
\(\emptyset, X_1, V({\mathfrak{m}}) = \left\{{{\mathfrak{m}}}\right\}\).
For
\(X_2\coloneqq\operatorname{Spec}K = \left\{{\left\langle{0}\right\rangle}\right\}\),
there is one points since it's a field. Use that
\(\iota: R \hookrightarrow K\), so define a morphism that does not come
from a ring morphism \(R\to K\):
\begin{align*}
(f, f^\#): (\operatorname{Spec}K, {\mathcal{O}}_{\operatorname{Spec}K}) &\to (\operatorname{Spec}R, {\mathcal{O}}_{\operatorname{Spec}R}) \\
\\
\\
f: \operatorname{Spec}K &\to \operatorname{Spec}R\\
0 &\mapsto {\mathfrak{m}}
\\
\\
f^\#: {\mathcal{O}}_{\operatorname{Spec}R} &\to f_* {\mathcal{O}}_{\operatorname{Spec}K} \\
\emptyset &\mapsto 0 \\
\operatorname{Spec}R & \mapsto R \\
\left\{{\left\langle{0}\right\rangle}\right\} &\mapsto K
.\end{align*}
using that
\(f^{-1}(\left\langle{0}\right\rangle) = \left\langle{m}\right\rangle\)
and we can realize the last assignment as a distinguished open mapping
to its stalk/localization. Then check
\begin{align*}
f_* {\mathcal{O}}_{\operatorname{Spec}K}(\emptyset) &= 0
f_* {\mathcal{O}}_{\operatorname{Spec}K}(\operatorname{Spec}R) &= K \\
f_* {\mathcal{O}}_{\operatorname{Spec}K}(\left\{{\left\langle{0}\right\rangle}\right\}) &= {\mathcal{O}}_{\operatorname{Spec}K}(f^{-1}(\left\langle{0}\right\rangle)) = {\mathcal{O}}_{\operatorname{Spec}K}(\emptyset) = 0 \\
.\end{align*}
This would induces a commutative diagram, showing this is a morphism of
ringed spaces:

\begin{center}
\begin{tikzcd}
    R && K \\
    \\
    K && 0 \\
    \\
    0 && 0
    \arrow["{f^\#(X)}", from=1-1, to=1-3]
    \arrow["{f^\#(\left\{{\left\langle{0}\right\rangle}\right\})}", from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-3, to=3-3]
    \arrow[from=3-3, to=5-3]
    \arrow[from=3-1, to=5-1]
    \arrow["{f^\#(\emptyset)}", from=5-1, to=5-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJSIl0sWzIsMCwiSyJdLFswLDIsIksiXSxbMiwyLCIwIl0sWzAsNCwiMCJdLFsyLDQsIjAiXSxbMCwxLCJmXlxcIyhYKSJdLFsyLDMsImZeXFwjKFxcdHN7XFxnZW5zIDB9KSJdLFswLDJdLFsxLDNdLFszLDVdLFsyLDRdLFs0LDUsImZeXFwjKFxcZW1wdHlzZXQpIl1d}{Link
to Diagram}
\end{quote}

\begin{question}

Is this a morphism of \emph{locally} ringed spaces?

\end{question}

The answer is \textbf{no}, since the induced morphism on stalks won't be
morphisms of \emph{local} rings. We can check
\begin{align*}
f^\#_{\left\langle{0}\right\rangle}: {\mathcal{O}}_{\operatorname{Spec}R, {\mathfrak{m}}} &\to {\mathcal{O}}_{\operatorname{Spec}K, \left\langle{0}\right\rangle} \\
f^\#_{\left\langle{0}\right\rangle}: R = R \left[ { \scriptstyle { {{\mathfrak{m}}}^{-1}}  } \right]  &\to K \left[ { \scriptstyle { {\left\langle{0}\right\rangle}^{-1}}  } \right]  = K
,\end{align*}
and
\((f^\#_{\left\langle{0}\right\rangle})^{-1}(\left\langle{0}\right\rangle) = \left\langle{0}\right\rangle \neq {\mathfrak{m}}\),
which is not the maximal ideal of \(R\).

On the other hand, using part (b) of the proposition, any
\(\phi\in \mathsf{Ring}(R, K)\) induces a morphism
\(\tilde \phi: \mathsf{Loc}\mathsf{RingSp}(\operatorname{Spec}K, \operatorname{Spec}R)\).
So \((f, f^\#)\) is not induced by any such ring map \(\phi\).

\end{example}

\begin{remark}

So the functor
\begin{align*}
\mathsf{Ring}&\to \mathsf{RingSp}\\
A &\mapsto (\operatorname{Spec}A, {\mathcal{O}}_{\operatorname{Spec}A})
\end{align*}
is not fully faithful, but restricting the essential image to
\(\mathsf{Loc}\mathsf{RingSp}\).

\end{remark}

\begin{remark}

Consider the heuristic
\(\operatorname{Spec}{\mathbf{C}} {\llbracket t \rrbracket } \sim {\mathbb{D}}\subseteq {\mathbf{C}}\)
and
\(\operatorname{Spec}{\mathbf{C}} { \left( {t} \right) } \sim {\mathbb{D}}\setminus\left\{{0}\right\}\).

\end{remark}

\hypertarget{wednesday-september-22}{%
\section{Wednesday, September 22}\label{wednesday-september-22}}

\begin{remark}

Today: how to build more schemes by gluing known ones together. Let
\((X_1, {\mathcal{O}}_{X_1})\) and
\((X_2, {\mathcal{O}}_{X_2}) \in {\mathsf{Sch}}\), i.e.~locally ringed
spaces locally isomorphic to
\((\operatorname{Spec}R, {\mathcal{O}}_{\operatorname{Spec}R})\). Let
\(U_i \subseteq X_i\), and suppose we're given an isomorphism in
\(\mathsf{Loc}\mathsf{RingSp}\):
\begin{align*}
\phi: 
(U_1, { \left.{{{\mathcal{O}}_{X_1}}} \right|_{{U_1}} }) \to 
(U_2, { \left.{{{\mathcal{O}}_{X_2}}} \right|_{{U_2}} }) 
.\end{align*}

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{42pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-09-22_11-41.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Define a locally ringed space as follows: set
\(X = X_1 {\textstyle\coprod}X_2/ x\sim \phi(x)\), and define
\begin{align*}
{\mathcal{O}}_X(U) \coloneqq
\left\{{ 
s_1 \in {\mathcal{O}}_{X_1}(X_1 \cap U), s_2\in {\mathcal{O}}_{X_2}(X_2 \cap U) {~\mathrel{\Big\vert}~}{ \left.{{s_1}} \right|_{{U_1 \cap U}} } = \phi^\#(U_2 \cap U)\qty{ { \left.{{s_2}} \right|_{{U_2 \cap U}} } }
}\right\} 
.\end{align*}

\end{remark}

\begin{example}[?]

Consider \(I \coloneqq(0, 1) \subseteq {\mathbf{R}}\) and take
\(X_1 = X_2 \coloneqq(I, C^\infty({-}; {\mathbf{R}}))\). Using these to
cover the circle, we can obtain \((S^1, C^\infty({-}; {\mathbf{R}}))\),
using smooth functions that agree on the overlap (here a disjoint union
of smaller intervals).

\end{example}

\begin{example}[A non-affine scheme]

Let
\(X_1 = X_2 \coloneqq{\mathbf{A}}^1_{/ {k}} \coloneqq\operatorname{Spec}k[x]\),
and set \(U_1 \subseteq X_1 \coloneqq D(x)\). Then take the clear
isomorphism
\begin{align*}
(U_1, {\mathcal{O}}_{X_1}{ \left.{{}} \right|_{{U_1}} }) \xrightarrow{\operatorname{id}} 
(U_2, {\mathcal{O}}_{X_2}{ \left.{{}} \right|_{{U_2}} })
,\end{align*}
since they're the same open subset of the same affine variety. Gluing
yields the following:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-09-22_11-56.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{example}

\begin{exercise}[?]

Prove this is not an affine scheme. Use that regular functions are
determined by their values on a Zariski open.

\end{exercise}

\begin{claim}

For \(X = \operatorname{Spec}R\), recall that
\(D(f) \coloneqq\left\{{ p\in \operatorname{Spec}R {~\mathrel{\Big\vert}~}p\not\ni f}\right\}\).
Then \((D(f), { \left.{{ {\mathcal{O}}_X }} \right|_{{D(f)}} })\) is an
affine scheme.

\end{claim}

\begin{proof}[?]

Consider
\(R \left[ { \scriptstyle { {f}^{-1}} } \right] \coloneqq\left\{{r/f^k {~\mathrel{\Big\vert}~}r\in R, k\geq 0}\right\}/\sim\).
There's a map \(R\to R \left[ { \scriptstyle { {f}^{-1}} } \right]\)
which induces a map
\(\operatorname{Spec}R \left[ { \scriptstyle { {f}^{-1}} } \right] \to \operatorname{Spec}R\),
and we claim it's the inclusion of \(D(f)\).

\begin{claim}

\(\operatorname{Spec}R \left[ { \scriptstyle { {f}^{-1}} } \right] = D(f)\)
as spaces.

\end{claim}

This uses that primes in the localization are primes in \(R\) not
intersecting the inverted set. So
\(\operatorname{Spec}R \left[ { \scriptstyle { {f}^{-1}} } \right] = \left\{{p\in \operatorname{Spec}R {~\mathrel{\Big\vert}~}p \cap\left\langle{f}\right\rangle = \emptyset}\right\}\),
then use that
\(p \cap\left\{{f^k}\right\} = \emptyset \iff f\not\in p\), since prime
ideals are radical. We now want to show
\({\mathcal{O}}_{\operatorname{Spec}R \left[ { \scriptstyle { {f}^{-1}} } \right] } = { \left.{{{\mathcal{O}}_{\operatorname{Spec}R}}} \right|_{{D(f)}} }\).
Check that
\begin{align*}
{\mathcal{O}}_{\operatorname{Spec}R \left[ { \scriptstyle { {f}^{-1}}  } \right] } = \left\{{s:U\to \coprod_{p\in U} (R \left[ { \scriptstyle { {f}^{-1}}  } \right] ) \left[ { \scriptstyle { {p}^{-1}}  } \right] }\right\}
\end{align*}
and
\begin{align*}
{ \left.{{{\mathcal{O}}_{\operatorname{Spec}R}}} \right|_{{D(f)}} }(U) = \left\{{s: U \to \coprod_{p\in U} R \left[ { \scriptstyle { {p}^{-1}}  } \right] }\right\}
,\end{align*}
but \((R_f)_p = R_p\) since \(f\not \in p\). This uses that
\((R \left[ { \scriptstyle { {S_1}^{-1}} } \right] ) \left[ { \scriptstyle { {S_2}^{-1}} } \right] = R \left[ { \scriptstyle { {S_2}^{-1}} } \right]\)
when \(S_1 \subseteq S_2\). The first are functions of the form
\((a/f^k)/(b/f^\ell) = f^\ell a/f^k b\), so anything locally a fraction
in \(R_f\) is locally a fraction in \(R\).

\end{proof}

\begin{example}[?]

Let \(X_1 = {\mathbf{A}}^1_{/ {k}}\) with
\(U_1 \coloneqq D(x) \subseteq X_2\) and
\(X_2 = {\mathbf{A}}^1_{/ {k}}\) with \(U_2 = D(y)\). Then
\begin{align*}
(U_1, {\mathcal{O}}_{X_1}{ \left.{{}} \right|_{{U_1}} }) \cong (k[x, x^{-1}], {\mathcal{O}}_{\operatorname{Spec}k[x, x^{-1}]}) 
\xrightarrow{(\phi, \phi^\#)} 
(U_2, {\mathcal{O}}_{X_2}{ \left.{{}} \right|_{{U_2}} }) \cong (k[y, y^{-1}], {\mathcal{O}}_{\operatorname{Spec}k[y, y^{-1}]})
.\end{align*}

Then \((\phi, \phi^\#)\) is an isomorphism in
\(\mathsf{Loc}\mathsf{RingSp}\) is given by a unique isomorphism
\begin{align*}
\iota: k[x, x^{-1}] &\to k[y, y^{-1}] \\
y &\mapsto x \\
y^{-1}&\mapsto x^{-1}
.\end{align*}
Note that there is another isomorphism:
\begin{align*}
\iota': k[x, x^{-1}] &\to k[y, y^{-1}] \\
y &\mapsto x^{-1}\\
y^{-1}&\mapsto x
.\end{align*}
So glue instead by the morphism \((\phi, \phi^\#)\) induced by
\(\iota'\). We'll then define \textbf{projective space} as
\begin{align*}
{\mathbf{P}}^1_{/ {k}}  \coloneqq{\mathbf{A}}^1_{/ {k}}  \coprod_{(\phi, \phi^\#)} {\mathbf{A}}^1_{/ {k}} 
.\end{align*}
Note that this works over any ring!

What does this do to closed points? The closed points of
\(\operatorname{Spec}k[x, x^{-1}]\) are
\(\left\{{(x-c) {~\mathrel{\Big\vert}~}c\neq 0}\right\}\) if
\(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\),
which corresponded to the closed points \(c\in {\mathbf{A}}^1_{/ {k}}\)
as a variety. Under \(x\mapsto y^{-1}\), we have
\((x-c)\mapsto (y^{-1}- c) = (y-c^{-1})\), thus the variety point \(c\)
gets sent to \(c^{-1}\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-09-22_12-21.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{example}

\hypertarget{projective-varieties-tuesday-september-28}{%
\section{Projective Varieties (Tuesday, September
28)}\label{projective-varieties-tuesday-september-28}}

\hypertarget{projective-space}{%
\subsection{Projective Space}\label{projective-space}}

\begin{definition}[Affine space]

Let \(R\) be a ring, then the \textbf{affine space of dimension \(n\)
over \(R\)} is defined as
\begin{align*}
{\mathbf{A}}^n_{/ {R}}  \coloneqq(\operatorname{Spec}S, {\mathcal{O}}_{\operatorname{Spec}S}) && S \coloneqq R[{ {x}_1, {x}_2, \cdots, {x}_{n}}]
.\end{align*}

\end{definition}

\begin{definition}[Slice schemes]

Let \(S\in {\mathsf{Sch}}\), then \(X\in {\mathsf{Sch}}_{/ {S}}\) is a
\textbf{scheme over \(S\)} iff \(X\) is a scheme equipped with a
morphism of schemes \(X\to S\). These form a category where morphisms
\(\phi\) are commuting triangles:

\begin{center}
\begin{tikzcd}
X 
  \ar[rd, "f_X"]
  \ar[rr, "\phi"] 
& 
& 
Y
  \ar[ld, "f_Y"] 
\\
& 
S 
& 
\end{tikzcd}
\end{center}

\end{definition}

\begin{remark}

Since \({\mathbf{Z}}\in \mathsf{CRing}\) is initial, there exists a
unique ring morphism \({\mathbf{Z}}\to {\mathbf{R}}\) for any
\(R\in \mathsf{CRing}\). Similarly,
\(\operatorname{Spec}{\mathbf{Z}}\in{\mathsf{Sch}}\) is final, so there
exist unique morphisms
\(\operatorname{Spec}R\to \operatorname{Spec}{\mathbf{Z}}\) for every
\(R\), and thus
\({\mathsf{Sch}}_{/ {\operatorname{Spec}{\mathbf{Z}}}} \cong {\mathsf{Sch}}\)
just recovers the category of schemes.

\end{remark}

\begin{remark}

Recall that if
\(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu \in \mathsf{Field}\)
is algebraically closed, then
\({\mathbf{P}}^n_{/ {k}} \coloneqq{\mathbf{A}}^{n+1}_{/ {k}} \setminus\left\{{0}\right\}/\sim\)
where \(\mathbf{x}\sim \lambda \mathbf{x}\) for every
\(\lambda \in k^{\times}\), or equivalently
\({\operatorname{Gr}}_k(k^{n+1})\), the space of lines through the
origin in \(k^{n+1}\) (regarded as a vector space). Let
\(f\in k[x_1, \cdots, x_{n}]_d^{ { \mathrm{homog} } }\) be homogeneous
of degree \(d\), so \(f(\lambda \mathbf{x}) = \lambda^d f(\mathbf{x})\).
Then its vanishing locus in projective space is well-defined:
\begin{align*}
V_p(f) \coloneqq\left\{{ \mathbf{x} = {\left[ {x_0: \cdots : x_n} \right]} \subseteq {\mathbf{P}}^n_{/ {k}}  {~\mathrel{\Big\vert}~}f(\mathbf{x}) = 0 }\right\}\subseteq {\mathbf{P}}^n_{/ {k}} 
.\end{align*}

\end{remark}

\hypertarget{graded-rings-and-homogeneous-ideals}{%
\subsection{Graded Rings and Homogeneous
Ideals}\label{graded-rings-and-homogeneous-ideals}}

\begin{definition}[Graded rings]

A ring \(R\in \mathsf{CRing}\) is \textbf{graded} if it admits a
decomposition as an abelian group \(R + \bigoplus _{d\geq 0} S_d\),
where \(S_d\) are the degree \(d\) pieces satisfying
\(S_a + S_b \subseteq S_{a+b}\).

\end{definition}

\begin{remark}

Note that \(S_d \in {}_{S_0}{\mathsf{Mod}}\) for any
\(S \in {\mathsf{gr}\,}\mathsf{CRing}\) and any degree \(d\).

\end{remark}

\begin{example}[?]

\(R\coloneqq k[x_1, \cdots, x_{n}]^ { \mathrm{homog} }\) is graded by
monomial degree:
\begin{align*}
1 + (x_0) + (x_0^2 +x_1^2) \in R_0 + R_1 + R_2
.\end{align*}

\end{example}

\begin{definition}[Homogeneous Ideals]

Let \(S\in {\mathsf{gr}\,}\mathsf{CRing}\) be a graded ring, then an
ideal \(I {~\trianglelefteq~}S\) is \textbf{homogeneous} if

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(I\) is generated by homogeneous elements, and
\item
  For all \(f\in I\), all homogeneous pieces \(f_i \in S_i\) such that
  \(f = \sum_{i\leq d} f_d\) lie in \(I\).
\end{enumerate}

\end{definition}

\begin{example}[?]

If \(f \coloneqq 1 + x_0 + x_0^2 + x_1^2 \in I\) is an element in a
homogeneous ideal, then \(1, x_0, x_0^2 + x_1\in I\) as well.

\end{example}

\begin{remark}

If \(I{~\trianglelefteq~}k[x_1, \cdots, x_{n}]\) is a homogeneous ideal,
say \(I = \left\langle{{ {f}_1, {f}_2, \cdots, {f}_{m}}}\right\rangle\)
with each \(f_i\) homogeneous of uniform degree \(d\), then \(V_p(I)\)
is a well-defined projective variety.

\end{remark}

\begin{example}[?]

\envlist

\begin{itemize}
\item
  Take \(V_p(y^2 - (x^3 + axz^2 + bz^3))\) for \(a, b\in k\) and
  \(4a^3-27b^2 \neq 0\). This defines an elliptic curve.
\item
  \({\mathbf{P}}^n_{/ {k}} = V_p(0)\).
\end{itemize}

\end{example}

\begin{definition}[Irrelevant Ideal]

We define \(S_+ \coloneqq\bigoplus _{d\geq 1} S_d\) to be the
\textbf{irrelevant ideal}.

\end{definition}

\hypertarget{projective-nullstellensatz}{%
\subsection{Projective
Nullstellensatz}\label{projective-nullstellensatz}}

\begin{remark}

We again define the Zariski topology on \(X = V_p(I)\) whose closed sets
are of the form \(V_p(J)\) for
\(J \subseteq (k[x_1, \cdots, x_{n}]/I)^{ { \mathrm{homog} } }\)

\end{remark}

\begin{theorem}[Projective Nullstellensatz]

Let \(k[X] \coloneqq k[x_1, \cdots, x_{n}]/I\) be the projective
coordinate ring of \(X \subset {\mathbf{P}}^n_{/ {k}}\) and
\(I = I(X)\). Then there is a bijection:
\begin{align*}
\left\{{\substack{
  \text{Points }x\in X
}}\right\}
&\rightleftharpoons
\left\{{\substack{
  \text{Homogeneous }I\in \operatorname{mSpec}S \\
  \text{with }I\not\supseteq S_+
}}\right\} \\
x &\mapsto I(x) \coloneqq\left\langle{ f\in k[x_1, \cdots, x_{n}]^ { \mathrm{homog} } {~\mathrel{\Big\vert}~}f(x) = 0}\right\rangle \\
V_p(I) &\mapsfrom I
\end{align*}

\end{theorem}

\begin{remark}

Note \(I\) doesn't contain \(S_+\) iff \(I\) is strictly contained in
\(S_+\).

\end{remark}

\hypertarget{proj}{%
\subsection{Proj}\label{proj}}

\begin{definition}[Proj]

Let \(S\in {\mathsf{gr}\,}\mathsf{CRing}\), then define
\begin{align*}
\mathop{\mathrm{Proj}}S \coloneqq\left\{{p\in  \operatorname{Spec}S \text{ homogeneous} {~\mathrel{\Big\vert}~}p\not\supseteq S_+}\right\}
,\end{align*}
where \(S_+ \coloneqq\bigoplus_{d\geq 1} S_d\) is the irrelevant ideal.

\end{definition}

\begin{remark}

We'll define \({\mathcal{O}}_{\mathop{\mathrm{Proj}}S}\) next class.

\end{remark}

\hypertarget{friday-october-01}{%
\section{Friday, October 01}\label{friday-october-01}}

\begin{remark}

Recall the Proj construction: for
\(S = \bigoplus_{d\geq 0} S_d \in {\mathsf{gr}\,}\mathsf{CRing}\) we
define the irrelevant ideal \(S_+ \coloneqq\bigoplus _{d\geq 1} S_d\)
and
\begin{align*}
\mathop{\mathrm{Proj}}S &\coloneqq\left\{{p\in \operatorname{Spec}S \text{ homog } {~\mathrel{\Big\vert}~}p\not\supseteq S_+}\right\} \\
{\mathcal{O}}_{\mathop{\mathrm{Proj}}S} &\coloneqq\left\{{s: U\to \coprod_{p\in U} S \left[ { \scriptstyle { { ({p}^c) }^{-1}}  } \right] {~\mathrel{\Big\vert}~}s(p) \in S \left[ { \scriptstyle { { ({p}^c) }^{-1}}  } \right], s \text{ locally a fraction}}\right\}
,\end{align*}
recalling that
\(S \left[ { \scriptstyle { { ({p}^c) }^{-1}} } \right] = \left\{{a /f {~\mathrel{\Big\vert}~}\deg a = \deg f, a,f\in S, f\not\in p}\right\}\).
We showed this was a locally ringed space using
\begin{align*}
( D(f), { \left.{{{\mathcal{O}}_{\mathop{\mathrm{Proj}}S}}} \right|_{{D(f)}} }  { \, \xrightarrow{\sim}\, }(\operatorname{Spec}S \left[ { \scriptstyle { {f}^{-1}}  } \right] , {\mathcal{O}}_{\operatorname{Spec}S \left[ { \scriptstyle { {f}^{-1}}  } \right] })
,\end{align*}
where
\(D(f) \coloneqq\left\{{p\in \mathop{\mathrm{proj}}S {~\mathrel{\Big\vert}~}f\not\in p}\right\}\),
and thus \(\mathop{\mathrm{Proj}}S \in {\mathsf{Sch}}\).

\end{remark}

\begin{exercise}[?]

Check that there is a natural map of schemes
\(\mathop{\mathrm{Proj}}S\to \operatorname{Spec}S_0\).

\end{exercise}

\begin{remark}

Consider
\begin{align*}
{\mathbf{P}}^n_{/ {R}}  \coloneqq\mathop{\mathrm{Proj}}R[x_0,\cdots, x_n] && R = k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu \in \mathsf{Field}
.\end{align*}
Then the closed points of \({\mathbf{P}}^n_{/ {k}}\) are of the form
\(\left\langle{a_i x_j - a_j x_i}\right\rangle \in \operatorname{mSpec}k[x_1, \cdots, x_{n}]\)
for points \([a_0: \cdots : a_n] \in k^n/\sim\) where
\(\mathbf{a} \sim \lambda \mathbf{a}\) for \(\lambda \in k^{\times}\).
Note that
\(D(x_i) = \left\{{p\in {\mathbf{P}}^n_{/ {k}} {~\mathrel{\Big\vert}~}x_i \not\in p}\right\}\)
-- what are the closed points? We discard the hyperplane \(a_i = 0\) in
\({\mathbf{P}}^n\) to obtain

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-01_11-43.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Then \(x_i \in {\mathfrak{m}}_q\) for \(q \coloneqq[a_0: \cdots : a_n]\)
iff \(a_i=0\), and
\begin{align*}
D(x_i) 
&= \operatorname{Spec}k[x_1, \cdots, x_{n}] \left[ { \scriptstyle { {x_i}^{-1}}  } \right]  \\
&= \left\{{f(x_0, \cdots, x_n)/x_i^d {~\mathrel{\Big\vert}~}\deg f = d}\right\} \\
&= \left\{{f\qty{ {x_0\over x_i }, \cdots, 1, \cdots, {x_n \over x_i }}}\right\} \\
&= k { \left[ \scriptstyle {{x_0\over x_i}, \cdots, {x_n \over x_i} } \right] }  \\
&\cong {\mathbf{A}}^n_{/ {k}} 
.\end{align*}
We claim that \(\bigcup_{i\geq 0} D(x_i) = {\mathbf{P}}^n_{/ {k}}\), or
equivalently
\(\emptyset = \bigcap_{i \geq 0} V(x_i) = V(\left\langle{x_0,\cdots, x_n}\right\rangle)\).
But this is true since
\(\left\langle{x_0, \cdots, x_n}\right\rangle = S^+\) is the irrelevant
ideal.

\end{remark}

\begin{proposition}[?]

Let
\(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu \in \mathsf{Field}\).
Then there is a fully faithful embedding of categories
\begin{align*}
F: {\mathsf{Var}}_{/ {k}}  \hookrightarrow{\mathsf{Sch}}_{/ {k}} 
.\end{align*}
Here \({\mathsf{Var}}_{/ {k}}\) are defined as topological spaces with
sheaves of rings of \emph{regular functions} which admitted an affine
cover of the form \(V(I) \subseteq {\mathbf{A}}^n_{/ {k}}\) (viewed as a
variety).

\end{proposition}

\begin{example}[Going from a variety to a scheme]

Consider \(X\coloneqq{\mathbf{A}}^2_{/ {k}}\) as a variety and
separately as a scheme \(X'\). As a variety,
\(X \coloneqq k{ {}^{ \scriptscriptstyle\times^{2} } }\) with the
Zariski topology, while as a scheme \(X' = \operatorname{Spec}k[x, y]\)
with the Zariski topology. Then there is an inclusion
\(X \hookrightarrow X'\) which is a bijection on closed points.

More generally, for \(X\in {\mathsf{Top}}\) any space, define \(t(X)\)
to be the set of irreducible closed subsets. Some facts:

\begin{itemize}
\tightlist
\item
  For \(Y \subseteq X\) closed, \(t(Y) \subseteq t(X)\),
\item
  \(t(Y_1 \cup Y_2) = t(Y_1) \cup t(Y_2)\),
\item
  \(t( \bigcap_i Y_i) = \bigcap_i t(Y_i)\).
\end{itemize}

Define a topology on \(t(X)\) by declaring closed sets to be any of the
form \(t(Y)\) for \(Y \subseteq X\) closed. Note that the scheme \(X'\)
has non-closed points, i.e.~irreducible subvarieties, i.e.~irreducible
closed subsets of \(X\) as a variety:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{35pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-01_12-02.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Then consider the map
\begin{align*}
\alpha: X &\to t(X) \\
p &\mapsto \left\{{p}\right\}
,\end{align*}
noting that this is only well-defined if points are closed in \(X\).

Now let \(V\in {\mathsf{Var}}_{/ {k}}\) with its sheaf of regular
functions \({\mathcal{O}}_V\) (i.e.~restrictions of polynomials). Define
a sheaf of rings on \(t(V)\) as \(\alpha_* {\mathcal{O}}_V\), using that
\(\alpha\) is continuous, and noting that
\(\alpha^{-1}(U) = U \cap\alpha(X)\).

To see this is a scheme, it suffices to check for \(V\) affine since
this entire construction is compatible with restriction and we can take
an affine cover. Letting \(I = I(V)\) for
\(V\in {\mathsf{Aff}}{\mathsf{Var}}_{/ {k}}\), then
\((t(V), \alpha_* {\mathcal{O}}_V) { \, \xrightarrow{\sim}\, }\operatorname{Spec}k[V] \coloneqq\operatorname{Spec}k[x_1, \cdots, x_{n}]/I\).
There is a bijection
\begin{align*}
t(V) &\rightleftharpoons\operatorname{Spec}k[V] \\
Y &\mapsto I(Y) \\
V(p) &\mapsfrom p
.\end{align*}
One can check that the topology on \(t(V)\) bijects with the Zariski
topology on \(\operatorname{Spec}k[V]\), and
\begin{align*}
\alpha_* {\mathcal{O}}_V(t(V)) = {\mathcal{O}}_V(V) = {\mathcal{O}}_{\operatorname{Spec}k[V]}(\operatorname{Spec}k[V]) = k[V]
.\end{align*}

\end{example}

\begin{exercise}[?]

Check this on open subsets of \(t(V)\).

\end{exercise}

\begin{remark}

\({\mathcal{O}}_X \in {\mathsf{Sh}}(X, {}_{k} \mathsf{Alg} )\) being a
sheaf of \(k{\hbox{-}}\)algebras means the following diagram commutes:

\begin{center}
\begin{tikzcd}
    k && {{\mathcal{O}}_X(U)} \\
    \\
    && {{\mathcal{O}}_X(V)}
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-1, to=3-3]
    \arrow["{\operatorname{res}_{UV}}", from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJrIl0sWzIsMCwiXFxPT19YKFUpIl0sWzIsMiwiXFxPT19YKFYpIl0sWzAsMV0sWzAsMl0sWzEsMiwiXFxyZXNfe1VWfSJdXQ==}{Link
to Diagram}
\end{quote}

This is the data of a morphism
\((X, {\mathcal{O}}_X) \to \operatorname{Spec}k\).

\end{remark}

\begin{remark}

What's the point of the theorem? Not everything of geometric interest is
in the essential image of \(F\). Consider
\(V(y-x^2) \subseteq {\mathbf{A}}^2_{/ {k}}\), and consider intersecting
it with lines \(y=t\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{42pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-01_12-19.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Letting \(I_1 = I_1(\left\langle{y-x^2}\right\rangle)\) and
\(I_2 = I(\left\langle{y-t}\right\rangle)\), intersecting in varieties
yields
\begin{align*}
V_1 \cap V_2 = V(I_1 + I_2) = V(\sqrt{I_1 + I_2})
.\end{align*}
One can check \(I_1 + I_2 = (x-\sqrt t, y-t)\cdot (x+\sqrt t, y-t)\),
and
\(\operatorname{Spec}k[x, y] / \left\langle{y-x^2, y}\right\rangle = \operatorname{Spec}k[x]/\left\langle{x^2}\right\rangle\)
when \(t=0\) (i.e.~when there's a tangency with multiplicity), since the
scheme intersection is
\(\operatorname{Spec}k[x, y] / \left\langle{I_1 + I_2}\right\rangle\).
Note that the regular functions on a point are just constant, so the
sheaf of regular functions on a point is \(k\) itself and thus doesn't
pick up the multiplicity of the intersection.

\end{remark}

\begin{remark}

There can be issues for \(\operatorname{Spec}R\) when \(R\) is finitely
generated but not reduced!

\end{remark}

\hypertarget{monday-october-04}{%
\section{Monday, October 04}\label{monday-october-04}}

\begin{remark}

We'll set up some properties for schemes. A scheme can be:

\begin{itemize}
\tightlist
\item
  Irreducible
\item
  Smooth
\item
  Reduced
\item
  Noetherian
\end{itemize}

\end{remark}

\begin{remark}

Recall that \(X\in {\mathsf{Top}}\) is \textbf{connected} iff whenever
\(X = X_1 {\textstyle\coprod}X_2\) with \(X_i\) closed, one of
\(X_i = X\) and the other is empty. \(X\) is \textbf{irreducible} iff
this holds in the weaker case when \(X = X_1 \cup X_2\) is not
necessarily disjoint. Note that irreducible implies connected.

\end{remark}

\begin{definition}[Connectedness and irreducibility for schemes]

\((X, {\mathcal{O}}_X) \in {\mathsf{Sch}}\) is \textbf{connected} (resp.
\textbf{irreducible}) iff
\({ {\left\lvert {X} \right\rvert} } \in {\mathsf{Top}}\) is connected
(resp. irreducible).

\end{definition}

\begin{example}[Connected and irreducible]

\(X\coloneqq{\mathbf{A}}^n_{/ {k}}\) is connected and irreducible. Write
\({\mathbf{A}}^n = \operatorname{Spec}k[x_1, \cdots, x_{n}]\), whose
closed sets are \(V(I) \coloneqq\left\{{p\supseteq I}\right\}\). Suppose
\(X = V(I) \cup V(J) = V(IJ)\), then all primes \(p\) satisfy
\(IJ \supseteq p\), and this
\(IJ \supseteq\cap_{p\in \operatorname{Spec}R} p = {\sqrt{0_{R}} } = 0\),
using a fact from commutative algebra. Then \(IJ = 0\) implies \(I=0\)
(wlog), so \(V(I) = X\).

\end{example}

\begin{example}[Connected but not irreducible]

Let
\(X \coloneqq\operatorname{Spec}k[x, y] / \left\langle{xy}\right\rangle\),
for \(k\) not necessarily algebraically closed. This is roughly the
coordinate axis in \(k^2\), except there are generic points
corresponding to irreducible closed subsets. Points
\(\left\langle{x-a, y-b}\right\rangle\) are closed and contained in
\(X\) when \((a, b)\) is on the axis. There are non-maximal primes
\(\left\langle{x}\right\rangle, \left\langle{y}\right\rangle\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-04_11-46.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Note
\(X \supseteq V(\left\langle{x}\right\rangle) = \left\{{p\in x}\right\} = \left\{{\left\langle{x}\right\rangle}\right\} \cup\left\{{\left\langle{x, y-b}\right\rangle}\right\}\),
and similarly,
\(V(\left\langle{y}\right\rangle) = \left\{{\left\langle{y}\right\rangle}\right\} \cup\left\{{\left\langle{x-a, y}\right\rangle}\right\}\).
However \(V(x) \cup V(y) = X\) but
\(V(x) \cap V(y) = \left\{{\left\langle{x, y}\right\rangle}\right\}\),
making \(X\) connected but not irreducible.

\end{example}

\begin{example}[Not connected]

Let
\(X\coloneqq\operatorname{Spec}( k[x]/\left\langle{x(x-1)}\right\rangle ) \cong \operatorname{Spec}( k[x]/\left\langle{x}\right\rangle ) \oplus \operatorname{Spec}(k[x]/\left\langle{x-1}\right\rangle) \cong \operatorname{Spec}(k \oplus k)\),
using the Chinese remainder theorem. Note that this has two prime
ideals, \(\left\{{0 \oplus k, k \oplus 0}\right\}\), making it a
discrete space and thus a disjoint union of its two closed points. Note
that \(0 \oplus 0\) isn't prime, consider \((a, 0) \oplus (0, b)\).

\end{example}

\begin{example}[Connected and irreducible]

Consider
\(X \coloneqq\operatorname{Spec}{ {\mathbf{Z}}_{\widehat{p}} }= \left\{{ \left\langle{0}\right\rangle, \left\langle{p}\right\rangle}\right\}\).
Then
\({ \operatorname{cl}} _X\qty{ \left\{{\left\langle{0}\right\rangle}\right\}} = X\),
since \(0\) is not a closed point. This is connected and irreducible,
and the picture is a point with a fuzzy point:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-04_11-55.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{example}

\begin{exercise}[$\spec \ZZ$ is connected and irreducible]

Show \(\operatorname{Spec}{\mathbf{Z}}\) is connected and irreducible.

\end{exercise}

\begin{definition}[Reduced schemes]

\((X, {\mathcal{O}}_X) \in {\mathsf{Sch}}\) is \textbf{reduced} iff
\({\mathcal{O}}_X(U)\) is a reduced ring for all open \(U \subseteq X\),
i.e.~contains no nilpotents. Equivalently, \({\mathcal{O}}_{X, p}\) is
reduced for all \(p\in X\).

\end{definition}

\begin{remark}

Note that this is a genuinely new concepts for schemes, since functions
valued in a field always yields a reduced ring.

\end{remark}

\begin{example}[$\spec R$ is reduced for $R$ reduced]

Let \(R\) be reduced and take \(X\coloneqq\operatorname{Spec}R\), then
for \(U \subseteq X\) with \(U = V(I)^c \supseteq D(f) \coloneqq V(f)\)
for any \(f\in I\). In fact \(U = \bigcup_{f\in I} D(f)\), and
\({\mathcal{O}}_X(U)\) by the sheaf property can be written as
\begin{align*}
{\mathcal{O}}_X(U) 
&= \left\{{\phi_f \in {\mathcal{O}}_X(D(f)) {~\mathrel{\Big\vert}~}{ \left.{{\phi_f}} \right|_{{D(f) \cap D(g) }} } = { \left.{{\phi_f}} \right|_{{D(f) \cap D(g)}} } }\right\} \\
&\subseteq \prod_{f\in I} {\mathcal{O}}_X(D(f)) \\
&= \prod_{f\in I} R \left[ { \scriptstyle { {f}^{-1}}  } \right] 
,\end{align*}
by the Dan-Changho theorem, and the claim is that
\(R \left[ { \scriptstyle { {f}^{-1}} } \right]\) is reduced for all
\(f\). This follows since \((a/f^k)^d \sim 0/1 \implies f^d a^d = 0\),
so \(f^{dm}a^d = (f^ma)^d = 0\), so \(f^ma=0\) since \(R\) is reduced
and this \(a/f^k \sim 0\), so the localization has no nilpotents. Then
\({\mathcal{O}}_X(U)\) is a subring of a reduced ring and thus reduced,
and \(\operatorname{Spec}R\) is a reduced scheme.

\end{example}

\begin{definition}[Integral schemes]

\((X, {\mathcal{O}}_X)\in {\mathsf{Sch}}\) is \textbf{integral} iff
\({\mathcal{O}}_X(U)\) is an integral domain for all \(U\).

\end{definition}

\begin{example}[$\spec R$ is integral for $R$ integral]

For \(R\) an integral domain, \(\operatorname{Spec}R\) is integral.
Supposing \(R\) is not an integral domain, so \(xy=0\) with
\(x,y\neq 0\). Then \(0\in p\) for any prime, so \(x\) or \(y\) is in
any prime ideal, so \(V(x) \cup V(y) = \operatorname{Spec}R\).
Alternatively, one can use
\({\mathcal{O}}_{\operatorname{Spec}R}(\operatorname{Spec}R) = R\).

\end{example}

\begin{proposition}[?]

\((X, {\mathcal{O}}_X)\) is integral \(\iff\) it is irreducible and
reduced.

\end{proposition}

\begin{proof}[of proposition]

\textbf{Integral \(\implies\) reduced}: Just use the domains have no
nilpotents.

\textbf{Integral \(\implies\) irreducible}: For the sake of
contradiction, suppose \(X = X_1 \cup X_2\) with \(X_i\) closed. Then
there exist \(U_i = X_i^c\) open and disjoint (using
\(X^c = X_1^c \cap X_2^c\)), so
\({\mathcal{O}}_(U_1 {\textstyle\coprod}U_2) = {\mathcal{O}}(U_1) \times {\mathcal{O}}(U_2)\)
by the sheaf property for \({\mathcal{O}}\). However, this is not an
integral domain.

\textbf{Irreducible and reduced \(\implies\) integral}: Next time!

\end{proof}

\hypertarget{wednesday-october-06}{%
\section{Wednesday, October 06}\label{wednesday-october-06}}

\begin{remark}

Recall: \(X\in {\mathsf{Sch}}\) is integral iff \(X\) is irreducible and
reduced, which are defined on sections in terms of rings.

\end{remark}

\begin{proof}[?]

\textbf{Irreducible and reduced \(\implies\) integral}: By
contrapositive, assume \({\mathcal{O}}_X(U)\) is not a domain, so
\(fg = 0\) in \({\mathcal{O}}_X(U)\). A local ring need not be domain.
However, the germ
\(f_p g_p \coloneqq\mathop{\mathrm{Res}}(U, {\mathfrak{p}})(fg) = 0\) in
the stalk \({\mathcal{O}}_{X, {\mathfrak{p}}}\). If
\((a/s)(b/t) = 0 \in {\mathfrak{p}}\), then either \(a/s\) or \(b/t\) is
in \({\mathfrak{p}}\), so \(f_p\) or \(g_p\) is in \({\mathfrak{p}}\).
Note that
\(U_1 \coloneqq\left\{{{\mathfrak{m}}\in U {~\mathrel{\Big\vert}~}f_p \in {\mathfrak{m}}}\right\} \subseteq U\)
is closed, as is
\(U_2 \coloneqq\left\{{{\mathfrak{m}}\in U {~\mathrel{\Big\vert}~}g_p \in {\mathfrak{m}}}\right\}\).

We can write \(U = U_1 \cup U_2\), so if \(X\) is irreducible then \(U\)
is irreducible, so some \(U_i = U\), say \(U_1\). So take an open affine
\(V \subseteq U_1\) with \({ \left.{{f}} \right|_{{V}} }\neq 0\), using
the sheaf property. Writing \(V = \operatorname{Spec}R\), we have
\({ \left.{{f}} \right|_{{V}} } \in {\mathcal{O}}_X(V) = R\), and the
stalk \(f_p\in p\) for all \(p\in R\). Then \(f\in p\) for all
\(p\in \operatorname{Spec}R\), thus in their intersection, and so
\(f\in {\sqrt{0_{R}} }\). Since \(f\neq 0\), this contradicts that \(X\)
is not reduced. \(\contradiction\)

\end{proof}

\begin{remark}

Recall that \emph{Noetherian rings} are those that satisfy the ACC, or
equivalently that all ideals are finitely generated (e.g.~a finitely
generated \(k{\hbox{-}}\)algebra).

A \emph{Noetherian space} is a space where every descending sequence of
closed sets stabilizes.

\end{remark}

\begin{definition}[Noetherian rings and spaces]

\(X\in {\mathsf{Sch}}\) is \textbf{locally Noetherian} iff there exists
an affine open cover with \(U_i = \operatorname{Spec}A_i\) for \(A_i\)
Noetherian. \(X\) is \textbf{(globally) Noetherian} if \(X\) is
additionally \emph{quasicompact}, i.e.~every open cover has a finite
subcover.

\end{definition}

\begin{example}[Non-Noetherian rings can produce Noetherian spaces]

The hypothesis of being a Noetherian space isn't enough in general.
Consider the ring of \emph{puiseux series} studied by Newton,
\(R = \bigcup_{n\geq 1} k {\llbracket t^{1\over n} \rrbracket }\). Then
\(\operatorname{Spec}R\) has 2 points:
\begin{align*}
\operatorname{Spec}R = \left\{{ {\mathfrak{p}}\coloneqq\left\langle{t^r}\right\rangle_{r \in {\mathbf{Q}}_{\geq 0}}, \left\langle{0}\right\rangle }\right\}
.\end{align*}
Here \(\left\langle{0}\right\rangle\) has closure containing
\({\mathfrak{p}}\), so \({\mathfrak{p}}\) is a generic point. This is a
valuation ring, just not a DVR, and is a Noetherian topological space
since there are only two closed sets. However, \(R\) is not Noetherian,
since there is an infinite chain of ideals:
\begin{align*}
\left\{{I_j}\right\}_{j\geq 1}: \quad \left\langle{t}\right\rangle \subsetneq \left\langle{t^{1\over 2}}\right\rangle \subsetneq \left\langle{t^{1\over 3}}\right\rangle \subsetneq \left\langle{t^{1\over 3}}\right\rangle \subsetneq \cdots
.\end{align*}
However, \(V(I_j) = V(I_{k})\) for all \(j, k\), and all equal to
\(V(\left\langle{p}\right\rangle)\), so \(\operatorname{Spec}R\) is a
Noetherian space!

Fun fact:
\(\operatorname{ff}(R) = \bigcup_{n\geq 1} k{\left(\left( t^{1\over n} \right)\right) } = \mkern 1.5mu\overline{\mkern-1.5muk{\left(\left( t \right)\right) }\mkern-1.5mu}\mkern 1.5mu\)
when
\(k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\).

\end{example}

\begin{remark}

There are many theorems of the form ``a scheme is locally
\emph{something}''. Here we required an open affine cover by
\(\operatorname{Spec}R\) for \(R\) Noetherian rings. The following two
conditions will thus be equivalent:

\begin{itemize}
\tightlist
\item
  A property \(P\) holds on every affine open \(U \subseteq X\),
\item
  There exists some affine cover \({\mathcal{U}}\rightrightarrows X\)
  satisfying property \(P\).
\end{itemize}

\end{remark}

\begin{theorem}[?]

\(X\in {\mathsf{Sch}}\) is locally Noetherian iff for any affine open
\(U = \operatorname{Spec}A \subseteq X\), \(A\) is a Noetherian ring.

\end{theorem}

\begin{proof}[of theorem]

\(\impliedby\): Definitional, just apply the hypothesis to some affine
open cover.

\(\implies\): The more nontrivial direction.

\begin{fact}[from ring theory]

The localization of any Noetherian ring is again Noetherian

\end{fact}

Let \(U \subseteq X\) be an affine open, so
\(U = \operatorname{Spec}B\), and let
\({\mathcal{U}}\rightrightarrows X\) be an affine cover:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-06_12-15.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Consider \(U \cap{\mathcal{U}}_i \subseteq {\mathcal{U}}_i\) open, which
can be covered by distinguished open sets. So write
\(U \cap{\mathcal{U}}_i = \bigcup_j V_{ij}\) with
\(V_{ij} = D(f_{ij}) \subseteq \operatorname{Spec}A_{i}\). Then \(U\) is
covered by
\(\operatorname{Spec}A_i \left[ { \scriptstyle { {f_{ij}}^{-1}} } \right]\),
i.e.~spectra of local Noetherian rings. Can we conclude that \(B\) is
Noetherian from this? This will follow from the fact that we can further
decompose \(V_{ij} = \bigcup W_{ijk}\) where \(W_{ijk} = D_B(f_{ijk})\).

So we want to show the following ring-theoretic statement: let
\(B\in \mathsf{Ring}\) and \(\left\{{g_i}\right\} \subseteq B\) be a
collection such that
\(\operatorname{Spec}B = \bigcup\operatorname{Spec}B \left[ { \scriptstyle { {g_i}^{-1}} } \right]\)
with each \(B \left[ { \scriptstyle { {g_i}^{-1}} } \right]\)
Noetherian, then \(B\) is necessarily Noetherian. Equivalently, we need
\(\left\langle{g_i}\right\rangle = \left\langle{1}\right\rangle\), which
corresponds to \(\cap V(g_i) = \emptyset\).

\end{proof}

\hypertarget{locally-noetherian-schemes-vs-noetherian-covers-friday-october-08}{%
\section{Locally Noetherian Schemes vs Noetherian Covers (Friday,
October
08)}\label{locally-noetherian-schemes-vs-noetherian-covers-friday-october-08}}

\hypertarget{proof-of-theorem}{%
\subsection{Proof of Theorem}\label{proof-of-theorem}}

Recall that we were proving the following:

\begin{theorem}[?]

\(X\in {\mathsf{Sch}}\) is locally Noetherian iff for any affine open
\(U = \operatorname{Spec}A \subseteq X\), \(A\) is a Noetherian ring.

\end{theorem}

\begin{remark}

Recall that we covered \(X\) by \(U_i\), had some affine open \(U\)
isomorphic to \(\operatorname{Spec}\) of a ring, and then covered each
intersection \(U \cap U_i\) by distinguished opens which were
\(\operatorname{Spec}R \left[ { \scriptstyle { {f_i}^{-1}} } \right] = D(f_i) = \left\{{{\mathfrak{p}}\not\ni f_i}\right\}\).
Then \(R \left[ { \scriptstyle { {f_i}^{-1}} } \right]\) is Noetherian
iff \(\bigcup_{i\in I} D(f_i) = \operatorname{Spec}R\), which implies
\(\bigcap_{i\in I} V(f_i) = \emptyset\), and thus
\(\not\exists p\in \operatorname{Spec}R\) prime with \(p\ni f_i\) for
all \(i\in I\). Then
\(\left\langle{f_i {~\mathrel{\Big\vert}~}i\in I}\right\rangle = \left\langle{1}\right\rangle\).

\end{remark}

\begin{proposition}[?]

\(\operatorname{Spec}R\) is quasicompact.

\end{proposition}

\begin{proof}[of proposition]

Let \({\mathcal{U}}\rightrightarrows\operatorname{Spec}R\), so
\(\operatorname{Spec}R = \bigcup_{i\in I} U_i\), then we want to find a
finite subcover. Take
\(\left\{{D(f_{ij})}\right\} \rightrightarrows U_i\); it suffices to
find a finite subcover of the refined cover by distinguished opens, so
reduce to \(U_i = D(f_i)\) for each \(i\). Using the argument from the
above remark,
\(\left\langle{f_i {~\mathrel{\Big\vert}~}i\in I}\right\rangle = \left\langle{1}\right\rangle\)
since this is a cover. But then there exists a finite sum
\(\sum_{j=1}^N a_j f_{ij} = 1\) for some \(a_j\in R\), so
\(\left\{{f_{ij} }\right\}_{j=1}^N = \left\langle{1}\right\rangle\)
which implies that \(\bigcup_{j=1}^N D(f_{ij}) = \operatorname{Spec}R\).

\end{proof}

\begin{remark}

Applying the proposition above, we can find a finite set
\(\left\{{f_i}\right\}\) such that
\(\left\langle{f_i {~\mathrel{\Big\vert}~}i\in I}\right\rangle = \left\langle{1}\right\rangle\)
with each \(R \left[ { \scriptstyle { {f_i}^{-1}} } \right]\)
Noetherian. We'll use the following:

\end{remark}

\begin{lemma}[?]

Let \(J{~\trianglelefteq~}R\) and
\(\phi_i: R\to R \left[ { \scriptstyle { {f_i}^{-1}} } \right]\) by the
canonical localization morphism. Setting
\(\left\langle{f_1,\cdots, f_s}\right\rangle = \left\langle{1}\right\rangle\),
\begin{align*}
J = \cap_{i=1}^s \phi_i \qty{ \phi_i^{-1}(J) R \left[ { \scriptstyle { {f_i}^{-1}}  } \right]  } 
.\end{align*}

\end{lemma}

\begin{proof}[?]

Note that
\(\phi_i\qty{ \phi_i^{-1}(J) R \left[ { \scriptstyle { {f}^{-1}} } \right] } \neq J\)
generally, e.g.~if \(f\in J\). So that \(J\) is contained in the
right-hand side is clear. For the reverse containment, let
\(b \in \bigcap_i \phi_i^{-1}(\phi_i(J) R \left[ { \scriptstyle { {f_i}^{-1}} } \right] )\).
Then
\(\phi_i(b)\in \phi_i(J) R \left[ { \scriptstyle { {f_i}^{-1}} } \right]\)
for all \(i\), so \(b \sim a_i / f_i^{n_i}\) in the localization for
some \(a_i \in J\).

Since \(\left\{{f_i}\right\}\) is finite, assume that that \(n_i = n\)
for some uniform \(n\), e.g.~\(n = \max \left\{{n_i}\right\}\). Then
\(b \sim a_i/f_i^N\), so there exist \(m_i\) such that
\(f_i^{m_i}(f_i^N b - a_i) = 0\) in the original ring \(R\). So now pick
\(M \coloneqq\max\left\{{m_i}\right\}\) to obtain
\(f_i^M(f_i^N - a_i) = 0\).

Now a trick: use that \(f_i^{M+N}b\in J\) for all \(i\), and the claim
is that
\(\left\langle{f_i^{M+N}}\right\rangle_{i\in I} = \left\langle{1}\right\rangle\).
More generally, raising all generators of a unit ideal to a fixed power
still generates the unit ideal. This follows from writing
\(1 = \sum_{i=1}^r c_i f_i \implies 1 = 1^M = \qty{\sum c_i f_i}^M\), so
choose \(M\) large enough so that some \(f_i\) occurs with an exponent
of at least \(m+n\), e.g.~choosing \(M = r(m+n)\).

\begin{example}[?]

If \(1 = \left\langle{x, y}\right\rangle\), then
\(\left\langle{x^2, y^2}\right\rangle = 1\) by taking \(ax + by = 1\)
and (say) expanding \((ax+by)^4=1\) and noting that any term is
divisible by either \(x^2\) or \(y^2\).

\end{example}

Now writing \(\sum_{i=1}^r c_i f_i^{m+n} = 1\), we get
\(\sum_{i=1}^r c_i f_i^{m+n} \in J\), and thus \(b\in J\).

\end{proof}

\begin{remark}

We now know that the \(R \left[ { \scriptstyle { {f_i}^{-1}} } \right]\)
are Noetherian. Let \(J_1 \subseteq J_2 \subseteq \cdots\) be an
ascending chain of ideals in \(R\), we'll show it stabilizes. Since the
\(R \left[ { \scriptstyle { {f_i}^{-1}} } \right]\) are Noetherian,
there is an ascending chain
\(J_1 R \left[ { \scriptstyle { {f_i}^{-1}} } \right] \subseteq J_2 R \left[ { \scriptstyle { {f_i}^{-1}} } \right] \subseteq \cdots\)
in \(R \left[ { \scriptstyle { {f_i}^{-1}} } \right]\), which is
Noetherian and thus stabilizes. So for some \(N = N(i) \gg 0\),
\(J_k R \left[ { \scriptstyle { {f_i}^{-1}} } \right] = J_{k+1} R \left[ { \scriptstyle { {=}^{-1}} } \right] \cdots\)
for all \(k\geq N\). But there are only finitely many \(f_i\), so we can
choose some uniformly large \(\tilde N\gg 0\) not depending on \(i\)
where \(J_k = J_{k+1} = \cdots\) for all \(k\geq \tilde N\) by applying
the lemma.

\end{remark}

\begin{remark}

On applying the lemma: use that
\begin{align*}
\phi_i^{-1}(J_k R \left[ { \scriptstyle { {f_i}^{-1}}  } \right] ) &= \phi_i^{-1}(J_{k+1}R \left[ { \scriptstyle { {f_i}^{-1}}  } \right] ) \quad \forall k\geq N \\
&\implies
\cap_i \phi_i^{-1}(J_k R \left[ { \scriptstyle { {f_i}^{-1}}  } \right] ) = \cap_i \phi_i^{-1}(J_{k+1}R \left[ { \scriptstyle { {f_i}^{-1}}  } \right] ) \quad \forall k\geq N \\
&\implies
J_{k} = J{k+1}\quad \forall k\neq N
.\end{align*}

\end{remark}

\hypertarget{other-properties}{%
\subsection{Other Properties}\label{other-properties}}

\begin{example}[A scheme that is not quasicompact]

Let \(X = {\mathbf{Z}}\) with the discrete topology (so every set is
open) and set \({\mathcal{O}}_X(U) = {\mathsf{Set}}(U, k)\) to be the
local ring of arbitrary set functions. Then for \(p\in X\), the stalks
are
\({\mathcal{O}}_{X, p} {\mathcal{O}}_X(\left\{{P}\right\}) = {\mathsf{Set}}(p, k) = k\),
which is local. This is a locally ringed space, since it's locally
isomorphic to \(\operatorname{Spec}R\): we can take an open cover of
such, or find a neighborhood where it holds, but
\(p \in \left\{{p}\right\}\) which is open and letting
\({\mathcal{F}}\coloneqq{ \left.{{{\mathcal{O}}_X}} \right|_{{\left\{{p}\right\}}} }\),
we have
\((\left\{{p}\right\}, {\mathcal{F}}) \cong \operatorname{Spec}k\). Then
\(X = \bigcup_{p\in X} \left\{{p}\right\}\) is an open cover with no
finite subcover.

So \({\mathbf{Z}}\) with the discrete topology is not
\(\operatorname{Spec}R\) with the Zariski topology for any ring.

\end{example}

\begin{exercise}[?]

\(X \coloneqq\operatorname{Spec}\mkern 1.5mu\overline{\mkern-1.5mu{\mathbf{Q}}\mkern-1.5mu}\mkern 1.5mu[t] = \left\{{\left\langle{0}\right\rangle}\right\} \cup\left\{{\left\langle{t-a_i}\right\rangle {~\mathrel{\Big\vert}~}i\in I}\right\}\)
where \(I\) is a countable enumeration of
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbf{Q}}\mkern-1.5mu}\mkern 1.5mu\).
Is this quasicompact?

\end{exercise}

\begin{exercise}[?]

Consider \(R \coloneqq\prod_{n\in {\mathbf{Z}}} {\mathbf{C}}\), then
there is a set map
\(\left\{{I{~\trianglelefteq~}R}\right\} { \, \xrightarrow{\sim}\, }{\mathcal{P}}({\mathbf{Z}})\),
given by sending any subset to the ideal
\({\mathbf{C}}\oplus {\mathbf{C}}\oplus \cdots\) which are zeroed out at
entries according to the complement of \(S\).

What is \(\operatorname{Spec}R\), and what is the topology? Is
\(\operatorname{Spec}R\) quasicompact?

\begin{quote}
Consider
\(I \coloneqq\left\{{(a_i) {~\mathrel{\Big\vert}~}a_i = 0\quad i\geq N \gg 0}\right\}\),
which forms an ideal. Is \(I\) prime? Are there prime ideals not
containing \(I\)?
\end{quote}

\end{exercise}

\hypertarget{monday-october-11}{%
\section{Monday, October 11}\label{monday-october-11}}

\begin{remark}

Last time: Noetherian isn't a purely topological property. Today:
another guiding principle in AG is that one can put properties on
schemes, or alternatively on morphisms, usually to
\(\operatorname{Spec}k\).

\end{remark}

\begin{definition}[Finite type morphisms]

A morphism \(X \xrightarrow{f} Y\) of schemes is \textbf{locally of
finite type} if there exists an affine open cover
\({\mathcal{V}}\rightrightarrows Y\) with
\(V_i = \operatorname{Spec}B_i\) such that there exists an affine open
cover \({\mathcal{U}}\rightrightarrows f^{-1}({\mathcal{V}})\) so
\(U_{ij}\supseteq f^{-1}(V_i)\) where
\(U_{ij} = \operatorname{Spec}A_{ij}\) such that the restricted function
\(f: U_{ij} \to V_{i}\) is induced by a ring morphism
\(B_{i } \to A_{ij}\) finitely generated over \(B_i\).

\end{definition}

\begin{remark}

This globalizes the notion of being a finitely generated ring,
essentially by covering the scheme morphisms by ring morphisms with the
desired property. As a special case, let
\(X \in {\mathsf{Sch}}_{/ {k}}\), so there is a morphism
\(X\to \operatorname{Spec}k\). Let \(X = \cup_j U_j\) with
\(U_j = \operatorname{Spec}A_j\), then we want that the map
\(\operatorname{Spec}A_{j}\to \operatorname{Spec}k\) is induced by a
finitely-generated ring morphism \(k\to A_j\), so \(A_j\) is a
finitely-generated \(k{\hbox{-}}\)algebra. So this is like having a
sheaf of \(k{\hbox{-}}\)algebras. As an abuse of notation/terminology,
we say that \(X\) is \textbf{finite type} over \(k\) (since the target
doesn't need to be covered).

\end{remark}

\begin{warnings}

Note that a subring of a finitely generated ring need not be finitely
generated!

\end{warnings}

\begin{example}[?]

Let \(A\in {}_{k} \mathsf{Alg} ^{{\mathrm{fg}}}\), then
\(\operatorname{Spec}A\) is finite type over \(k\). One can change the
definition from ``there exists an open cover'' to ``for all open
covers'' -- this amounts to checking localizations of ring maps.

\end{example}

\begin{example}[?]

Consider
\(X \coloneqq\mathop{\mathrm{Proj}}k[x, y] = {\mathbf{P}}^1_{/ {k}}\),
then recall that
\((D(f), { \left.{{{\mathcal{O}}_X}} \right|_{{D(f)}} }) \cong \operatorname{Spec}R \left[ { \scriptstyle { {f}^{-1}} } \right]\).
Then
\begin{align*}
{\mathbf{P}}^1_{/ {k}}  = D(x) {\textstyle\coprod}_f D(y) \cong {\mathbf{A}}^1_{/ {k}} {\textstyle\coprod}_f {\mathbf{A}}^1_{/ {k}} 
,\end{align*}
glued along inversion. Then
\(k[x,y] \left[ { \scriptstyle { {x}^{-1}} } \right] \cong k { \left[ {y\over x} \right] }\)
and
\(k[x, y] \left[ { \scriptstyle { {y}^{-1}} } \right] \cong k { \left[ {x\over y} \right] }\).
One can check that \({\mathbf{P}}^1_{/ {k}} \to \operatorname{Spec}k\)
is finite type, and this works for \({\mathbf{P}}^n\) as well.

Note that if \(S\coloneqq k[x_1, \cdots, x_{n}]/I\) for \(I\) a
homogeneous ideal, then \(\mathop{\mathrm{Proj}}S\) is also finite type
over \(k\). We can write
\(\mathop{\mathrm{Proj}}S = \bigcup_{i=0}^n D(\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_i)\),
since taking complements yields
\(\emptyset = \bigcap_{i=0}^n V(\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_i)\),
which is the set of homogeneous prime ideals
\(p\in \operatorname{Spec}S\) with
\(p\supseteq\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_i\)
for all \(i\) and \(p\not\supseteq S^+\) the irrelevant ideal, which is
empty. So \(S \left[ { \scriptstyle { {x_i}^{-1}} } \right]\) is a
finitely generated ring, with generators of the form \(x_j/x_i\).

\end{example}

\begin{example}[a non-example]

\(\operatorname{Spec}k {\llbracket t \rrbracket } \to \operatorname{Spec}k\)
is not a finite type morphism, i.e.~\(k{\llbracket t \rrbracket }\) is
not a finitely generated \(k{\hbox{-}}\)algebra. Toward a contradiction
suppose
\(k{\llbracket t \rrbracket } = \left\langle{f_1, \cdots, f_r}\right\rangle\),
so there is a ring map
\(k { \left[ \scriptstyle {f_1,\cdots, f_r} \right] } \twoheadrightarrow k{\llbracket t \rrbracket }\).
Take \(k= {\mathbf{Q}}\), or any countable field, then the LHS is
countable but the right-hand side is not.

\end{example}

\begin{definition}[Finite morphisms]

Recall that a morphism \(\phi:B\to A \in \mathsf{CRing}\) is a
\textbf{finite morphism} if \(A\) is finitely-generated as a
\(B{\hbox{-}}\)module. A morphism
\(X \xrightarrow{f} Y\in {\mathsf{Sch}}\) is \textbf{finite} iff there
exists an affine open cover \({\mathcal{V}}\rightrightarrows Y\) with
\(V_i = \operatorname{Spec}B_i\) and
\(f^{-1}(V_i) = \operatorname{Spec}A_i\), and the induced ring maps
\(B_i\to A_i\) are finite.

\end{definition}

\begin{remark}

Here the module structure is \(b\cdot a \coloneqq\phi(b)a\). Note that
finite type required finite generation as rings, so
\(B(g_1,\cdots, g_r)\twoheadrightarrow A\), but here we require that any
\(a\in A\) is of the form \(a = \sum_{i=1}^r \phi(b_i) a_i\) for some
\(b_i\).

\end{remark}

\begin{example}[?]

Consider
\(X \coloneqq\operatorname{Spec}R \coloneqq\operatorname{Spec}{\mathbf{C}}[x, y]/ \left\langle{y^2-f(x)}\right\rangle\)
where \(f\) has no repeated roots, which yields a hyperelliptic curve.
This is a reduced ring, so \(X\) is the scheme associated to a variety.
Letting \(r_i\) be the roots of \(f\), we have the following:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-11_12-10.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Consider the function \(f:X\to Y\) induced by the following ring map:
\begin{align*}
{\mathbf{C}}[x] &\to R \\
x &\mapsto x
,\end{align*}
which is projection onto the axis. Note that
\(R \cong {\mathbf{C}}[x] \left\langle{1}\right\rangle \oplus {\mathbf{C}}[x] \left\langle{y}\right\rangle\)
as a \({\mathbf{C}}[x]{\hbox{-}}\)module.

\end{example}

\begin{example}[?]

Consider
\(\operatorname{Spec}{\mathbf{C}}[x,y,z]/\left\langle{xyz - 1}\right\rangle \to \operatorname{Spec}{\mathbf{C}}[x]\),
whose real locus yields a hyperboloid:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-11_12-18.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Note that finite type should approximately be spec of finite type
\(k{\hbox{-}}\)algebras, i.e.~essentially varieties, where for
finiteness the fibers should be finite.

\end{example}

\begin{example}[?]

Consider
\(\operatorname{Spec}{\mathbf{C}}[x, x^{-1}] \to \operatorname{Spec}{\mathbf{C}}[x]\),
i.e.~\({\mathbf{G}}_m \hookrightarrow{\mathbf{A}}^1\). However,
\({\mathbf{C}}[x, x^{-1}]\) is not finitely-generated as a
\({\mathbf{C}}[x]\) modules, even though it has finite fibers. Given any
finite set of generators, one can take
\({\mathbf{C}}[x]\left\langle{f_i\over x^{k_i}}\right\rangle\) which
doesn't contain \(1/x^{\max k_i + 1}\).

\end{example}

\begin{remark}

We'll define subschemes soon.

\end{remark}

\hypertarget{wednesday-october-13}{%
\section{Wednesday, October 13}\label{wednesday-october-13}}

\begin{remark}

Result from last time: there doesn't exist a surjection
\(k { \left[ {f_1,\cdots, f_m} \right] } \twoheadrightarrow k {\llbracket x \rrbracket }\)
for any finite collection \(\left\{{f_i}\right\}_{i=1}^m\) of
polynomials. This can be checked by just considering their dimension as
a \(k{\hbox{-}}\)modules, where \(\dim_k LHS = \# {\mathbb{N}}\) and
\(\dim_k RHS = \# {\mathbf{R}}\).

We said that a morphism \(X \xrightarrow{f} Y\) is \textbf{locally
finite type} if it is locally modeled on
\(\operatorname{Spec}A\to \operatorname{Spec}B\) with \(B\to A\) a
finitely-generated ring morphism, and is \textbf{finite} if locally
modeled on \(B \to A\) module-finite. Note that in the first case, we
require \(f^{-1}(U) \supseteq U\to V\), but in the latter
\(U = f^{-1}(V)\).

\end{remark}

\begin{example}[?]

As an example, the map \(D(x) \to {\mathbf{A}}^1_{/ {k}}\) was not
finite since \(k[x] \to k[x, x^{-1}]\) is not module-finite, despite the
fact that this geometrically corresponds to
\({\mathbf{A}}^1\setminus\left\{{0}\right\}\hookrightarrow{\mathbf{A}}^1\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-13_11-41.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{example}

\begin{example}[Not finite type: $\spec$ of a local ring of a variety]

Let \(p \in V\) be a \(k{\hbox{-}}\)variety for \(k\) an infinite field,
which we can assume to be affine (so \(k[x_1, \cdots, x_{n}]/I\) for
\(I\) reduced). Then \({\mathcal{O}}_{V, p}\) is not not
finitely-generated as a \(k{\hbox{-}}\)algebra, and
\(\operatorname{Spec}{\mathcal{O}}_{V, p}\to \operatorname{Spec}k\) is
not of finite type. Consider the local ring of
\({\mathbf{A}}^1_{/ {k}}\) at the prime ideal
\(p \coloneqq\left\langle{x}\right\rangle\), then
\(k[x] \left[ { \scriptstyle { { ({p}^c) }^{-1}} } \right] = \left\{{f/g {~\mathrel{\Big\vert}~}g\not\in p}\right\}\),
so \(g(0) \neq 0\) for such \(g\). Note that this is not
\(k[x] \left[ { \scriptstyle { {x}^{-1}} } \right]\)!

Idea: there are only finitely many denominators: if
\begin{align*}
\phi: k { \left[ {f_1/g_1, \cdots, f_r/g_r} \right] }  \twoheadrightarrow k[x] \left[ { \scriptstyle { { ({p}^c) }^{-1}}  } \right]
,\end{align*}
then \(\operatorname{im}\phi\) contains contains those \(f/g\) such that
\(V(g) \subseteq \cup V(g_i)\), so such a \(\phi\) can not exist. Note
that this is still true for \(k\) a finite field, just not by this
proof.

\end{example}

\hypertarget{openclosed-subschemes}{%
\subsection{Open/Closed Subschemes}\label{openclosed-subschemes}}

\begin{definition}[Open subschemes]

Given \(X\in {\mathsf{Sch}}\), an \textbf{open subscheme} of \(X\) is an
open subset \(U \subseteq { {\left\lvert {X} \right\rvert} }\) with
structure sheaf
\({\mathcal{O}}_U = { \left.{{{\mathcal{O}}_X}} \right|_{{U}} }\).

\end{definition}

\begin{remark}

Why does \((U, {\mathcal{O}}_U) \in \mathsf{Loc}\mathsf{RingSp}\) form a
scheme? One needs to check that it's locally isomorphic to the spectrum
of a ring: let \(\left\{{X_i}\right\}\rightrightarrows X\) be an open
affine cover, then \(U_i\coloneqq U \cap X_i\) is open in \(U\) and in
\(X_i\), so can be covered by distinguished opens \(V_{ij}\). But then
\(\left\{{V_{ij}}\right\}\rightrightarrows U\) is a cover by affine
schemes. The inclusion
\((U, {\mathcal{O}}_U) \hookrightarrow(X, {\mathcal{O}}_X)\) is clearly
a morphism in \(\mathsf{Loc}\mathsf{RingSp}\).

\end{remark}

\begin{definition}[Open Immersion]

The inclusion above is an \textbf{open immersion}.

\end{definition}

\begin{remark}

A small dictionary

\begin{longtable}[]{@{}ll@{}}
\toprule
AG & Rest of Math \\
\midrule
\endhead
Immersion & Embedding \\
Nothing! & Immersion \\
\bottomrule
\end{longtable}

\end{remark}

\begin{remark}

Here I write
\({ {\left\lvert {X} \right\rvert} } \coloneqq\mathrm{sp}(X)\) as an
alternative for Hartshorne's notation.

\end{remark}

\begin{definition}[Closed immersion]

A \textbf{closed immersion} is a morphism
\(X \xrightarrow{f} Y\in {\mathsf{Sch}}\) such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  The induced map
  \({ {\left\lvert {X} \right\rvert} }\to { {\left\lvert {Y} \right\rvert} } \in {\mathsf{Top}}\)
  is a homeomorphism onto a closed subset of
  \({ {\left\lvert {Y} \right\rvert} }\),
\item
  \(f^\#: {\mathcal{O}}_Y \twoheadrightarrow f_* {\mathcal{O}}_X \in {\mathsf{Sh}}(Y)\)
  is surjective.
\end{enumerate}

\end{definition}

\begin{remark}

Set \(U = D(f) \subseteq \operatorname{Spec}A\) defines an open
immersion
\(\operatorname{Spec}A \left[ { \scriptstyle { {f}^{-1}} } \right] \to \operatorname{Spec}A\).
So this corresponds to the ring map
\(A\hookrightarrow A \left[ { \scriptstyle { {f}^{-1}} } \right]\) since
\(\operatorname{Spec}A \left[ { \scriptstyle { {f}^{-1}} } \right] \subseteq \operatorname{Spec}A\)
are those ideals not containing \(f\).

\end{remark}

\begin{example}[?]

Consider
\(U \coloneqq{\mathbf{A}}^2_{/ {k}} \setminus\left\{{{\left[ {x, y} \right]}}\right\} \hookrightarrow{\mathbf{A}}^2_{/ {k}}\).
Then \(\left\{{{\left[ {x, y} \right]}}\right\} = V(x, y)\), and this is
a subscheme of an affine scheme which is not itself affine. One can use
that \(\dim D(f)^c \geq 1\)

\end{example}

\begin{exercise}[?]

Prove that this is not affine. Hint: check
\({\mathcal{O}}_U = k[x,y]\),\footnote{This says that any regular
  function on \(U\) actually extends to all of \({\mathbf{A}}^2\)} and
use that for any \(X \in {\mathsf{Aff}}{\mathsf{Sch}}\) we have
\({\mathcal{O}}_X(X) = R\). However,
\(\operatorname{Spec}k[x, y] = {\mathbf{A}}^2\neq U\).

\end{exercise}

\begin{quote}
Check ``affinization''? It fills in holes of at codimension at most 2,
and satisfies a universal property. Consider
\(X = {\mathbf{A}}^1\times {\mathbf{P}}^1\).
\end{quote}

\begin{remark}

All examples are locally of this form:
\(F:X\hookrightarrow Y = \operatorname{Spec}A\) where
\({ {\left\lvert {X} \right\rvert} }\to U \subseteq { {\left\lvert {Y} \right\rvert} }\)
maps homeomorphically onto a closed subset. Recall that the closed
subsets are of the form \(U = V(I)\), and here we need
\(f^\#: {\mathcal{O}}_Y \to f_* {\mathcal{O}}_X\) surjective. Let
\(X = \operatorname{Spec}A/I\), and recall that every surjective ring
map is of the form \(A\to A/I\). Here \(q: A\twoheadrightarrow A/I\)
where \({\mathfrak{p}}\mapsfrom q^{-1}({\mathfrak{p}})\), so we get some
map \(\operatorname{Spec}A/I\to \operatorname{Spec}A\), and this is
homeomorphism onto \(V(I) \subseteq \operatorname{Spec}A\):
\begin{align*}
\operatorname{Spec}A/I &\to V(I) \\
{\mathfrak{p}}&\mapsto q^{-1}({\mathfrak{p}}) \supseteq I \\
q({\mathfrak{q}}) &\mapsfrom {\mathfrak{q}}
.\end{align*}
We also get an induced map
\(A \left[ { \scriptstyle { {g}^{-1}} } \right] \to (A/I) \left[ { \scriptstyle { {g}^{-1}} } \right]\),
which is precisely
\begin{align*}
f^\#(D(g)): {\mathcal{O}}_Y(D(g)) \to {\mathcal{O}}_X(f^{-1}(D(g)))
\end{align*}
and is thus surjective. Since it's surjective on a basis, by gluing
it'll be surjective on the entire space.

\end{remark}

\hypertarget{friday-october-15}{%
\section{Friday, October 15}\label{friday-october-15}}

\begin{remark}

Last time: open and closed subschemes, where openness was easy since we
required \(f:X\to Y\) to be a homeomorphism onto a closed subset of
\(Y\) with \(f^\#\) surjective. Any example of a closed subscheme is
locally of the following form:
\(V(I) \coloneqq\operatorname{Spec}A/I \to \operatorname{Spec}A\)
induced by some \(A\twoheadrightarrow A/I\) in rings. Here
\(A \left[ { \scriptstyle { {g}^{-1}} } \right] \twoheadrightarrow(A/I) \left[ { \scriptstyle { {g}^{-1}} } \right]\)
implies that \(f^(D(g))\) surjective for every distinguished open, so
\(f^\#\) is a surjective sheaf map. However, this need not be surjective
on global sections.

\end{remark}

\begin{example}[?]

Recall that
\({\mathbf{P}}^1_{/ {k}} = {\mathbf{A}}^1 \coprod_{x\mapsto x^{-1}}{\mathbf{A}}^1\)
and \({\mathcal{O}}_{{\mathbf{P}}^1}({\mathbf{P}}^1) = k\) where we
glued \(k[t] \cap k[s] = k\) along \(s=1/t\). Consider the closed
subscheme of \({\mathbf{A}}^1\) given by
\(X\coloneqq\operatorname{Spec}{\mathbf{C}}[t]/ t^2\) and the global
restriction map
\begin{align*}
f^\#({\mathbf{P}}^1): {\mathcal{O}}_{{\mathbf{P}}^1}({\mathbf{P}}^1) &\to {\mathcal{O}}_X(X) \\
{\mathbf{C}}&\to{\mathbf{C}}[t]/t^2
,\end{align*}
which is not surjective.

\end{example}

\begin{example}[?]

Consider \({\mathbf{A}}^2_{/ {k}}\) for
\(k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\),
how many closed subschemes are homeomorphic onto the origin \(0\)
corresponding to
\(\left\langle{x, y}\right\rangle\in \operatorname{Spec}k[x, y]\). Since
they're all locally of the form \(V(I)\), these correspond to ideals
\(I\) where \(V(I) = 0\). These are ideals \(I\) where
\(\left\langle{x, y}\right\rangle\) is the only ideal containing \(I\),
so we can write this as
\(\left\{{I {~\mathrel{\Big\vert}~}\sqrt I = \left\langle{x, y}\right\rangle}\right\}\),
i.e.~the primary decomposition of \(I\) has only 1 prime, namely
\(\left\langle{x, y}\right\rangle\). Some ideals of this form:

\begin{itemize}
\tightlist
\item
  \(\left\langle{x, y}\right\rangle^k\) for any \(k\geq 0\),
\item
  \(\left\langle{x^ay^b, x^cy^d}\right\rangle\) where
  \(\operatorname{det}{ \begin{bmatrix}  {a} & {b} \\  {c} & {d} \end{bmatrix} } \neq 0\),
  e.g.~\(\left\langle{x^2, y}\right\rangle\).
\item
  \(\left\langle{(x+y)^2, y}\right\rangle\)
\item
  \(\left\langle{f, g}\right\rangle\) where \(V(f) \cap V(g) = 0\) as a
  set, e.g.~two curves only intersecting at the origin.
\end{itemize}

\end{example}

\begin{remark}

What kinds of schemes are these? For example, considering \(V(x-y^2)\)
and \(V(y)\), we have
\(\left\langle{y-x^2, y}\right\rangle = \left\langle{x^2, y}\right\rangle\),
yielding a non-reduced scheme. We have
\(k[x,y]/\left\langle{x^2, y}\right\rangle= k \bigoplus kx\in {}_{k}{\mathsf{Mod}}\),
thought of as functions as the tangent vector at 0 pointing
horizontally. Similarly,
\(k[x,y] = \left\langle{x^2, xy, y^2}\right\rangle = k \oplus kx \oplus ky\),
which can be thought of as functions on
\({\mathbf{T}}_p {\mathbf{A}}^2\) for \(p=0\). The rough idea: we want
\({\mathbf{T}}_0 {\mathbf{A}}^2 \cong \operatorname{Spec}k[x,y]/\left\langle{x, y}\right\rangle^2\).

\end{remark}

\begin{definition}[Reduced subscheme structures]

Let \(Z \subseteq { {\left\lvert {Y} \right\rvert} }\) be closed, then
there exists a unique scheme structure \(X\) on \(Z\) such that
\({ {\left\lvert {X} \right\rvert} } = Z\), the \textbf{reduced
subscheme structure on \(Z\)}. Affine locally, for
\(Z \subseteq { {\left\lvert {\operatorname{Spec}A} \right\rvert} }\)
given by \(V(I)\) for some ideal \(I\), and we define this as
\(\operatorname{Spec}A/\sqrt{I}\). This will glue because passing to
reduction will commute with localization,
i.e.~\((A_{ \text{red} }) \left[ { \scriptstyle { {f_{ \text{red} }}^{-1}} } \right] = (A \left[ { \scriptstyle { {f}^{-1}} } \right] )_{ \text{red} }\)
where \(A_{ \text{red} }= A/\sqrt{0}\).

\end{definition}

\begin{example}[?]

Take \(0\in {\left\lvert {{\mathbf{A}}^2_{/ {k}} } \right\rvert}\), then
its reduced subscheme structure is
\(\operatorname{Spec}k[x, y] / \left\langle{x, y}\right\rangle\).

\end{example}

\begin{remark}

Any closed subscheme structure along \(Z\) is locally given by
\(\operatorname{Spec}A/I\) with \(V(I) = Z\), and there's always a map
\(\operatorname{Spec}A/\sqrt{I} \to \operatorname{Spec}A/I\) dual to the
reduction map \(A/I \to (A/I)_{ \text{red} }\). For any closed subscheme
\(X \subseteq Y\), we define \(X_{ \text{red} }\) as the reduced
subscheme associated to \({\left\lvert {X} \right\rvert}\), and there is
a morphism \(X_{ \text{red} }\to X\).

\begin{quote}
Idea: this is a space such that all of its functions kill nilpotents.
\end{quote}

\end{remark}

\begin{proposition}[?]

\(X_{ \text{red} }\) is well-defined

\end{proposition}

\begin{proof}[?]

Let \(Y\in {\mathsf{Sch}}\) and
\(Z \subset { {\left\lvert {Y} \right\rvert} }\in {\mathsf{Top}}\)
closed. Take a cover \({\mathcal{U}}\to Y\) with
\(U_i = \operatorname{Spec}A_i\), then
\(Z \cap{ {\left\lvert {U_i} \right\rvert} }\) is closed and thus equal
to some \(V(I_i)\). Define a reduced scheme
\(X_i \coloneqq\operatorname{Spec}(A_i / \sqrt{I_i})\), which we'll try
to glue to define \(X_{ \text{red} }\). Note that we can write
\begin{align*}
\sqrt{I_i} = \bigcap_{{\mathfrak{p}}\in Z \cap{ {\left\lvert {U_i} \right\rvert} }} {\mathfrak{p}}
,\end{align*}
which generalizes
\({\sqrt{0_{R}} } = \bigcap_{{\mathfrak{p}}\in \operatorname{Spec}R} {\mathfrak{p}}\).

To give a gluing amounts to defining isomorphisms:
\begin{align*}
f_{ij}: X_i \cap U_{j} \to X_j \cap U_i
.\end{align*}

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-15_12-13.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

So pass to an open affine cover. We'll have
\((A_i) \left[ { \scriptstyle { {f}^{-1}} } \right] = (A_j) \left[ { \scriptstyle { {g}^{-1}} } \right]\)
for some \(f, g\), and this will induce isomorphisms
\begin{align*}
(A_i) \left[ { \scriptstyle { {f}^{-1}}  } \right]  / \sqrt{I_i}  { \, \xrightarrow{\sim}\, }{\mathcal{O}}_{X_j \cap U_i}(Z \cap D(f))
.\end{align*}

\end{proof}

\hypertarget{monday-october-18}{%
\section{Monday, October 18}\label{monday-october-18}}

\hypertarget{dimension}{%
\subsection{Dimension}\label{dimension}}

\begin{question}

If \(X = \operatorname{Spec}A\) is affine and
\(U \subset { {\left\lvert {X} \right\rvert} }\) is open, is the
inclusion \(U \hookrightarrow X\), represented say by
\(\operatorname{Spec}A' \hookrightarrow\operatorname{Spec}A\),
represented by a ring map \(A\to A'\)?

\end{question}

\begin{definition}[Dimension]

For \(X\in {\mathsf{Sch}}\), write
\(\dim X \coloneqq\dim_{\mathsf{Top}}{ {\left\lvert {X} \right\rvert} }\)
as the topological dimension of the underlying space, which is the
length of the longest chain of irreducible closed subsets
\begin{align*}
\emptyset \subsetneq Z_0 \subseteq Z_1 \subsetneq \cdots \subsetneq Z_n \subseteq { {\left\lvert {X} \right\rvert} }
,\end{align*}
where equality at the end is possible if
\({ {\left\lvert {X} \right\rvert} }\) is irreducible.

\end{definition}

\begin{example}[?]

\envlist

\begin{itemize}
\tightlist
\item
  \(\dim \operatorname{Spec}k = 0\)
\item
  \(\dim \operatorname{Spec}{ {\mathbf{Z}}_{\widehat{p}} }\): consider
  \(\emptyset \subsetneq {\operatorname{pt}}\subseteq \operatorname{Spec}{ {\mathbf{Z}}_{\widehat{p}} }\),
  where \({\operatorname{pt}}\) is a generic point, so
  \(\dim \operatorname{Spec}{ {\mathbf{Z}}_{\widehat{p}} }= 1\).
\item
  \(\dim {\mathbf{P}}^n_{/ {k}} = \dim {\mathbf{A}}^n_{/ {k}} = n\).
\end{itemize}

\end{example}

\begin{example}[?]

If \(X = \operatorname{Spec}A\) is affine for \(A\) then
\(\dim X = \operatorname{krulldim}A\) is the Krull dimension of the ring
\(A\). This follows because irreducible closed subsets of
\(\operatorname{Spec}A\) biject with prime ideals of \(A\). Why is this
true?

\(\impliedby\):

Suppose \(p\subseteq A\) is prime, then note that
\(V(p) = \left\{{q \in \operatorname{Spec}A {~\mathrel{\Big\vert}~}q \supseteq p}\right\}\).
If \(V(p) = V(I) \cup V(J) = V(IJ)\), then \(p \supseteq IJ\) so \(p\)
contains one of \(I, J\). But then \(V(p) = V(I)\) wlog, so \(V(p)\) is
an irreducible closed subset.

\(\implies\): We can reverse almost all of these implications:

\begin{itemize}
\tightlist
\item
  \(V(p) = V(IJ)\)
\item
  \(\iff p\supseteq IJ\)
\item
  \(\iff p \subseteq I\) or \(p \subseteq J\)
\item
  \(\iff V(p) = V(I)\) or \(V(J)\).
\end{itemize}

Note that bijections preserve strict containments, so we have
correspondences on chains:
\begin{align*}
\emptyset \subsetneq Z_0 \subsetneq \cdots \subsetneq Z_n \subset X = \operatorname{Spec}A \\
\iff \\
\left\langle{1}\right\rangle \supsetneq p_0 \supsetneq \cdots \supsetneq p_n
.\end{align*}

\end{example}

\begin{remark}

So we can use that \(\operatorname{krulldim}k[x_1, \cdots, x_{n}]= n\)
to show \(\dim {\mathbf{A}}^n_{/ {k}} = n\). For
\({\mathbf{P}}^n_{/ {k}}\), use that any maximal chain contains a point
\(\left\{{z_0}\right\}\), so choosing such a point and intersecting
\(z_i\) with the embedded copy of
\({\mathbf{A}}^n_{/ {k}} \hookrightarrow{\mathbf{P}}^n_{/ {k}}\). Then
use that there is a chain
\(\left\langle{0}\right\rangle \subsetneq \left\langle{x_1}\right\rangle \subsetneq \left\langle{x_1, x_2}\right\rangle \cdots \subsetneq \left\langle{x_1,\cdots, x_n}\right\rangle\),
so \(\dim X \geq n\). For the reverse inequality: this is hard! See
Atiyah-MacDonald's discussion of regular systems of parameters.

\end{remark}

\begin{definition}[Codimension]

The \textbf{codimension} \(\operatorname{codim}(Z, X)\) for
\(Z \subseteq X\) a closed irreducible subset is the length of the
longest chain starting at \(Z\):
\begin{align*}
Z = Z_0 \subsetneq Z_1 \subsetneq \cdots \subsetneq Z_n \subset X
.\end{align*}

\end{definition}

\begin{fact}

For \(X = \operatorname{Spec}A\) and
\(A\in {}_{k} \mathsf{Alg} ^{\mathrm{fg}}\), there is a formula
\begin{align*}
\dim(Z) + \operatorname{codim}(X, Z) = \dim (X)
.\end{align*}

\end{fact}

\begin{remark}

This is not true in general, even for Noetherian rings -- see
\textbf{catenary rings}, where any chain of prime ideals can be extended
to a chain of fixed maximal length \(n\). Without this, one can extend
chains to maximal chains of differing lengths.

\end{remark}

\begin{example}[?]

\(\dim \operatorname{Spec}{\mathbf{Z}}= 1\), instead of having dimension
zero! This is because there's always a chain
\(0 \to \left\langle{p}\right\rangle \to {\mathbf{Z}}\) for any prime.
An analogy here is a curve
\(\operatorname{Spec}k[x, y] / \left\langle{f(x, y)}\right\rangle\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-18_12-16.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

One can similarly do this for \({\mathcal{O}}_K\) the ring of integers
in a number field \(K\) and get
\(\dim \operatorname{Spec}{\mathcal{O}}_K = 1\). This leads to a good
theory of divisors (free modules on codimension 1 subvarieties) and the
Picard group, so a useful geometrization of number theory.

\end{example}

\hypertarget{fiber-products}{%
\subsection{Fiber Products}\label{fiber-products}}

\begin{remark}

Perhaps the most important construction in schemes! Picks up
intersection multiplicities.

\end{remark}

\begin{definition}[Fiber products]

Let \(X, Y \in {\mathsf{Sch}}_{/ {S}}\) then
\(X \underset{\scriptscriptstyle {S} }{\times} Y \in {\mathsf{Sch}}_{/ {S}}\)
is an \(S{\hbox{-}}\)scheme equipped with morphisms of
\(S{\hbox{-}}\)schemes onto \(X, Y\) satisfying a universal property.
For any \(Z\) with maps to \(X\) and \(Y\), there is a unique \(\theta\)
making the following diagram commute:

\begin{center}
\begin{tikzcd}
    & Z \\
    & {} \\
    & {X \underset{\scriptscriptstyle {S} }{\times}  Y} \\
    Y && X \\
    & S
    \arrow[from=4-1, to=5-2]
    \arrow[from=4-3, to=5-2]
    \arrow[from=1-2, to=4-1]
    \arrow[from=1-2, to=4-3]
    \arrow["{\exists !\theta}"{description}, dashed, from=1-2, to=3-2]
    \arrow[from=3-2, to=4-1]
    \arrow[from=3-2, to=4-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMSwwLCJaIl0sWzAsMywiWSJdLFsyLDMsIlgiXSxbMSw0LCJTIl0sWzEsMV0sWzEsMiwiWFxcZnB7U30gWSJdLFsxLDNdLFsyLDNdLFswLDFdLFswLDJdLFswLDUsIlxcZXhpc3RzICFcXHRoZXRhIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzUsMV0sWzUsMl1d}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{remark}

Note that on the ring side, this yields a tensor product over \(S\).

\end{remark}

\hypertarget{wednesday-october-20}{%
\section{Wednesday, October 20}\label{wednesday-october-20}}

\begin{remark}

Today: only the most important property of schemes, the existence of
fiber products! Let \(X, Y\in {\mathsf{Sch}}_{/ {S}}\), then the fiber
product
\(X \underset{\scriptscriptstyle {S} }{\times} Y \in {\mathsf{Sch}}_{/ {S}}\)
is an object satisfying a universal property:

\begin{center}
\begin{tikzcd}
    {\forall Z} \\
    \\
    && {X \underset{\scriptscriptstyle {S} }{\times}  Y} && X \\
    &&& {} \\
    && Y && Z
    \arrow[from=5-3, to=5-5]
    \arrow[from=3-5, to=5-5]
    \arrow["{\exists! \theta}"{description}, dashed, from=1-1, to=3-3]
    \arrow["\beta"{description}, from=1-1, to=3-5]
    \arrow["\alpha"{description}, from=1-1, to=5-3]
    \arrow["p"{description}, from=3-3, to=5-3]
    \arrow["q"{description}, from=3-3, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMiwyLCJYXFxmcHtTfSBZIl0sWzMsM10sWzQsMiwiWCJdLFsyLDQsIlkiXSxbNCw0LCJaIl0sWzAsMCwiXFxmb3JhbGwgWiJdLFszLDRdLFsyLDRdLFs1LDAsIlxcZXhpc3RzISBcXHRoZXRhIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzUsMiwiXFxiZXRhIiwxXSxbNSwzLCJcXGFscGhhIiwxXSxbMCwzLCJwIiwxXSxbMCwyLCJxIiwxXV0=}{Link
to Diagram}
\end{quote}

Note that needing the square involving \(Z\) and
\(X \underset{\scriptscriptstyle {S} }{\times} Y\) to commute is
automatic, since we're working in \({\mathsf{Sch}}_{/ {S}}\) instead of
just \({\mathsf{Sch}}\). Note that
\(X\times Y = X \underset{\scriptscriptstyle {\operatorname{Spec}{\mathbf{Z}}} }{\times} Y\)
recovers the product.

\end{remark}

\begin{question}

Do fiber products exist? They're unique up to isomorphism if they do, so
we just need to construct it.

\end{question}

\begin{theorem}[Existence of fiber products]

Fiber products exist and are unique up to isomorphism.

\end{theorem}

\hypertarget{the-7-step-proof}{%
\subsection{The 7-Step Proof}\label{the-7-step-proof}}

\begin{proof}[Step 1: Prove for $X, Y, S$ are affine.]

Write
\(X = \operatorname{Spec}A, Y= \operatorname{Spec}B, S = \operatorname{Spec}R\).
We start in \(\mathsf{Ring}\), and noting contravariance of
\(\operatorname{Spec}({-})\), the claim is that
\(\operatorname{Spec}(A\otimes_R B) \cong X \underset{\scriptscriptstyle {S} }{\times} Y\).
Use that \(\operatorname{Spec}: \mathsf{Ring}\to {\mathsf{Sch}}\) is
fully faithful, which almost allows just reversing arrows if some care
is taken. A map \(Z\to \operatorname{Spec}A\) is the same data as a map
\(A\to {\mathcal{O}}_Z(Z)\). Let \({\mathcal{U}}\rightrightarrows Z\)
with \(U_i = \operatorname{Spec}C_i\), then a morphism
\(Z\to \operatorname{Spec}A \in {\mathsf{Sch}}_{/ {S}}\) is equivalently
a collection of compatible morphisms \(A\to C_i \in \mathsf{Ring}\) by
the sheaf condition, so the restrictions to
\({\mathcal{O}}_{Z_i \cap Z_j}\) are compatible. So we can interchange
any two diagrams of the following form:

\begin{center}
\begin{tikzcd}
    Z & Y && {{\mathcal{O}}_Z(Z)} & B \\
    X & S && A & R
    \arrow[from=1-1, to=1-2]
    \arrow[""{name=0, anchor=center, inner sep=0}, from=1-2, to=2-2]
    \arrow[from=1-1, to=2-1]
    \arrow[from=2-1, to=2-2]
    \arrow[from=2-5, to=1-5]
    \arrow[from=1-5, to=1-4]
    \arrow[from=2-5, to=2-4]
    \arrow[""{name=1, anchor=center, inner sep=0}, from=2-4, to=1-4]
    \arrow[shorten <=14pt, shorten >=14pt, Rightarrow, from=0, to=1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMCwwLCJaIl0sWzAsMSwiWCJdLFsxLDEsIlMiXSxbMSwwLCJZIl0sWzMsMCwiXFxPT19aKFopIl0sWzQsMCwiQiJdLFszLDEsIkEiXSxbNCwxLCJSIl0sWzAsM10sWzMsMl0sWzAsMV0sWzEsMl0sWzcsNV0sWzUsNF0sWzcsNl0sWzYsNF0sWzksMTUsIiIsMCx7InNob3J0ZW4iOnsic291cmNlIjoyMCwidGFyZ2V0IjoyMH19XV0=}{Link
to Diagram}
\end{quote}

Now the universal property of \(A\otimes_R B \in \mathsf{Ring}\) yields
a unique map \(A\otimes_R B \xrightarrow{\theta*} {\mathcal{O}}_Z(Z)\),
so equivalently
\(Z \xrightarrow{\theta} \operatorname{Spec}(A\otimes_R B)\).

\end{proof}

For step 2, the universal property will imply uniqueness if it exists,
which we'll need for gluing.

\begin{proof}[Step 3: Gluing morphisms on covers]

A morphism \(X \to Y\in {\mathsf{Sch}}\) is equivalently the data of
\({\mathcal{U}}\rightrightarrows X\) and morphism
\(U_i \xrightarrow{f_i} Y \in {\mathsf{Sch}}\) with
\({ \left.{{f_i}} \right|_{{U_i \cap U_j}} } = { \left.{{f_j}} \right|_{{U_i \cap U_j}} }\).
This is true more generally for any \(X\in {\mathsf{Top}}\) and
\(F\in {\mathsf{Sh}}_{/ {X}} (\mathsf{C})\) with values in any category.

\end{proof}

\begin{proof}[Step 4: Passing to open subsets of a factor]

Let \(X, Y\in {\mathsf{Sch}}_{/ {S}}\) be arbitrary and
\(U \subseteq X\) open. If
\(X \underset{\scriptscriptstyle {S} }{\times} Y\) exists, it is
equipped with morphisms \(q\) to \(X\) and \(p\) to \(Y\). Note that
every open subset has a canonical open subscheme structure.

\begin{claim}

\(U \underset{\scriptscriptstyle {S} }{\times} Y \cong p^{-1}(U)\),
noting that we don't yet know that fibers are schemes.

\end{claim}

\begin{proof}[?]

Let \(Z\in {\mathsf{Sch}}_{/ {S}}\) such that we have the following:

\begin{center}
\begin{tikzcd}
    Z && U && X \\
    \\
    Y
    \arrow["\iota", hook, from=1-3, to=1-5]
    \arrow["\alpha", from=1-1, to=1-3]
    \arrow["\beta"', from=1-1, to=3-1]
    \arrow["{\alpha'}", curve={height=-30pt}, from=1-1, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJaIl0sWzIsMCwiVSJdLFs0LDAsIlgiXSxbMCwyLCJZIl0sWzEsMiwiXFxpb3RhIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMCwxLCJcXGFscGhhIl0sWzAsMywiXFxiZXRhIiwyXSxbMCwyLCJcXGFscGhhJyIsMCx7ImN1cnZlIjotNX1dXQ==}{Link
to Diagram}
\end{quote}

If \(X \underset{\scriptscriptstyle {S} }{\times} Y\) exists, then
\(\exists ! \theta: Z\to X \underset{\scriptscriptstyle {S} }{\times} Y\).
Then \(\Theta(Z) \subseteq p^{-1}(U)\) since
\(p \circ \Theta = \alpha' \coloneqq\iota \circ \alpha\), so
\(\operatorname{im}(p \circ \Theta) \subseteq U\). So
\(\theta:Z\to p^{-1}(U)\) is unique, making
\(p^{-1}(U) \cong U \underset{\scriptscriptstyle {S} }{\times} Y\).

\end{proof}

So if \(X \underset{\scriptscriptstyle {S} }{\times} Y\) exists then
\(U \underset{\scriptscriptstyle {S} }{\times} Y\) exists.

\end{proof}

\begin{proof}[Step 5: Gluing fiber products from an open cover]

Let \(X, Y\in {\mathsf{Sch}}_{/ {S}}\) and suppose
\({\mathcal{X}}\rightrightarrows X\) where
\(X_i \underset{\scriptscriptstyle {S} }{\times} Y\) exists, then the
claim is that \(X \underset{\scriptscriptstyle {S} }{\times} Y\) exists.
Define \(X_{ij} \coloneqq X_i \cap X_j\), then by step 4
\(p_i^{-1}(X_{ij})\) is a fiber product
\(X_{ij} \underset{\scriptscriptstyle {S} }{\times} Y\), and similarly
\(p_j^{-1}(X_{ij}) = X_{ij} \underset{\scriptscriptstyle {S} }{\times} Y\).
By uniqueness in step 2, there is a unique isomorphism
\(\phi_{ij}: p_i^{-1}(X_{ij})\to p_j^{-1}(X_{ij})\) of fiber products.
Furthermore, the cocycle condition is satisfied since \(\phi_{ik}\) is
unique, so \(\phi_{ij} \circ \phi_{jk} = \phi_{ik}\). These are schemes
(or more generally any ringed space), so we can glue to get \emph{some}
scheme which we'll suggestively write
\(X \underset{\scriptscriptstyle {S} }{\times} Y\). The claim is that
this satisfies the correct universal property. First: there are
morphisms
\(X \underset{\scriptscriptstyle {S} }{\times} Y \xrightarrow{p} X\) and
\(X \underset{\scriptscriptstyle {S} }{\times} Y \xrightarrow{q} Y\).
Note that the \(X \underset{\scriptscriptstyle {S} }{\times} Y\) cover
\(X_i \underset{\scriptscriptstyle {S} }{\times} Y\) and
\(X = {\textstyle\coprod}X_i/\sim\) Define the following maps:

\begin{center}
\begin{tikzcd}
    && {Z_i \coloneqq\alpha^{-1}(X_i)} && {X_i} \\
    \\
    Y && Z && X
    \arrow["\alpha", from=3-3, to=3-5]
    \arrow["\beta"', from=3-3, to=3-1]
    \arrow[hook, from=1-3, to=3-3]
    \arrow[hook, from=1-5, to=3-5]
    \arrow["{{ \left.{{\alpha}} \right|_{{Z_i}} }}", from=1-3, to=1-5]
    \arrow["{{ \left.{{\beta}} \right|_{{Z_i}} }}"', curve={height=30pt}, from=1-3, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMiwyLCJaIl0sWzQsMiwiWCJdLFswLDIsIlkiXSxbMiwwLCJaX2kgXFxkYSBcXGFscGhhXFxpbnYoWF9pKSJdLFs0LDAsIlhfaSJdLFswLDEsIlxcYWxwaGEiXSxbMCwyLCJcXGJldGEiLDJdLFszLDAsIiIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzQsMSwiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMyw0LCJcXHJve1xcYWxwaGF9e1pfaX0iXSxbMywyLCJcXHJve1xcYmV0YX17Wl9pfSIsMix7ImN1cnZlIjo1fV1d}{Link
to Diagram}
\end{quote}

Then
\(\exists ! \Theta_i: Z_i\to X_i \underset{\scriptscriptstyle {S} }{\times} Y\)
where the \(\Theta_i\) agree on overlaps \(Z_{ij}\) as morphisms
\(Z_{ij} \to X_{ij} \underset{\scriptscriptstyle {S} }{\times} Y\). By
step 3, these glue to a unique
\(\Theta: Z\to X \underset{\scriptscriptstyle {S} }{\times} Y\), since
the gluing is defined as
\(X \underset{\scriptscriptstyle {S} }{\times} Y = \coprod_i (X_i \underset{\scriptscriptstyle {S} }{\times} Y)/\phi_{ij}(p) \sim p\).
Why does it make the above diagram commute?

\begin{center}
\begin{tikzcd}
    Z \\
    \\
    && {X \underset{\scriptscriptstyle {S} }{\times}  Y} && X \\
    \\
    && Y && S
    \arrow[from=3-5, to=5-5]
    \arrow[from=5-3, to=5-5]
    \arrow[from=3-3, to=5-3]
    \arrow[from=3-3, to=3-5]
    \arrow["\Theta"{description}, from=1-1, to=3-3]
    \arrow["\alpha", from=1-1, to=3-5]
    \arrow["\beta", from=1-1, to=5-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMiwyLCJYXFxmcHtTfSBZIl0sWzIsNCwiWSJdLFs0LDQsIlMiXSxbNCwyLCJYIl0sWzAsMCwiWiJdLFszLDJdLFsxLDJdLFswLDFdLFswLDNdLFs0LDAsIlxcVGhldGEiLDFdLFs0LDMsIlxcYWxwaGEiXSxbNCwxLCJcXGJldGEiXV0=}{Link
to Diagram}
\end{quote}

This commutes because such a map is determined on an open cover, and we
have commutativity in the following:

\begin{center}
\begin{tikzcd}
    {Z_i} \\
    \\
    && {X_i \underset{\scriptscriptstyle {S} }{\times}  Y} && X
    \arrow["{\Theta_i}", from=1-1, to=3-3]
    \arrow["{p_i}", from=3-3, to=3-5]
    \arrow["{\alpha_i}", from=1-1, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJaX2kiXSxbMiwyLCJYX2lcXGZwe1N9IFkiXSxbNCwyLCJYIl0sWzAsMSwiXFxUaGV0YV9pIl0sWzEsMiwicF9pIl0sWzAsMiwiXFxhbHBoYV9pIl1d}{Link
to Diagram}
\end{quote}

So \(X \underset{\scriptscriptstyle {S} }{\times} Y\) exists if
\(X_i \underset{\scriptscriptstyle {S} }{\times} Y\) exists for
\({\mathcal{X}}\rightrightarrows X\).

\end{proof}

\begin{proof}[Step 6: Affine base]

Let \(S\in {\mathsf{Aff}}{\mathsf{Sch}}\), then by step 1
\(X_i \underset{\scriptscriptstyle {S} }{\times} Y_j\) exists, to
\(X_i \underset{\scriptscriptstyle {S} }{\times} Y\) exists by step 5,
which implies \(X \underset{\scriptscriptstyle {S} }{\times} Y\) exists

\end{proof}

\begin{proof}[Step 7: Arbitrary]

Let \(X,Y,S\in {\mathsf{Sch}}_{/ {S}}\) be arbitrary. Take
\({\mathcal{S}}\rightrightarrows S\), and set \(X_i = p^{-1}(S_i)\) and
\(Y_i = q^{-1}(S_i)\). Then
\(X_i \underset{\scriptscriptstyle {S_i} }{\times} Y_i\) exists, and the
claim is that there is an isomorphism
\begin{align*}
X_i \underset{\scriptscriptstyle {S_i} }{\times}  Y_i \cong X \underset{\scriptscriptstyle {S} }{\times}  Y_i \in {\mathsf{Sch}}_{/ {S}} 
.\end{align*}

Then there exist \(Z\to Y_i\), and \(\operatorname{im}(Z\to S)\) must
lie in \(S\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-20_12-32.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{proof}

\hypertarget{fiber-products-friday-october-22}{%
\section{Fiber Products (Friday, October
22)}\label{fiber-products-friday-october-22}}

\begin{remark}

Last time: we defined and proved the existence of fiber products in
\({\mathsf{Sch}}_{/ {S}}\), and for
\(X,Y,S \in {\mathsf{Aff}}{\mathsf{Sch}}\) equal to
\(\operatorname{Spec}A, \operatorname{Spec}B, \operatorname{Spec}R\)
respectively,
\begin{align*}
X{ \operatorname{fp} }{S} Y = \operatorname{Spec}(A\otimes_R B)
.\end{align*}

\end{remark}

\begin{definition}[Residue field]

For \(X \xrightarrow{f} Y \in {\mathsf{Sch}}\) and \(p\in Y\), define
the \textbf{residue field}
\begin{align*}
k(p) \coloneqq{\mathcal{O}}_{Y, p} / {\mathfrak{m}}_{Y, p}
.\end{align*}

\end{definition}

\begin{remark}

There is a closed immersion
\(\operatorname{Spec}k(p) \hookrightarrow Y\) if \(p\) is a closed point
(since it came from a quotient map), and we can take a fiber product

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}k(p)  \underset{\scriptscriptstyle {Y} }{\times}  \operatorname{Spec}k(p)} && X \\
    \\
    {\operatorname{Spec}k(p)} && Y
    \arrow[from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgayhwKSBcXGZpYmVycHJvZHtZfSBcXHNwZWMgayhwKSJdLFsyLDAsIlgiXSxbMCwyLCJcXHNwZWMgayhwKSJdLFsyLDIsIlkiXSxbMSwzXSxbMiwzXSxbMCwyXSxbMCwxXSxbMCwzLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{example}[?]

Consider
\(\operatorname{Spec}k[x,y,t]/ \left\langle{xy-t}\right\rangle \to \operatorname{Spec}k[t]\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-22_11-41.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

What is the fiber over \(p\coloneqq\left\langle{t-1}\right\rangle\) or
\(q= \left\langle{t}\right\rangle\)?

\begin{itemize}
\tightlist
\item
  \(k(p) = k[t] / \left\langle{t-1}\right\rangle \cong k\),
\item
  \(k(q) = k[t] / \left\langle{t}\right\rangle \cong k\),
\end{itemize}

so they are abstractly isomorphic. We have the following tensor product
in rings:

\begin{center}
\begin{tikzcd}
    {k[x,y,t]/\left\langle{xy=t}\right\rangle \otimes_{k[t]} k} && {k[x,y,t]/\left\langle{xy-t}\right\rangle} \\
    \\
    {k\cong k[t]/\left\langle{t-1}\right\rangle} && {k[t]}
    \arrow[from=3-3, to=1-3]
    \arrow[from=3-3, to=3-1]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMiwwLCJrW3gseSx0XS9cXGdlbnN7eHktdH0iXSxbMCwwLCJrW3gseSx0XS9cXGdlbnN7eHk9dH0gXFx0ZW5zb3Jfe2tbdF19IGsiXSxbMCwyLCJrXFxjb25nIGtbdF0vXFxnZW5ze3QtMX0iXSxbMiwyLCJrW3RdIl0sWzMsMF0sWzMsMl0sWzEsMl0sWzEsMF0sWzEsMywiIiwxLHsic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV1d}{Link
to Diagram}
\end{quote}

Generally, pulling back over \(k[t] / \left\langle{t-c}\right\rangle\)
has the effect of setting \(t=c\) in the tensor product, and thus the
fiber products are given by

\begin{itemize}
\tightlist
\item
  \(k(p) \underset{\scriptscriptstyle {\operatorname{Spec}k[t]} }{\times} X = \operatorname{Spec}k[x, y] / \left\langle{xy-1}\right\rangle\)
\item
  \(k(q) \underset{\scriptscriptstyle {\operatorname{Spec}k[t]} }{\times} X = \operatorname{Spec}k[x, y] / \left\langle{xy}\right\rangle\)
\end{itemize}

\end{example}

\begin{example}[Fiber products aren't quite set products]

Consider
\begin{align*}
X \coloneqq{\mathbf{A}}^1_{/ {k}}  \underset{\scriptscriptstyle {\operatorname{Spec}k} }{\times}  {\mathbf{A}}^1_{/ {k}}  \cong \operatorname{Spec}(k[s] \otimes_k k[t]) \cong \operatorname{Spec}(k[s, t]) \cong {\mathbf{A}}^2_{/ {k}} 
.\end{align*}
However, \(X\) is not the set-theoretic product of the two constituent
sets, although it does contain the product. Why? Consider
\(p \coloneqq\left\langle{y^2 - x^3}\right\rangle \in \operatorname{Spec}{\mathbf{A}}^2_{/ {k}}\),
which is prime (check irreducibility in each variable!) and thus yields
a point which is not the product of any two points in
\({\mathbf{A}}^1_{/ {k}}\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-22_12-00.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{example}

\begin{example}[Reduction mod $p$]

Let
\(X\in {\mathsf{Sch}} { \, \xrightarrow{\sim}\, }{\mathsf{Sch}}_{/ {\operatorname{Spec}{\mathbf{Z}}}}\)
with structure map \(X\to \operatorname{Spec}{\mathbf{Z}}\), and let
\(p = \left\langle{P}\right\rangle \in \operatorname{Spec}{\mathbf{Z}}\).
Then \(k(p) = {\mathbf{Z}}/p = { \mathbf{F} }_p\), so consider the fiber
over \(p\):

\begin{center}
\begin{tikzcd}
    {X\times \operatorname{Spec}{ \mathbf{F} }_p} && X \\
    \\
    {\operatorname{Spec}{ \mathbf{F} }_p} && {\operatorname{Spec}{\mathbf{Z}}}
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJYXFx0aW1lcyBcXHNwZWMgXFxGRl9wIl0sWzIsMCwiWCJdLFswLDIsIlxcc3BlYyBcXEZGX3AiXSxbMiwyLCJcXHNwZWMgXFxaWiJdLFsxLDNdLFswLDJdLFsyLDNdLFswLDFdLFswLDMsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==}{Link
to Diagram}
\end{quote}

Call this the \textbf{reduction mod \(p\)}, denoted
\(X_{{ \mathbf{F} }_p}\). If \(X = \operatorname{Spec}R\), then
\(X_{{ \mathbf{F} }_p} = \operatorname{Spec}(R \otimes_{\mathbf{Z}}{\mathbf{Z}}/p) = \operatorname{Spec}(R/\left\langle{p}\right\rangle)\).

\end{example}

\begin{example}[?]

Take
\(X \coloneqq\operatorname{Spec}{\mathbf{Z}}[x,y,z]/\left\langle{x^5+y^5=z^5}\right\rangle\),
so nontrivial \({\mathbf{Z}}{\hbox{-}}\)points yield counterexamples to
Fermat. Then
\(X_{{ \mathbf{F} }_p} = { \mathbf{F} }_p[x,y,z] / \left\langle{x^5 + y^5 + z^5}\right\rangle\),
which reduces the coefficients of the equations.

\end{example}

\begin{quote}
How are these related to models of a scheme?
\end{quote}

\begin{example}[?]

Take \(X = \operatorname{Spec}{\mathbf{C}}\) to get
\({\mathbf{C}}\otimes_{{ \mathbf{F} }_p} {\mathbf{Z}}\) -- what is this
ring? One has to regard \({\mathbf{C}}\) as a ring over \({\mathbf{Z}}\)
first, so write
\({\mathbf{C}}= { \mkern 1.5mu\overline{\mkern-1.5mu \mathbf{Q} \mkern-1.5mu}\mkern 1.5mu }(T)\)
where \(T\) is an uncountable basis of transcendental elements. So this
yields
\(\mkern 1.5mu\overline{\mkern-1.5mu{ \mathbf{F} }_p\mkern-1.5mu}\mkern 1.5mu(T)\).

\end{example}

\begin{remark}

Consider \(X \in {\mathsf{Var}}_{/ {{\mathbf{C}}}}\),
e.g.~\(X = \operatorname{Spec}{\mathbf{C}}[x,y,z]/\left\langle{x=\sqrt{2} y, y^2=\pi z^3}\right\rangle\).
Then consider the (much smaller) subring generated by the coefficients
of the defining equations, so
\(R\coloneqq{\mathbf{Z}}[\sqrt 2, \pi] \cong {\mathbf{Z}}[\sqrt 2][t]\)
and consider
\(\operatorname{Spec}R[x,y,z]/ \left\langle{x=\sqrt y, y^2 = \pi z^3}\right\rangle\).
This has the exact same equations but is now defined of
\(\operatorname{Spec}R\). Note that having finitely many equations
yields a finitely generated as a \({\mathbf{Z}}{\hbox{-}}\)module.

Since \(R\hookrightarrow{\mathbf{C}}\) we get a morphism
\(\operatorname{Spec}{\mathbf{C}}\to \operatorname{Spec}R\), and we get
a diagram

\begin{center}
\begin{tikzcd}
    {Y \coloneqq R[x,y,z]/\left\langle{x=\sqrt 2 y, y^2 = \pi z^3}\right\rangle} && X \\
    \\
    {\operatorname{Spec}R} && \operatorname{Spec}{\mathbf{C}}
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJZIFxcZGEgUlt4LHksel0vXFxnZW5ze3g9XFxzcXJ0IDIgeSwgeV4yID0gXFxwaSB6XjN9Il0sWzIsMCwiWCJdLFswLDIsIlxcc3BlYyBSIl0sWzIsMiwiXFxzcGVjXFxDQyJdLFsyLDNdLFswLDJdLFsxLDNdLFswLDFdLFswLDMsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{warnings}

So in the literature, \emph{reduction of \(X\) mod \(p\)} generally
means \(Y_{{ \mathbf{F} }_p}\) and not \(X_{{ \mathbf{F} }_p}\).

\end{warnings}

\begin{definition}[Base Change]

Given \(X, Y\in {\mathsf{Sch}}_{/ {S}}\) with structure maps \(f, g\)
respectively, the \textbf{base change of \(f\) along \(g\)} is defined
as the fiber product
\(X \underset{\scriptscriptstyle {S} }{\times} Y \in {\mathsf{Sch}}_{/ {Y}}\).
So there is a functor
\begin{align*}
{-} \underset{\scriptscriptstyle {S} }{\times}  Y: {\mathsf{Sch}}_{/ {S}}  &\to {\mathsf{Sch}}_{/ {Y}} 
.\end{align*}

\end{definition}

\begin{quote}
What is the adjoint? Probably the forgetful functor given by composing
along \(g\).
\end{quote}

\hypertarget{monday-october-25}{%
\section{Monday, October 25}\label{monday-october-25}}

\hypertarget{length}{%
\subsection{Length}\label{length}}

\begin{remark}

A correction from last time: we said
\({\mathbf{C}}= { \mkern 1.5mu\overline{\mkern-1.5mu \mathbf{Q} \mkern-1.5mu}\mkern 1.5mu }(t_j {~\mathrel{\Big\vert}~}j\in J)\)
for some uncountable set of generators \(J\). Noting that
\(R \otimes_{\mathbf{Z}}{ \mathbf{F} }_p = R/pR\), which is zero if
\({1\over p}\in R\), so
\({\mathbf{C}}\otimes_{\mathbf{Z}}{ \mathbf{F} }_p = 0\). However, this
doesn't happen for
\({ \mkern 1.5mu\overline{\mkern-1.5mu {\mathbf{Z}}\mkern-1.5mu}\mkern 1.5mu }(t_j {~\mathrel{\Big\vert}~}j\in J)\),
so passing to a ring given by adjoining coefficients of equations is
still a reasonable thing to do.

Last time: for \(X \xrightarrow{f} Y\), the fiber over \(p\in Y\) was
\(X \underset{\scriptscriptstyle {Y} }{\times} \operatorname{Spec}k(p)\)
where \(k(p) \coloneqq{\mathcal{O}}_{Y, p}/{\mathfrak{m}}_p\), sometime
denoted \(f^{-1}(p)\), the \emph{scheme-theoretic fiber}.

\end{remark}

\begin{example}[?]

Consider intersecting a parabola with a family of lines:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-25_11-36.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Then there is a map
\(\operatorname{Spec}{\mathbf{C}}[x, y]/\left\langle{y-x^2}\right\rangle \to \operatorname{Spec}{\mathbf{C}}[y]\)
corresponding to a ring map
\({\mathbf{C}}[y] \to {\mathbf{C}}[x,y]/\left\langle{y-x^2}\right\rangle\).
We showed the scheme theoretic fiber over \(y=c_0\) is precisely
\(\operatorname{Spec}{\mathbf{C}}[x] / \left\langle{c_0 - x^2}\right\rangle \cong {\mathbf{C}}{ {}^{ \scriptscriptstyle\oplus^{2} } }\)
if \(c_0\neq 0\), and
\(\operatorname{Spec}{\mathbf{C}}[x]/\left\langle{x^2}\right\rangle\) if
\(c_0 = 0\). The former has no nilpotents while the latter does, so the
fibers are reduced away from \(c_0 = 0\).

\end{example}

\begin{definition}[Length]

If \(R\in {}_{k} \mathsf{Alg} ^{\mathrm{fg}}\) with
\(\operatorname{krulldim}(R) = 0\), then
\(\mathop{\mathrm{length}}(R) \coloneqq\dim_k R < \infty\).
\(\operatorname{Spec}R\) is called a length \(l\) scheme over \(k\).

\end{definition}

\begin{remark}

Note that \(\dim_{\mathbf{C}}{\mathbf{C}}[x, y] = \infty\), but has
Krull dimension 2. Most \(k{\hbox{-}}\)algebras will have infinite
\(k{\hbox{-}}\)dimension in this setting.

\end{remark}

\begin{remark}

If \(R\) is reduced and
\(k = { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\),
one can prove that \(R = k{ {}^{ \scriptscriptstyle\oplus^{\ell} } }\)
and \(\operatorname{Spec}R = \coprod_{i\leq \ell} \operatorname{Spec}k\)
where \(\ell\) is the number of reduced points.

\end{remark}

\begin{example}[?]

Take
\begin{align*}
\operatorname{Spec}{\mathbf{C}}\oplus {\mathbf{C}}[x]/\left\langle{x^3}\right\rangle \oplus {\mathbf{C}}[x, y]/\left\langle{x^2, xy, y^3}\right\rangle
.\end{align*}
The terms have dimension \(1,3,4\) respectively, yielding a
0-dimensional length 8 scheme. Note that
\(\operatorname{Spec}R = \operatorname{Spec}R_{ \text{red} }\), and
reducing this ring yields
\({\mathbf{C}}{ {}^{ \scriptscriptstyle\oplus^{3} } }\).

\end{example}

\begin{remark}

Let \(Y\in {\mathsf{Sch}}^{\mathrm{irr}}_{/ {k}}\) and
\(X \xrightarrow{f} Y\), then \(Y\) has a generic point
\(\sqrt 0 \in \operatorname{Spec}A_i\) for some cover
\(\left\{{\operatorname{Spec}A_i}\right\}\rightrightarrows Y\). This
corresponds to the irreducible closed subset \(Y\) itself, and yields a
unique open generic point \(Y_{\mathrm{gen}}\) One can then take the
fiber \(f^{-1}(Y_{\mathrm{gen}})\) -- what fiber product is this? Check
that
\(Y_{\mathrm{gen}}= A_i \left[ { \scriptstyle { {\sqrt 0}^{-1}} } \right] / \sqrt 0 = \operatorname{ff}(A_i)\)
is exactly the fraction field. Note that if
\(\operatorname{Spec}B_{ij} \subseteq \operatorname{Spec}A_i\) a
distinguished open, we have
\(\operatorname{ff}(B_{ij}) \subseteq \operatorname{ff}(A_i)\), so this
doesn't depend on the choice of the affine open
\(\operatorname{Spec}A_i\).

\end{remark}

\begin{example}[?]

Consider
\(\operatorname{Spec}{\mathbf{Z}}[x, y]/\left\langle{y^2 = x^3-1}\right\rangle \in {\mathsf{Sch}}_{/ {{\mathbf{Z}}}}\),
which comes equipped with a map to \(\operatorname{Spec}{\mathbf{Z}}\).
The generic fiber is the base change to
\(\operatorname{Spec}{\mathbf{Q}}\):

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}{\mathbf{Q}}[x, y] / \left\langle{y^2=x^3-1}\right\rangle} && X \\
    \\
    {\operatorname{Spec}{\mathbf{Q}}} && {\operatorname{Spec}{\mathbf{Z}}}
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMiwwLCJYIl0sWzAsMCwiXFxzcGVjIFxcUVFbeCwgeV0gLyBcXGdlbnN7eV4yPXheMy0xfSJdLFswLDIsIlxcc3BlYyBcXFFRIl0sWzIsMiwiXFxzcGVjIFxcWloiXSxbMiwzXSxbMSwyXSxbMCwzXSxbMSwwXSxbMSwzLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=}{Link
to Diagram}
\end{quote}

This can be done more generally to base change from a number field to
its fraction field. Consider a degree 2 field extension
\(K \to K[x]/\left\langle{x^2-y}\right\rangle\), then for example if
\(K = {\mathbf{C}}(y)\) we can construct the following:

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}{\mathbf{C}}[x, y]/\left\langle{y-x^2}\right\rangle} && {\operatorname{Spec}{\mathbf{C}}[y]} \\
    \\
    {\operatorname{Spec}{\mathbf{C}}(y)[x]/\left\langle{y-x^2}\right\rangle} && {\operatorname{Spec}{\mathbf{C}}(y) = \left\{{{\operatorname{pt}}}\right\}}
    \arrow[from=3-3, to=1-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=3-1, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgXFxDQ1t4LCB5XS9cXGdlbnN7eS14XjJ9Il0sWzAsMiwiXFxzcGVjIFxcQ0MoeSlbeF0vXFxnZW5ze3kteF4yfSJdLFsyLDIsIlxcc3BlYyBcXENDKHkpID0gXFx0c3tcXHB0fSJdLFsyLDAsIlxcc3BlYyBcXENDW3ldIl0sWzIsM10sWzEsMl0sWzAsM10sWzEsMF1d}{Link
to Diagram}
\end{quote}

Note that \(\operatorname{Spec}{\mathbf{C}}(y)\) is just a single point!
So this doesn't quite pick up that any specific choice of generic point
splits into two components, since \(x^2-y\) doesn't split unless
\(\sqrt y \in K\). One can remedy this by passing to
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbf{C}}(y)\mkern-1.5mu}\mkern 1.5mu\)
in this case. For \(f\) a finite morphism to an irreducible \(Y\), one
can define the \textbf{degree} of \(f\) as the degree of the extension
associated to a generic point.

\end{example}

\begin{example}[?]

Consider two lines projecting onto the \(y\) axis, say
\((x-1)(x+1) = 0\), then this splits/factors over the generic point.

\end{example}

\hypertarget{separatedproper-morphisms}{%
\subsection{Separated/Proper
Morphisms}\label{separatedproper-morphisms}}

\begin{definition}[Diagonal]

Let \(X\in {\mathsf{Sch}}_{/ {S}}\) with structure map
\(X \xrightarrow{f} S\), then the \textbf{diagonal}
\(\Delta: X\to X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {S} }{\times} ^{2} } }\)
is the following induced map:

\begin{center}
\begin{tikzcd}
    X \\
    & {X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {S} }{\times} ^{2} }  }} && X \\
    \\
    & X && S
    \arrow["f", from=4-2, to=4-4]
    \arrow["f"', from=2-4, to=4-4]
    \arrow[from=2-2, to=2-4]
    \arrow[from=2-2, to=4-2]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=4-4]
    \arrow["\Delta"{description}, from=1-1, to=2-2]
    \arrow[from=1-1, to=2-4]
    \arrow[from=1-1, to=4-2]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMSwxLCJYXFxmaWJlcnBvd2Vye1N9ezJ9Il0sWzMsMSwiWCJdLFsxLDMsIlgiXSxbMywzLCJTIl0sWzAsMCwiWCJdLFsyLDMsImYiXSxbMSwzLCJmIiwyXSxbMCwxXSxbMCwyXSxbMCwzLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbNCwwLCJcXERlbHRhIiwxXSxbNCwxXSxbNCwyXV0=}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{definition}[Separated]

A structure map \(X \xrightarrow{f} S\) is \textbf{separated} if the
diagonal
\(\Delta: X\to X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {S} }{\times} ^{2} } }\)
is a closed embedding. \(X\) itself is separated if
\(\Delta: X\to X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {\operatorname{Spec}{\mathbf{Z}}} }{\times} ^{2} } }\)
is separated.

\end{definition}

\begin{warnings}

The usual ``Hausdorff iff diagonal is closed'' depends on a separation
axiom! Which will often not hold in AG: for example,
\(\operatorname{Spec}R\) is separated for any ring but never Hausdorff.

\end{warnings}

\begin{proposition}[?]

Any morphism in \({\mathsf{Aff}}{\mathsf{Sch}}\) is separated.

\end{proposition}

\begin{proposition}[?]

Consider:

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}B} \\
    \\
    && {\operatorname{Spec}(B{ {}^{ \scriptstyle\otimes_{A}^{2} }  })} && {\operatorname{Spec}B} \\
    \\
    && {\operatorname{Spec}B} && {\operatorname{Spec}A}
    \arrow[from=3-5, to=5-5]
    \arrow[from=5-3, to=5-5]
    \arrow[from=3-3, to=3-5]
    \arrow[from=3-3, to=5-3]
    \arrow[from=1-1, to=3-5]
    \arrow[from=1-1, to=5-3]
    \arrow["\Delta"{description}, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMiwyLCJcXHNwZWMgKEJcXHRlbnNvcnBvd2Vye0F9ezJ9KSJdLFs0LDIsIlxcc3BlYyBCIl0sWzIsNCwiXFxzcGVjIEIiXSxbNCw0LCJcXHNwZWMgQSJdLFswLDAsIlxcc3BlYyBCIl0sWzEsM10sWzIsM10sWzAsMV0sWzAsMl0sWzQsMV0sWzQsMl0sWzQsMCwiXFxEZWx0YSIsMV1d}{Link
to Diagram}
\end{quote}

Then there is a ring morphism
\begin{align*}
\Delta^* B{ {}^{ \scriptstyle\otimes_{A}^{2} }  } &\to B && \in \mathsf{CRing}\\
b_1\otimes b_2 &\mapsto b_1 b_2
.\end{align*}
Since this is surjective, \(\Delta\) is a closed immersion.

\end{proposition}

\begin{example}[A classic non-example]

Let \(X\) be \({\mathbf{A}}^1\) with the doubled origin, so
\(X = {\mathbf{A}}^1{\textstyle\coprod}_f {\mathbf{A}}^1\) where for
\(U\coloneqq{\mathbf{A}}^1\setminus\left\{{0}\right\}\), we glue by
\(\operatorname{id}_U\). Taking the product
\(X \underset{\scriptscriptstyle {\operatorname{Spec}k} }{\times} X\)
yields the following:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-25_12-22.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Note that \((0, 0') \not\in \Delta(X)\) is not closed, but
\((0, 0')\in \mkern 1.5mu\overline{\mkern-1.5mu\Delta(X)\mkern-1.5mu}\mkern 1.5mu\)
is in its closure.

\end{example}

\hypertarget{wednesday-october-27}{%
\section{Wednesday, October 27}\label{wednesday-october-27}}

\begin{remark}

Recall that if \(X\in {\mathsf{Sch}}_{/ {S}}\) with \(f:X\to S\), there
is an induced diagonal map
\begin{align*}
\Delta: X\to X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {S} }{\times} ^{2} }  } \\
,\end{align*}
which is induced by
\((\operatorname{id}_X, \operatorname{id}_X): X\to X{ {}^{ \scriptscriptstyle\times^{2} } }\).
We said \(f\) is \textbf{separated} if \(\Delta\) is a closed immersion,
which in particular is a homeomorphism onto a closed subset.

\end{remark}

\begin{example}[?]

An example: any morphism of affine schemes
\(f\in {\mathsf{Sch}}(\operatorname{Spec}B, \operatorname{Spec}A)\).

A non-example:
\({\mathbf{A}}^1_{/ {k}} \coprod_{{\mathbf{A}}^1_{/ {k}} \setminus\left\{{0}\right\}} {\mathbf{A}}^1_{/ {k}}\),
the line with the doubled origin. We saw
\((0, 0')\in {{\partial}}\Delta(X) = \mkern 1.5mu\overline{\mkern-1.5mu\Delta(X)\mkern-1.5mu}\mkern 1.5mu\setminus\Delta(X)\),
and in fact
\(X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {\operatorname{Spec}k} }{\times} ^{2} } } = \cup_{i\leq 4} {\mathbf{A}}^2_{/ {k}}\).

\end{example}

\begin{proposition}[?]

\(f\in {\mathsf{Sch}}(X, S)\) is separated
\(\iff \operatorname{im}(\Delta)\) is closed.

\end{proposition}

\begin{proof}[?]

\(\implies\): This is definitional.

\(\impliedby\): First show \(\Delta\) is a homeomorphism onto its image.
Use the universal property to get
\(p_1 \circ \delta = \operatorname{id}_X\) where
\(p_i: X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {S} }{\times} ^{2} } }\to X\)
are the projections. Since both are continuous, \(\Delta\) is a
homeomorphism onto its image, which is closed.

It then suffices to show
\(\Delta^\#: {\mathcal{O}}_{X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {S} }{\times} ^{2} } }} \to \Delta_* {\mathcal{O}}_{X}\)
is a surjective map in
\({\mathsf{Sh}}(X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {S} }{\times} ^{2} } })\)
to get a closed immersion. For any
\(p\in X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {S} }{\times} ^{2} } }\),
we need to show that there exists an open \(N\ni p\) such that there is
a surjection on sections
\begin{align*}
\Delta^\#(N): {\mathcal{O}}_{X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {S} }{\times} ^{2} }  }}(N) \twoheadrightarrow\Delta_* {\mathcal{O}}_X(N) \in \mathsf{CRing}
.\end{align*}

Observe that if \(p\not\in \operatorname{im}\Delta\) and
\(N= (\operatorname{im}\Delta)^c\) is open, then
\(\Delta_* {\mathcal{O}}_X(N) = {\mathcal{O}}_X(\Delta^{-1}(N)) = {\mathcal{O}}_X(\emptyset) = 0\)
and anything surjects onto the zero ring. So if
\(p\not\in \mathop{\mathrm{supp}}\Delta_* {\mathcal{O}}_X\), this is
surjective. For \(p\in \operatorname{im}\Delta\), write
\(p = \Delta(q)\) using the \(\Delta\) is a homeomorphism onto its image
and thus \(q\) is unique, and choose \(U\ni q\) an affine open in \(X\).
Then \(f(U) \subseteq V\) is contained in an affine open in \(S\). The
gluing construction of fiber products yields that
\(U{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {V} }{\times} ^{2} } } \subseteq X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {S} }{\times} ^{2} } }\)
is again an affine open, and \(N\ni p\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-27_11-56.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Then \(\Delta^*(N)\) is surjective since \(U, V\) are affine and thus
\({ \left.{{f}} \right|_{{U}} }:U\to V\) is separated.

\end{proof}

\begin{example}[?]

A morphism of schemes can be a homeomorphism onto a closed subset but
\emph{not} a closed immersion. Consider
\(k\hookrightarrow k[x]/\left\langle{x^2}\right\rangle\), inducing
\(\operatorname{Spec}k[x]/\left\langle{x^2}\right\rangle \to \operatorname{Spec}k\).
This is not a surjective map of rings, and on affines this is equivalent
to being a closed immersion.

Note that even though \(f\) is a closed immersion here, \(\Delta\) is
not: the fiber product is given by
\begin{align*}
\operatorname{Spec}k[x]/x^2  \underset{\scriptscriptstyle {\operatorname{Spec}k} }{\times}  \operatorname{Spec}k[y]/y^2 = \operatorname{Spec}k[x, y] / \left\langle{x^2, y^2}\right\rangle
.\end{align*}

Note that taking \(k[x]/x^2\to k\) where \(x\to 0\) \emph{is} a closed
immersion.

\end{example}

\begin{quote}
To google: if \(f\) is a homeomorphism onto its image and satisfies some
condition for the induced map on Zariski tangent spaces, is it
necessarily a closed immersion?
\end{quote}

\hypertarget{valuative-criterion-of-separatedness}{%
\subsection{Valuative Criterion of
Separatedness}\label{valuative-criterion-of-separatedness}}

\begin{remark}

A map from a punctured curve should extend uniquely!

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-27_12-17.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{remark}

\begin{theorem}[?]

Let \(f\in {\mathsf{Sch}}(X, Y)\) with \(X\) Noetherian, then \(f\) is
separated iff for any \(R\in \mathsf{DVR}\) and
\(K \coloneqq\operatorname{ff}(R)\), if the following lift exists, it is
unique:

\begin{center}
\begin{tikzcd}
    K && {\operatorname{Spec}K} && X \\
    \\
    R && {\operatorname{Spec}R} && Y
    \arrow[from=1-5, to=3-5]
    \arrow[from=3-3, to=3-5]
    \arrow["{!\theta}", dashed, from=3-3, to=1-5]
    \arrow[hook, from=3-1, to=1-1]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-3, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\begin{align*}Link to Diagram\end{align*}
(https://q.uiver.app/?q=WzAsNixbMiwwLCJcXHNwZWMgSyJdLFsyLDIsIlxcc3BlYyBSIl0sWzQsMCwiWCJdLFs0LDIsIlkiXSxbMCwyLCJSIl0sWzAsMCwiSyJdLFsyLDNdLFsxLDNdLFsxLDIsIiFcXHRoZXRhIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzQsNSwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMCwxXSxbMCwyXV0=
\end{quote}

\end{theorem}

\begin{example}[?]

For \(\operatorname{Spec}R = \left\{{{\mathfrak{m}}, 0}\right\}\),
\(\operatorname{Spec}K = \left\{{0}\right\}\) is just the generic point:

\includegraphics{figures/2021-10-27_12-23-53.png}

And moreover the generic points must be mapped to each other.

\end{example}

\begin{example}[?]

Consider
\(R \coloneqq{\mathbf{C}}[t] \left[ { \scriptstyle { {t}^{-1}} } \right]\),
which is the stalk
\({\mathcal{O}}_{{\mathbf{A}}^1_{/ {{\mathbf{C}}}} , 0}\), and
\(K = {\mathbf{C}}(t)\).

\end{example}

\begin{slogan}

\(\operatorname{Spec}R\) for \(R\in \mathsf{DVR}\) is like a small piece
of a curve.

\end{slogan}

\hypertarget{monday-november-01}{%
\section{Monday, November 01}\label{monday-november-01}}

\begin{remark}

Recall that \(f:X\to Y\) is \textbf{separated} if
\(\Delta X\to X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {Y} }{\times} ^{2} } }\)
is a closed immersion, or equivalently
\(\Delta(X) \subseteq X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {Y} }{\times} ^{2} } }\)
is closed. We discussed the valuative criterion of separatedness, which
is slightly more useful when proving things, but only holds for
Noetherian (quasicompact, admits a a finite cover of affines) schemes:
\(X\) is separated iff any diagram admitting a lift \(\Theta\) of the
following form admits a \emph{unique} lift:

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}K} && X \\
    \\
    {\operatorname{Spec}R, R\in \mathsf{DVR}} && Y
    \arrow["f", from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow["{!\Theta}"{description}, dashed, from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJcXHNwZWMgUiwgUlxcaW4gXFxEVlIiXSxbMCwwLCJcXHNwZWMgSyJdLFsyLDAsIlgiXSxbMiwyLCJZIl0sWzIsMywiZiJdLFswLDNdLFsxLDBdLFsxLDJdLFswLDIsIiFcXFRoZXRhIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link
to Diagram}
\end{quote}

Here \(R\in \mathsf{DVR}\) and \(K\in \operatorname{ff}(R)\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-11-01_11-40.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{remark}

\begin{example}[?]

Consider mapping to the line with two origins:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{42pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-11-01_11-42.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Then given \(f:X\to Y\) and \(\operatorname{Spec}R\to Y\) there is an
induced map
\(k[t] \left[ { \scriptstyle { {t}^{-1}} } \right] \leftarrow k[t]\).
But note that there are two distinct extensions
\(\operatorname{Spec}R\to X\), say \(\Theta_1, \Theta_2\), and there is
an extension of the following form:

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}k(t)} && {\operatorname{Spec}k[t]} \\
    \\
    {\operatorname{Spec}k[t] \left[ { \scriptstyle { {t}^{-1}}  } \right] }
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow["{\exists \Theta_1}", shift left=1, from=3-1, to=1-3]
    \arrow["{\exists \Theta_2}"', shift right=1, from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXHNwZWMgayh0KSJdLFsyLDAsIlxcc3BlYyBrW3RdIl0sWzAsMiwiXFxzcGVjIGtbdF1cXGxvY2FsaXple3R9Il0sWzAsMl0sWzAsMV0sWzIsMSwiXFxleGlzdHMgXFxUaGV0YV8xIiwwLHsib2Zmc2V0IjotMX1dLFsyLDEsIlxcZXhpc3RzIFxcVGhldGFfMiIsMix7Im9mZnNldCI6MX1dXQ==}{Link
to Diagram}
\end{quote}

\end{example}

\begin{remark}

Taking fraction fields corresponds to throwing out everything but the
generic point.

\end{remark}

\begin{proof}[of valuative criterion]

Omitted, see Hartshorne. We'll discuss one key idea:
\textbf{specialization}.

\end{proof}

\hypertarget{specialization}{%
\subsection{Specialization}\label{specialization}}

\begin{remark}

Consider the data of a morphism from \(\operatorname{Spec}K\):

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}K} && X \\
    \\
    {\operatorname{Spec}R} && Y
    \arrow["\psi", from=3-1, to=3-3]
    \arrow["f"', from=1-3, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow["\phi", from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgSyJdLFsyLDAsIlgiXSxbMiwyLCJZIl0sWzAsMiwiXFxzcGVjIFIiXSxbMywyLCJcXHBzaSJdLFsxLDIsImYiLDJdLFswLDNdLFswLDEsIlxccGhpIl1d}{Link
to Diagram}
\end{quote}

This is the data of a point \(p\in X\), so
\(\phi( \left\langle{0}\right\rangle) = p\in { {\left\lvert {X} \right\rvert} }\),
and (it suffices to have) a pushforward \(\phi_p^\sharp\) which is a
morphism of local rings inducing a diagram:

\begin{center}
\begin{tikzcd}
    {{\mathcal{O}}_{X, p}} && K \\
    \\
    {\kappa(p)}
    \arrow["{\operatorname{mod}{\mathfrak{m}}_p}"', from=1-1, to=3-1]
    \arrow["{\phi^\sharp_p}", from=1-1, to=1-3]
    \arrow[from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXE9PX3tYLCBwfSJdLFsyLDAsIksiXSxbMCwyLCJcXGthcHBhKHApIl0sWzAsMiwiXFxtb2QgXFxtZm1fcCIsMl0sWzAsMSwiXFxwaGleXFxzaGFycF9wIl0sWzIsMV1d}{Link
to Diagram}
\end{quote}

Here \(\phi^\sharp_p({\mathfrak{m}}_p) = \left\langle{0}\right\rangle\)
and \(\kappa(p)\) is the residue field at \(p\).

\end{remark}

\begin{remark}

What data is needed to specify \(\psi: \operatorname{Spec}R \to Y\)? We
need two points \(p_0, p_1\in { {\left\lvert {X} \right\rvert} }\) with
\(p_1 = \psi(\left\langle{0}\right\rangle) \in \psi(\operatorname{Spec}K)\)
and \(p_0 = \psi({\mathfrak{m}})\). Since
\(\psi ^{-1}\qty{\mkern 1.5mu\overline{\mkern-1.5mu\left\{{p_1}\right\}\mkern-1.5mu}\mkern 1.5mu}\)
is closed, we also need
\(p_0 \in \mkern 1.5mu\overline{\mkern-1.5mu\left\{{p_1}\right\}\mkern-1.5mu}\mkern 1.5mu\),
so \(\kappa(p_1) \subseteq K\).

Consider
\(Z \coloneqq\mkern 1.5mu\overline{\mkern-1.5mu\left\{{p_1}\right\}\mkern-1.5mu}\mkern 1.5mu \ni p_0\)
with its structure as a reduced closed subscheme of \(X\). This yields a
map \(\operatorname{Spec}R\to Z\), and we need an injective (dominant)
ring map \({\mathcal{O}}_{Z, p_0}\to R\). Why does this produce a map
\(\operatorname{Spec}R\to Y\)? We have a closed immersion
\(\operatorname{Spec}R\hookrightarrow\operatorname{Spec}{\mathcal{O}}_{Z, p_0} \to Z \hookrightarrow X\)

\end{remark}

\begin{definition}[Specialization]

A point \(p_0\) is a \textbf{specialization} of \(p_1\) relative to
\(R\in \mathsf{DVR}\) and \(K = \operatorname{ff}(R)\) if \(p_1\) is a
\(K{\hbox{-}}\)point, so \(\kappa(p_1) \subseteq K\) such that
\(p_0 \in \mkern 1.5mu\overline{\mkern-1.5mup_1\mkern-1.5mu}\mkern 1.5mu\)
and
\({\mathcal{O}}_{\mkern 1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu_1, p} \twoheadrightarrow R\)

\end{definition}

\begin{example}[?]

Take \(R\coloneqq k {\llbracket t \rrbracket }\) and
\(\operatorname{ff}(R) = k((t))\) for \(k\in \mathsf{Field}\). Consider
\(\operatorname{Spec}k\to {\mathbf{A}}^1_{/ {k}}\) corresponding to
\(k[t] \hookrightarrow k((t))\). Setting
\(p_1 = \operatorname{im}\left\langle{0}\right\rangle = \left\langle{0}\right\rangle \in k[t]\)
to be the generic point of \({\mathbf{A}}^1_{/ {k}}\), we have
\(\mkern 1.5mu\overline{\mkern-1.5mup_1\mkern-1.5mu}\mkern 1.5mu = {\mathbf{A}}^1_{/ {k}}\).
Set \(p_0 = \left\langle{t}\right\rangle\), and note that
\(p_0 \in \mkern 1.5mu\overline{\mkern-1.5mup_1\mkern-1.5mu}\mkern 1.5mu\),
we then want a ring map
\({\mathcal{O}}_{\mkern 1.5mu\overline{\mkern-1.5mup_1\mkern-1.5mu}\mkern 1.5mu, p_0} \to R = k[[t]]\).
Note that
\({\mathcal{O}}_{\mkern 1.5mu\overline{\mkern-1.5mup_1\mkern-1.5mu}\mkern 1.5mu, p_0} = k[t] \left[ { \scriptstyle { {t}^{-1}} } \right]\),
and there is a ring map
\(\left\{{f(t)/g(t) {~\mathrel{\Big\vert}~}g(0) \neq 0}\right\} \to k[[t]]\).
This is injective, yielding a domination of rings.

\end{example}

\hypertarget{wednesday-november-03}{%
\section{Wednesday, November 03}\label{wednesday-november-03}}

\begin{remark}

Let \(R\in \mathsf{DVR}, k = \operatorname{ff}(R)\), then a
\(k{\hbox{-}}\)point of \(X\) is a morphism
\(\operatorname{Spec}k \to X\) and is given by the data of an inclusion
\(p_1\in { {\left\lvert {X} \right\rvert} }\) and an inclusion
\(\kappa(p_1) \hookrightarrow k\).

\end{remark}

\begin{example}[?]

Why these are called \(k{\hbox{-}}\)points? Given
\(\operatorname{Spec}{\mathbf{Q}}\to \operatorname{Spec}S\coloneqq\operatorname{Spec}{\mathbf{Z}}[x,y,z]/ \left\langle{x^5 +y^5-z^5}\right\rangle\),
then there is a ring map \(S\to {\mathbf{Q}}\) where
\(x,y,z \mapsto x_0,y_0, z_0\) satisfying \(x_0^2 +y_0^2=z_0^2\). So
these are rational solutions to the defining equations.

\end{example}

\begin{remark}

Lifting a \(k{\hbox{-}}\)point to an \(R{\hbox{-}}\)point
\(\operatorname{Spec}R \to X\) requires
\(p_0 \in { \operatorname{cl}} (\left\{{p_1}\right\})\) and a domination
\({\mathcal{O}}_{{ \operatorname{cl}} (\left\{{p_1}\right\}), p_0} \to R\)
inducing the \(k{\hbox{-}}\)point in the sense that the generic point of
\(\operatorname{Spec}R\) maps to the generic point of
\(\operatorname{Spec}{\mathcal{O}}_{{ \operatorname{cl}} \left\{{p_1}\right\}, p_0}\)
corresponding to an inclusion of fields. So we get a morphism of local
rings.

\end{remark}

\begin{remark}

We saw that \(f:X\to Y\) is separated iff
\(\Delta(X) \hookrightarrow X{ {}^{ \scriptscriptstyle \underset{\scriptscriptstyle {Y} }{\times} ^{2} } }\)
is closed iff any \(k{\hbox{-}}\)point of \(X\) has at most one
specialization over a given \(R{\hbox{-}}\)point of \(Y\). Idea: rules
out two lifts.

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-11-03_11-40.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Note that this needs \(X\) to be Noetherian.

\begin{quote}
Is separatedness local? Perhaps \(X\) only locally Noetherian would
suffice.
\end{quote}

\end{remark}

\begin{remark}

Review the difference between ``of finite type'' and ``\emph{locally} of
finite type''.

\end{remark}

\begin{definition}[Closed and universally closed]

A morphism \(f:X\to Y\) is \textbf{closed} if the underlying map
\({ {\left\lvert {f} \right\rvert} }: { {\left\lvert {X} \right\rvert} } \to{ {\left\lvert {Y} \right\rvert} } \in {\mathsf{Top}}\)
is a closed continuous map (so images of closed sets are closed), and
\(f\) is \textbf{universally closed} if for all \(Y' \to Y\) the change
\(f': X \underset{\scriptscriptstyle {Y} }{\times} Y'\to Y'\) is closed.

\end{definition}

\begin{example}[?]

Identity maps \(\operatorname{id}_X:X\to X\) are closed, using that
\(Y \underset{\scriptscriptstyle {X} }{\times} X \cong Y\) and pulling
back \(\operatorname{id}_X\) yields \(\operatorname{id}_Y\).

\end{example}

\begin{example}[A non-example]

Consider \({\mathbf{A}}^1_{/ {k}} \to \operatorname{Spec}k\), which is
closed since \(\operatorname{Spec}k = {\operatorname{pt}}\) and has the
discrete topology. This is not universally closed, since we have

\begin{center}
\begin{tikzcd}
    {{\mathbf{A}}^2_{/ {k}} } && {{\mathbf{A}}^1_{/ {k}} } \\
    \\
    {{\mathbf{A}}^1 _{/ {k}} } && {\operatorname{Spec}k}
    \arrow["f", from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow["{f'}"', from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXEFBXjJcXHNsaWNlIGsiXSxbMCwyLCJcXEFBXjEgXFxzbGljZSBrIl0sWzIsMiwiXFxzcGVjIGsiXSxbMiwwLCJcXEFBXjFcXHNsaWNlIGsiXSxbMywyLCJmIl0sWzEsMl0sWzAsMSwiZiciLDJdLFswLDNdXQ==}{Link
to Diagram}
\end{quote}

Consider \(Z\coloneqq V(xy-1) \subseteq {\mathbf{A}}^2_{/ {k}}\), then
\(f'(Z) = {\mathbf{A}}^2_{/ {k}} \setminus\left\{{0}\right\}\) is
projection onto the \(x{\hbox{-}}\)axis and is not closed in
\({\mathbf{A}}^2_{/ {k}}\). What this is a projection onto the
\(x{\hbox{-}}\)axis: this comes from the map
\(f: k[x] \hookrightarrow k[x, y] \cong k[x] \otimes_k k[y]\) where
\(f^{-1}(\left\langle{x-x_0, y-y_0}\right\rangle) = \left\langle{x-x_0}\right\rangle\),
so geometrically this yields the \((x_0, y_0) \to x_0\).

\end{example}

\begin{example}[?]

Consider
\({\mathbf{P}}^1_{/ {k}} \coloneqq\mathop{\mathrm{proj}}k[x,y]\) and
consider \({\mathbf{P}}^1_{/ {k}} \to \operatorname{Spec}k\). Here
\({\mathbf{P}}^1_{/ {k}}\) is supposed to be ``compact'' in the sense
that graphs of all functions are closed.

\end{example}

\begin{exercise}[?]

Are compact spaces universally closed in \({\mathsf{Top}}\)?

\end{exercise}

\begin{definition}[Proper]

A morphism \(f:X\to Y\) is \textbf{proper} if

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(f\) is of finite type,
\item
  \(f\) is separated,
\item
  \(f\) is \emph{universally closed}
\end{enumerate}

\end{definition}

\begin{remark}

This ranges over all possible base changes, so it's quite hard to
actually check! The following result gives an easier way:

\end{remark}

\begin{theorem}[Valuative criterion of properness]

Let \(f:X\to Y\) be a finite type morphism with \(X\) Noetherian. Then
\(f\) is proper \(\iff\) there exists unique lifts \(\Theta\) of the
following form:

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}k} && X \\
    \\
    {\operatorname{Spec}R} && Y
    \arrow["f", from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-1, to=3-1]
    \arrow["{!\Theta}"{description}, dashed, from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgayJdLFswLDIsIlxcc3BlYyBSIl0sWzIsMiwiWSJdLFsyLDAsIlgiXSxbMywyLCJmIl0sWzEsMl0sWzAsM10sWzAsMV0sWzEsMywiIVxcVGhldGEiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=}{Link
to Diagram}
\end{quote}

\end{theorem}

\begin{remark}

Most spaces in practice are separated and of finite type, unless you're
working with moduli of K3 surfaces!

\end{remark}

\begin{proof}[$\implies$]

Suppose \(f\) is proper, then \(f\) is separated and we have uniqueness
for any lifts by the valuative criterion for separatedness. This uses
that \(X\) is Noetherian. It then suffices to show existence of
\(\Theta\), using that \(f\) is universally closed. Consider the base
change
\(X_R \coloneqq X \underset{\scriptscriptstyle {Y} }{\times} \operatorname{Spec}R\),
then using commutativity we get a morphism
\(s: \operatorname{Spec}k \to X_R\). Let
\(p_1 = s(\left\langle{0}\right\rangle) \subseteq X_R\), we'll then try
to specialize \(p_1\). Let
\(Z \coloneqq{ \operatorname{cl}} \left\{{p_1}\right\} \subseteq X_R\),
then since \(f\) is proper and \(Z\) is closed in \(X_R\), \(f_R(Z)\) is
closed:

\begin{center}
\begin{tikzcd}
    {X_R} && X \\
    \\
    {\operatorname{Spec}R} && Y
    \arrow["f", from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow["{f_R}"', from=1-1, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJYX1IiXSxbMCwyLCJcXHNwZWMgUiJdLFsyLDAsIlgiXSxbMiwyLCJZIl0sWzIsMywiZiJdLFsxLDNdLFswLDJdLFswLDEsImZfUiIsMl1d}{Link
to Diagram}
\end{quote}

We can compose
\(\operatorname{Spec}k \to Z \xrightarrow{f_R} \operatorname{Spec}R\) to
get \(\tilde f\), which is an inclusion of the generic point:

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}k} \\
    & {X_R} && X \\
    \\
    & {\operatorname{Spec}R} && Y
    \arrow["f", from=2-4, to=4-4]
    \arrow[from=4-2, to=4-4]
    \arrow[from=1-1, to=2-4]
    \arrow["{\tilde f}"', from=1-1, to=4-2]
    \arrow["{!\Theta}"{description}, dashed, from=4-2, to=2-4]
    \arrow[from=1-1, to=2-2]
    \arrow[from=2-2, to=2-4]
    \arrow["{f_R}"{description}, from=2-2, to=4-2]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJcXHNwZWMgayJdLFsxLDMsIlxcc3BlYyBSIl0sWzMsMywiWSJdLFszLDEsIlgiXSxbMSwxLCJYX1IiXSxbMywyLCJmIl0sWzEsMl0sWzAsM10sWzAsMSwiXFx0aWxkZSBmIiwyXSxbMSwzLCIhXFxUaGV0YSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDRdLFs0LDNdLFs0LDEsImZfUiIsMV1d}{Link
to Diagram}
\end{quote}

Then \(f_R(Z) = \operatorname{Spec}R\) and so there exists \(p_0 \in Z\)
with \(f_R(p_0) = {\mathfrak{m}}\), the closed point in
\(\operatorname{Spec}R\). So we get
\begin{align*}
g: Z &\to \operatorname{Spec}R \\
{ \operatorname{cl}} \left\{{p_1}\right\}\ni p_0 &\mapsto {\mathfrak{m}}\\
p_1 &\mapsto \left\langle{0}\right\rangle
.\end{align*}

Taking stalks yields a local ring morphism
\(g^\sharp_{p_0}: R\to {\mathcal{O}}_{Z, p_0}\), and this completes to a
diagram:

\begin{center}
\begin{tikzcd}
    R && {{\mathcal{O}}_{Z, p_0}} \\
    \\
    k && {\operatorname{ff}{\mathcal{O}}_{Z, p_0} = \kappa(p_1)}
    \arrow[from=1-1, to=3-1]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJSIl0sWzIsMCwiXFxPT197WiwgcF8wfSJdLFswLDIsImsiXSxbMiwyLCJcXGZmIFxcT09fe1osIHBfMH0gPSBcXGthcHBhKHBfMSkiXSxbMCwyXSxbMiwzXSxbMSwzXSxbMCwxXV0=}{Link
to Diagram}
\end{quote}

But \(R\) is final with respect to domination for local rings \(R'\) in
\(k\) with \(\operatorname{ff}R' = k\), and if final objects admit
morphisms to other objects, those objects must also be final, so
\(R = {\mathcal{O}}_{Z, p_0}\). This yields a domination
\({\mathcal{O}}_{Z, p_0} \to R\), which corresponds to a lift
\(\operatorname{Spec}R \to X\).

\end{proof}

\hypertarget{friday-november-05}{%
\section{Friday, November 05}\label{friday-november-05}}

\begin{remark}

Last time: valuative criterion for properness. A morphism
\(f:X\to Y\in {\mathsf{Sch}}\) is proper \(\iff\)

\begin{itemize}
\tightlist
\item
  \(f\) is separated,
\item
  \(f\) is of finite type,
\item
  \(f\) is universally closed (closed to closed, and preserved under
  base change)
\end{itemize}

If \(X\) is Noetherian and \(f\) is of finite type, then \(f\) is proper
\(\iff\) for \(R \in \mathsf{DVR}, K = \operatorname{ff}(R)\), we have
lifts:

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}K} && X \\
    \\
    {\operatorname{Spec}R} && Y
    \arrow["f", from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=3-1, to=3-3]
    \arrow["{\exists !\Theta}"{description}, from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgSyJdLFswLDIsIlxcc3BlYyBSIl0sWzIsMiwiWSJdLFsyLDAsIlgiXSxbMywyLCJmIl0sWzAsM10sWzAsMV0sWzEsMl0sWzEsMywiXFxleGlzdHMgIVxcVGhldGEiLDFdXQ==}{Link
to Diagram}
\end{quote}

We proved that \(f\) proper implies \(\exists \Theta\). Erratum: we said
\(R \subseteq K\) is final with respect to local rings contained in
\(K\) with fraction field \(K\), but rather it's maximal. As an example,
\({\mathbf{Z}}{ \scriptsize {}_{ \left[ { \scriptstyle { {p}^{-1}} } \right] } }{2}, {\mathbf{Z}}{ \scriptsize {}_{ \left[ { \scriptstyle { {p}^{-1}} } \right] } }{3} \hookrightarrow{\mathbf{Q}}\)
but there's no common ring they map to. Proof of \(\impliedby\): see
Hartshorne.

\end{remark}

\begin{corollary}[?]

Some applications/corollaries of the valuative criterion for properness:

\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\columnwidth - 2\tabcolsep) * \real{0.45}}
  >{\raggedright\arraybackslash}p{(\columnwidth - 2\tabcolsep) * \real{0.54}}@{}}
\toprule
Separated & Proper \\
\midrule
\endhead
Open or closed immersions & Closed immersions\footnote{Very rarely
  proper! Only if inclusion of a connected component.} \\
Compositions & Compositions \\
Stable under base change & Stable under base change \\
Products & Products \\
Local on base & Local on base \\
\(X \xrightarrow{f} Y \xrightarrow{g} Z\) with \(g \circ f\) separated
\(\implies f\) separated & \(X \xrightarrow{f} Y \xrightarrow{g} Z\)
with \(g \circ f\) proper and \(g\) separated \(\implies f\)
separated \\
\bottomrule
\end{longtable}

\begin{itemize}
\tightlist
\item
  ``Stable under base change'' means that whenever
  \(X \xrightarrow{f} Y\) has a property \(P(f)\), any fiber product
  along \(Y' \to Y\) yields the same property \(P(f')\):
\end{itemize}

\begin{center}
\begin{tikzcd}
    {X'} && X \\
    \\
    {Y'} && Y
    \arrow[from=3-1, to=3-3]
    \arrow["f", color={rgb,255:red,92;green,92;blue,214}, from=1-3, to=3-3]
    \arrow["{f'}", color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJYJyJdLFswLDIsIlknIl0sWzIsMiwiWSJdLFsyLDAsIlgiXSxbMSwyXSxbMywyLCJmIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXX0sWzI0MCw2MCw2MCwxXV0sWzAsMSwiZiciLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdfSxbMjQwLDYwLDYwLDFdXSxbMCwzXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=}{Link
to Diagram}
\end{quote}

\begin{itemize}
\item
  A \emph{product} of morphisms in \({\mathsf{Sch}}_{/ {S}}\) is the
  product in \({\mathsf{Sch}}_{/ {S}}\), or equivalently the fiber
  product over \(S\). So given \(f:X\to Y\) and \(f':X'\to Y'\), the
  product is
  \((f, f'): X \underset{\scriptscriptstyle {S} }{\times} X' \to Y \underset{\scriptscriptstyle {S} }{\times} Y'\).
  So here ``Products'' means that if \(P(f), P(f')\) holds, then
  \(P(f, f')\) holds.
\item
  \(P\) is \emph{local on the base} if whenever \(P\) holds for
  \(X \xrightarrow{f} Y\) then for all open \(U \subseteq Y\), the
  restriction \({ \left.{{f}} \right|_{{f^{-1}(U)}} }: f^{-1}(U) \to U\)
  also satisfies \(P\)
\end{itemize}

\end{corollary}

\begin{proof}[stability under base change]

Diagram chases involving the valuative criteria and universal properties
of the fiber product. For example, we'll do stability under base change:
let \(X \xrightarrow{f} Y\) be separated and \(Y'\to Y\), we'll show
\(X' \xrightarrow{f'} Y'\) is separated where
\(X' \coloneqq X \underset{\scriptscriptstyle {Y} }{\times} Y'\). We
need to show an extension \(\Theta\) of the following form is unique if
it exists:

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}K} && {X'} && X \\
    \\
    {\operatorname{Spec}R} && {Y'} && Y
    \arrow["{f'}", from=1-3, to=3-3]
    \arrow["\alpha"', from=3-1, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-1, to=3-1]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
    \arrow["{\Theta_1}", shift left=1, dashed, from=3-1, to=1-3]
    \arrow[from=3-3, to=3-5]
    \arrow["f"{description}, from=1-5, to=3-5]
    \arrow["\beta"{description}, from=1-3, to=1-5]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=3-5]
    \arrow["{\Theta_2}"', shift right=1, dashed, from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJcXHNwZWMgSyJdLFswLDIsIlxcc3BlYyBSIl0sWzIsMCwiWCciXSxbMiwyLCJZJyJdLFs0LDAsIlgiXSxbNCwyLCJZIl0sWzIsMywiZiciXSxbMSwzLCJcXGFscGhhIiwyXSxbMCwyXSxbMCwxXSxbMCwzLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbMSwyLCJcXFRoZXRhXzEiLDAseyJvZmZzZXQiOi0xLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMyw1XSxbNCw1LCJmIiwxXSxbMiw0LCJcXGJldGEiLDFdLFsyLDUsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dLFsxLDIsIlxcVGhldGFfMiIsMix7Im9mZnNldCI6MSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link
to Diagram}
\end{quote}

Note that \(\beta \circ \theta_1 = \beta \circ \Theta_1\), since \(f\)
is separated, using the valuative criterion for separatedness. Since
\(X'\) is a fiber product, by the universal property there exists a
unique product morphism
\((\beta \circ \Theta_1, \alpha) = (\beta \circ \Theta_2, \alpha)\). So
\(\Theta_1 = \Theta_2\) and \(f'\) is separated by the valuative
criterion of separatedness.

\end{proof}

\begin{proof}[of products]

We want to show that if \(f:X\to Y, f':X'\to Y'\) are proper then
\((f, f'): X \underset{\scriptscriptstyle {S} }{\times} X' \to Y \underset{\scriptscriptstyle {S} }{\times} Y'\)
is proper. We can produce a diagram:

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,92;green,92;blue,214}{\operatorname{Spec}K} && {X \underset{\scriptscriptstyle {S} }{\times}  X'} && {X'} \\
    & \textcolor{rgb,255:red,92;green,92;blue,214}{\operatorname{Spec}R} && {Y \underset{\scriptscriptstyle {S} }{\times}  Y'} && {Y'} \\
    && X && S \\
    &&& Y && S
    \arrow[from=2-4, to=4-4]
    \arrow[from=4-4, to=4-6]
    \arrow[from=2-6, to=4-6]
    \arrow[from=2-4, to=2-6]
    \arrow[from=1-5, to=3-5]
    \arrow[from=3-5, to=4-6]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-3, to=3-3]
    \arrow[from=3-3, to=3-5]
    \arrow[from=1-3, to=2-4]
    \arrow[from=3-3, to=4-4]
    \arrow[from=1-5, to=2-6]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=1-3]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=2-2, to=2-4]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=2-2]
    \arrow["{\text{WTS: } \exists ! \Theta}"{description}, dashed, from=2-2, to=1-3]
    \arrow[curve={height=30pt}, dotted, from=1-1, to=3-3]
    \arrow[curve={height=30pt}, dotted, from=2-2, to=4-4]
    \arrow["{\exists !}"{description}, color={rgb,255:red,214;green,92;blue,92}, squiggly, from=2-2, to=3-3]
    \arrow["{\exists !}"{description, pos=0.7}, color={rgb,255:red,214;green,92;blue,92}, squiggly, from=2-2, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

Here we get existence of unique maps
\(\operatorname{Spec}R\to X, \operatorname{Spec}R\to X'\), which thus
yields a unique map
\(\operatorname{Spec}R \to X \underset{\scriptscriptstyle {S} }{\times} X'\).

\end{proof}

\begin{proof}[locality on base]

Suppose \(X \xrightarrow{f} Y\) is proper, we'll show that
\({ \left.{{f}} \right|_{{f^{-1}(U)}} }:f^{-1}(U) \to U\) is proper for
\(U \subseteq Y\). We use that \(f^{-1}(U)\) is a fiber product and
apply the universal property:

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}K} && {f^{-1}(U)} && X \\
    \\
    {\operatorname{Spec}R} && U && Y
    \arrow["f", from=1-5, to=3-5]
    \arrow[from=1-1, to=3-1]
    \arrow[from=3-3, to=3-5]
    \arrow[from=1-3, to=1-5]
    \arrow["{{ \left.{{f}} \right|_{{f^{-1}(U)}} }}"{pos=0.8}, from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=3-5]
    \arrow["{\exists !}"{description, pos=0.4}, curve={height=-6pt}, from=3-1, to=1-5]
    \arrow["{\therefore \exists!}"{description}, dashed, from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbNCwyLCJZIl0sWzQsMCwiWCJdLFswLDAsIlxcc3BlYyBLIl0sWzAsMiwiXFxzcGVjIFIiXSxbMiwyLCJVIl0sWzIsMCwiZlxcaW52KFUpIl0sWzEsMCwiZiJdLFsyLDNdLFs0LDBdLFs1LDFdLFs1LDQsIlxccm97Zn17ZlxcaW52KFUpfSIsMCx7ImxhYmVsX3Bvc2l0aW9uIjo4MH1dLFszLDRdLFsyLDVdLFs1LDAsIiIsMCx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dLFszLDEsIlxcZXhpc3RzICEiLDEseyJsYWJlbF9wb3NpdGlvbiI6NDAsImN1cnZlIjotMX1dLFszLDUsIlxcdGhlcmVmb3JlIFxcZXhpc3RzISIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

For the converse, it suffices to check properness on an open cover
\({\mathcal{U}}\rightrightarrows Y\). Why? It follows if for any diagram
of the following form, there exists an open \(U_i \subseteq Y\) such
that \(\operatorname{im}\operatorname{Spec}R \subseteq U_i\):

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}K} && X \\
    \\
    {\operatorname{Spec}R} && Y
    \arrow[from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-1, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgSyJdLFswLDIsIlxcc3BlYyBSIl0sWzIsMiwiWSJdLFsyLDAsIlgiXSxbMywyXSxbMSwyXSxbMCwzXSxbMCwxXV0=}{Link
to Diagram}
\end{quote}

Note that \(\operatorname{Spec}R\) has two points, so this is not
completely trivial. Consider the closed point
\({\mathfrak{m}}\in \operatorname{Spec}R\) and let
\(p_0 = \operatorname{im}({\mathfrak{m}})\).

\end{proof}

\hypertarget{monday-november-08}{%
\section{Monday, November 08}\label{monday-november-08}}

\begin{remark}

Let \(f:X\to Y \in {\mathsf{Sch}}\), then \(f\) is \textbf{proper} iff

\begin{itemize}
\tightlist
\item
  \(f\) is separated,
\item
  \(f\) is of finite type,
\item
  \(f\) is universally closed, so for all \(Y'\to Y\), the base change
  morphisms \(X \underset{\scriptscriptstyle {Y} }{\times} Y' \to Y'\)
  is closed.
\end{itemize}

The valuative criterion of properness stated that if \(R\) is a
valuation ring and \(K \coloneqq\operatorname{ff}(R)\), so
\(\operatorname{Spec}K\to \operatorname{Spec}R\), there are unique lifts
of the following form:

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}K} && X \\
    \\
    {\operatorname{Spec}R} && Y
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow[from=3-1, to=3-3]
    \arrow["\Theta"{description}, dashed, from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgSyJdLFswLDIsIlxcc3BlYyBSIl0sWzIsMiwiWSJdLFsyLDAsIlgiXSxbMywyXSxbMCwxXSxbMCwzXSxbMSwyXSxbMSwzLCJcXFRoZXRhIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{definition}[Projective space over a scheme]

Let \(Y \in {\mathsf{Sch}}\), then define \textbf{projective space over
\(Y\)} as
\({\mathbf{P}}^n_{/ {Y}} \coloneqq{\mathbf{P}}^n_{/ {{\mathbf{Z}}}} \times Y \coloneqq{\mathbf{P}}^n_{/ {{\mathbf{Z}}}} \underset{\scriptscriptstyle {\operatorname{Spec}{\mathbf{Z}}} }{\times} Y\).

\end{definition}

\begin{remark}

This is analogous to
\({\mathbf{P}}^n_{/ {R}} \coloneqq\mathop{\mathrm{Proj}}R[x_0,\cdots, x_n]\)
for \(R\in \mathsf{CRing}\), and these two constructions turn out to be
the same. Note that
\(R[x_0, \cdots, x_n] \cong {\mathbf{Z}}[x_0,\cdots, x_n] \otimes_{\mathbf{Z}}R\).

\end{remark}

\begin{definition}[Projective morphisms]

A morphism \(X \xrightarrow{f} Y\in {\mathsf{Sch}}\) is
\textbf{projective} iff \(f\) factors as
\(X \hookrightarrow{\mathbf{P}}^n_{/ {Y}} \twoheadrightarrow Y\), a
closed immersion into projective space followed by projection onto
\(Y\).

\end{definition}

\begin{example}[?]

Let \(S\in {\mathsf{gr}\,}_{\mathbf{Z}}\mathsf{CRing}\) be a graded
ring, then \(S_0 \leq S\) is a subring. Suppose \(S\) is finitely
generated over \(S_0\) by \(S_1\), so there exists a finite generating
set \(\left\{{x_i}\right\}_{i\leq n}\) and a surjective map
\(S_0[x_0, \cdots, x_n] \twoheadrightarrow S\) sending \(x_i\) to
elements of \(S_1\). Note that this preserves the grading, since \(S_0\)
elements have degree zero and \(x_i\) have degree 1 on both sides, so we
get a map
\(\psi: \mathop{\mathrm{Proj}}S \to \mathop{\mathrm{Proj}}S_0[x_0, \cdots, x_n] \coloneqq{\mathbf{P}}^n_{/ {S_0}}\).
So the ring maps on affine opens will be surjective, since they are
localizations of \(\psi\), so this yields a closed immersion and thus a
projective morphism
\(\mathop{\mathrm{Proj}}S \to \operatorname{Spec}S_0\).

\end{example}

\begin{example}[?]

If \(S = k \in \mathsf{Field}\) with
\(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\),
then if \(R \in {\mathsf{gr}\,}_{\mathbf{Z}}\mathsf{CRing}\) is finitely
generated in degree 1, then
\(\mathop{\mathrm{proj}}R\to \operatorname{Spec}k\) is projective, since
these are exactly quotients of \(k[x_1, \cdots, x_{n}]\) by a
homogeneous ideal. If this homogeneous ideal is radical, then these
correspond to projective varieties over \(k\).

\end{example}

\begin{remark}

We'll show that projective implies proper, which will furnish many
examples of proper maps.

\end{remark}

\begin{theorem}[?]

Any projective morphism \(f:X\to Y\) is proper.

\end{theorem}

\begin{exercise}[Hartshorne 3.13, checking when a morphism is finite type]

See Hartshorne, try it!

\end{exercise}

\begin{proof}[?]

We know that base changes of proper morphisms are proper, using the
valuative criterion. With the above example and exercise, it suffices to
show \({\mathbf{P}}^n_{/ {{\mathbf{Z}}}}\) is proper. Why? We have a
closed immersion
\(X\to {\mathbf{P}}^n_{/ {Y}} \coloneqq{\mathbf{P}}^n_{/ {{\mathbf{Z}}}} \times Y\)
by the definition of \(f:X\to Y\) being proper. Then the projection
\({\mathbf{P}}^n_{/ {{\mathbf{Z}}}} \times Y\to Y\) is proper, and is a
base change of
\({\mathbf{P}}^n_{/ {{\mathbf{Z}}}} \to \operatorname{Spec}{\mathbf{Z}}\).
So if we know the latter is proper, compositions of proper maps are
proper and thus \(f\) is proper.

\todo[inline]{todo, try to form a diagram here.}

Idea: clear denominators in a minimal way. We can cover
\({\mathbf{P}}^n_{/ {{\mathbf{Z}}}}\) by affine opens:
\begin{align*}
D(x_i) \coloneqq\left\{{ {\mathfrak{p}}\in {\mathbf{Z}}[x_0,\cdots, x_n]_{ { \mathrm{homog} } } {~\mathrel{\Big\vert}~}z_i\not\in {\mathfrak{p}}}\right\}
\cong \operatorname{Spec}{\mathbf{Z}} { \left[ \scriptstyle { {x_0 \over x_1}, \cdots, {x_n \over x_1}} \right] } 
.\end{align*}
Then
\(\left\{{D(x_i)}\right\}_{1\leq i \leq n+1} \rightrightarrows{\mathbf{P}}^n_{/ {{\mathbf{Z}}}}\),
making \({\mathbf{P}}^n_{/ {{\mathbf{Z}}}}\) finite type since it admits
a finite cover by \(\operatorname{Spec}R_i\) for
\(R_i\in \mathsf{Alg} _{/ {{\mathbf{Z}}}} ^{\mathrm{fg}}\). It thus
suffices to verify the valuative criterion of properness, since this
will imply separatedness. So we'll show any morphism
\(g: \operatorname{Spec}K\to {\mathbf{P}}^n_{/ {{\mathbf{Z}}}}\) lifts
uniquely to a morphism
\(\operatorname{Spec}K\to {\mathbf{P}}^n_{/ {{\mathbf{Z}}}}\). Given
\(g\), \(g(0)\) is a single point, and by a linear change of coordinates
we can ensure \(g(0) \in \bigcap_i D(x_i)\). So we have
\begin{align*}
g(0) \in D(x_0, \cdots, x_n) \cong \operatorname{Spec}T \coloneqq\operatorname{Spec}{\mathbf{Z}} { \left[ \scriptstyle {\left\{{x_i\over x_j}\right\}_{0\leq j\leq n}} \right] }  &&\forall i
.\end{align*}
So \(g\) factors through the open immersion
\(\operatorname{Spec}T \hookrightarrow{\mathbf{P}}^n_{/ {{\mathbf{Z}}}}\),
and is thus a map of affine schemes and equivalently the data of a ring
map \(\phi: \operatorname{Spec}T\to K \in \mathsf{CRing}\). Let
\(\phi_{ij} = \phi\qty{x_i\over x_j}\), and note that
\(\phi_{ij} \in K^{\times}\) for every \(i, j\). These satisfy a cocycle
condition \(\phi_{ij} \phi_{jk} = \phi_{ik}\), so letting
\(v_i\coloneqq v(\phi_{i, 0})\) for \(v\) the valuation, there is some
minimal \(v_i\).

\begin{quote}
Continued next time.
\end{quote}

\end{proof}

\hypertarget{wednesday-november-10}{%
\section{Wednesday, November 10}\label{wednesday-november-10}}

\begin{remark}

We defined projective space
\({\mathbf{P}}^n_{/ {Y}} \coloneqq{\mathbf{P}}^n_{/ {{\mathbf{Z}}}} \underset{\scriptscriptstyle {\operatorname{Spec}{\mathbf{Z}}} }{\times} Y\),
and a projective morphism as one that factors as a closed immersion into
\({\mathbf{P}}^n_{/ {Y}}\) for some \(Y\) followed by projection onto
\(Y\). Continuing the proof from last time: we reduced to
\({\mathbf{P}}^n_{/ {{\mathbf{Z}}}}\) and produced a map
\(\operatorname{Spec}R\to {\mathbf{P}}^n_{/ {{\mathbf{Z}}}}\).

\end{remark}

\begin{proof}[continued]

We noted that \(\phi_{ij}\in k^{\times}\) since
\(\phi_{ij}\phi{ji} = 1\), and more generally
\(\phi_{ij}\phi_{jk} = \phi_{ik}\). We chose
\(v_i = {\operatorname{val}}(\phi_{i, 0}) \in {\mathbf{Z}}\) minimally,
so assume without loss of generality by relabeling that it is \(v_1\),
so \(v_i\geq v_1\) for all \(i\). Then use that
\(\phi_{ij} = \phi_{i1}/\phi_{j1} \implies v(\phi_{i1}) = v(\phi_{i0}) - v(\phi_{10}) = v_2 -v_1 \geq 0\).
Writing
\(R = \left\{{\phi \in k\ {~\mathrel{\Big\vert}~}v(\phi) \geq 0}\right\}\),
we have \(\phi_{i1} \in R \subseteq k\). Consider the ring map
\begin{align*}
{\mathbf{Z}}\left[ {{x_0 \over x_1}, \cdots, {x_n \over x_1}} \right] &\to R \\
{x_i \over x_1} &\mapsto \phi_{i1}
.\end{align*}
This yields a map
\(\operatorname{Spec}R\to {\mathbf{A}}^n_{/ {{\mathbf{Z}}}} \cong D(x_1)\),
which restricts to a map
\(\operatorname{Spec}K\to D(x_0, \cdots, x_n)\), a smaller open set.

\begin{quote}
Fire alarm! Class canceled.
\end{quote}

\end{proof}

\hypertarget{friday-november-12}{%
\section{Friday, November 12}\label{friday-november-12}}

\begin{remark}

Continuing from last time: this is equivalent to
\(\Delta({\mathbf{P}}^n_{/ {{\mathbf{Z}}}} )\) being closed. Since every
affine scheme is separated,
\(\Delta({\mathbf{A}}^n_{/ {{\mathbf{Z}}}} )\) is closed for every
\(D(x_i)\). Suppose a closed point \((P, P')\) lies in the closure of
\(\Delta({\mathbf{P}}^n_{/ {{\mathbf{Z}}}} )\), then if
\(P, P'\in D(x_i)\) for some \(i\) then \(P = P'\) since \(D(x_i)\) is
separated. We can ensure this is possible by potentially taking a linear
change of coordinates. Introducing new variables \(x_i' = \sum n_i x_i\)
with
\(N \coloneqq(n_i)_{i\in I} \in \operatorname{GL}_{n+1}({\mathbf{Z}})\).
Then there is an isomorphism of graded rings
\({\mathbf{Z}}[x_0, \cdots, x_n] \to {\mathbf{Z}}[x_0', \cdots, x_n']\)
inducing an isomorphism
\({\mathbf{P}}^n_{/ {{\mathbf{Z}}}} {\circlearrowleft}\). By doing this
we can replace \(x_0\) with any linear combination of \(x_i\)s, and we
need to show that there exists a linear map \(L\) such that
\(L(x) = \sum n_i x_i\) for which \(L(x)\neq P, L(x)\neq P'\), so
\(P, P' \in D(L(x))\). Consider the image of \(L(x)\) in
\({\mathbf{Z}}[x_0, \cdots, x_n]/P\) and similarly for \(P'\). These are
(finite) fields since \(P, P'\) are maximal. Any map
\({\mathbf{Z}}{ {}^{ \scriptscriptstyle\times^{n+1} } }\to { \mathbf{F} }_q\)
is a group morphism, and the kernel is a finite index sublattice. One
can always find an element of
\({\mathbf{Z}}{ {}^{ \scriptscriptstyle\times^{n+1} } }\) which isn't on
the union of two strict sublattices, i.e.~\(1/a + 1/b - 1/ab < 1\).

\end{remark}

\begin{example}[?]

A projective variety over \(k\) is proper over \(\operatorname{Spec}k\).
These are of the form \(\mathop{\mathrm{Proj}}k[x_1, \cdots, x_{n}]/ I\)
for \(I\) a homogeneous ideal, and thus come with a closed immersion
into \({\mathbf{P}}^n_{/ {k}}\).

\end{example}

\begin{example}[The main class of examples]

If \(X \xrightarrow{f} Y \in \mathop{\mathrm{Proj}}{\mathsf{Var}}\) or
\({\mathsf{Sch}}_{/ {k}}\). Then the maps \(X\to \operatorname{Spec}k\)
and \(Y\to \operatorname{Spec}k\) are proper, and the second is
separated. Peeling off the compositions shows \(f\) is proper.

\end{example}

\begin{example}[?]

Let \(X \xrightarrow{f} Y\) be any morphism from a projective scheme to
a separated scheme of finite type over \(k\). This is also proper, and
thus universally closed, and its image in \(Y\) is also proper using
that closed subschemes of separated schemes are separated and of finite
type, and morphisms factor through their images.

\end{example}

\begin{corollary}[?]

Any regular function on a projective (or even proper) variety is locally
constant.

\end{corollary}

\begin{proof}[?]

A regular function on projective \(X\) is a morphism
\(X \xrightarrow{f} {\mathbf{A}}^1\), so consider the open immersion
\({\mathbf{A}}^1\hookrightarrow{\mathbf{P}}^1\). The composition
\(x\circ f: X\to {\mathbf{P}}^1\) is projective, thus proper, so
\((i \circ f )(X) \subseteq {\mathbf{A}}^1 \subseteq {\mathbf{P}}^1\) is
closed, but the only such closed sets are finite. Thus \(i \circ f\) and
thus \(f\) is constant on a connected component of \(X\).

\end{proof}

\begin{corollary}[?]

Any morphism from a proper variety to an affine variety is locally
constant.

\end{corollary}

\begin{proof}[?]

If \(X \xrightarrow{f} Y\) with \(X\) proper and \(Y\) affine, then
there is an open immersion \(\iota: Y\hookrightarrow{\mathbf{A}}^n\).
The composition of \(\iota \circ f\) is locally constant on each
coordinate by the previous corollary, making \(f\) locally constant.

\end{proof}

\begin{example}[?]

For \(X\) any variety and \(Y\) proper, \(X\times Y\to X\) is proper
because it is the base change along \(Y\to \operatorname{Spec}k\). So
e.g.~\({\mathbf{P}}^1\times {\mathbf{A}}^1 \to {\mathbf{A}}^1\) is
proper. Another example: blow up a projective variety at an ideal.

\end{example}

\begin{slogan}

Proper means compact fibers.

\end{slogan}

\hypertarget{monday-november-22}{%
\section{Monday, November 22}\label{monday-november-22}}

\begin{remark}

Last time: a local ring \(R\) is \textbf{regular} iff the number of
generators of \({\mathfrak{m}}_R\) is equal to the Krull dimension of
\(R\). There is a slightly weaker notion: a scheme is regular in
codimension 1 iff every local ring \({\mathcal{O}}_{X, x}\) of dimension
1 is regular. Note that this is locally the generic point associated to
a height 1 prime ideal.

\end{remark}

\begin{remark}

Note that not every local ring is a domain,
e.g.~\({\mathcal{O}}_{V(xy), \mathbf{0}}\). Algebra fact: a
1-dimensional regular local ring is a DVR, since this forces
\({\mathfrak{m}}\) to be generated by 1 element and thus principal.

\end{remark}

\begin{remark}

A new standing assumption: \(X\in {\mathsf{Sch}}\) is

\begin{itemize}
\tightlist
\item
  Noetherian,
\item
  Integral (covered by spec of integral domains, equivalently reduced
  and irreducible)
\item
  Separated,
\item
  Regular in codimension 1.
\end{itemize}

\end{remark}

\begin{example}[?]

Examples are smooth projective varieties, but may include singular
varieties, e.g.~\(V(xy - z^2) \subseteq {\mathbf{A}}^3_{/ {k}}\). Note
that the partials vanish at \(\mathbf{0}\), an singularity is picked up
by the fact that
\(k[x,y]/\left\langle{f}\right\rangle \left[ { \scriptstyle { {\left\langle{ x, y}\right\rangle }^{-1}} } \right]\)
and this is conical hyperboloid:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-11-22_11-43.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Note that some curves may be singular, namely those passing through
\(0\), but generically they are nonsingular. An equation of a line on
\(X\) might be \(V(x, z)\), so consider
\({\mathfrak{p}}= \left\langle{x, y}\right\rangle \in k[x,y,z]/\left\langle{xy-z^2}\right\rangle\).

\end{example}

\begin{exercise}[A good one]

Check that
\(R \left[ { \scriptstyle { {{\mathfrak{p}}}^{-1}} } \right]\) is
regular, and
\({\mathfrak{p}}R \left[ { \scriptstyle { {{\mathfrak{p}}}^{-1}} } \right]\)
is principal.

\end{exercise}

\begin{definition}[Prime divisors]

A \textbf{prime divisor} on \(X\) is an integrable subscheme of
codimension 1.

\end{definition}

\begin{example}[?]

Take \(V(y^2-x^3) \subseteq {\mathbf{A}}^2_{/ {k}}\), or
\(V(x, z) \subseteq \operatorname{Spec}R\). Generally, if \(f\in R\) is
irreducible then \(V(f) \subseteq \operatorname{Spec}R\) is a prime
divisor. Note that the second example is codimension 1 in
\({\mathbf{A}}^2_{/ {k}}\).

\end{example}

\begin{remark}

If \(X=\operatorname{Spec}R\), then the prime divisors are in 1-to-1
correspondence with height 1 prime ideals in \(R\). Check that
\(\left\langle{0}\right\rangle \subsetneq {\mathfrak{p}}\) since \(R\)
is a domain, and no prime ideal can fit between these. Note that in the
above example,
\(\left\langle{0}\right\rangle \subsetneq \left\langle{x, z}\right\rangle \subseteq k[x,y,z]/\left\langle{xy-z^2}\right\rangle\),
where e.g.~\(\left\langle{x}\right\rangle\) isn't prime because
\(xy\in \left\langle{x}\right\rangle \implies z^2\in \left\langle{x}\right\rangle \implies z\in \left\langle{x}\right\rangle\).

\end{remark}

\begin{example}[?]

Some examples of prime divisors:

\begin{itemize}
\tightlist
\item
  For \(X\) a nice variety: the irreducible subvarieties of codimension
  1.
\item
  For \(X = \operatorname{Spec}{\mathbf{Z}}\): closed points, i.e.~any
  maximal ideal.
\item
  For
  \(X=\operatorname{Spec}{\mathbf{Z}} { \left[ \scriptstyle {\sqrt{-5}} \right] }\):
  an example might be \(\left\langle{2, 1 + \sqrt{-5}}\right\rangle\).
\item
  For \(X= { \mathbf{F} }_3[t]\), consider
  \(\left\langle{t-a_i}\right\rangle\) for \(a_i=0,1,2\) and
  \(\left\langle{t^2-2}\right\rangle\). Note that being a prime ideal is
  not preserved under base change, e.g.
\end{itemize}

\begin{center}
\begin{tikzcd}
    {X' = \operatorname{Spec}{ \mathbf{F} }_3[t]\otimes{ \mathbf{F} }_q = \operatorname{Spec}{ \mathbf{F} }_q[t]} && {X = \operatorname{Spec}{ \mathbf{F} }_3[t]} \\
    \\
    {\operatorname{Spec}{ \mathbf{F} }_q} && {\operatorname{Spec}{ \mathbf{F} }_3}
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-1, to=3-1]
    \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=45}, draw=none, from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJYJyA9IFxcc3BlYyBcXEZGXzNbdF1cXHRlbnNvciBcXEZGX3EgPSBcXHNwZWMgXFxGRl9xW3RdIl0sWzIsMCwiWCA9IFxcc3BlYyBcXEZGXzNbdF0iXSxbMiwyLCJcXHNwZWMgXFxGRl8zIl0sWzAsMiwiXFxzcGVjIFxcRkZfcSJdLFszLDJdLFsxLDJdLFswLDFdLFswLDNdLFswLDIsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==}{Link
to Diagram}
\end{quote}

\end{example}

\begin{definition}[Weil Divisor]

The \textbf{Weil divisors} on \(X\) is the free
\({\mathbf{Z}}{\hbox{-}}\)module on the prime divisors, and is denoted
\(\operatorname{Div}(X)\).

\end{definition}

\begin{example}[?]

\begin{itemize}
\tightlist
\item
  \(1[\left\langle{x, y}\right\rangle] \in \operatorname{Div}(\operatorname{Spec}k[x,y,z] / \left\langle{xy-z^2}\right\rangle)\).
\item
  \(2[\left\langle{z}\right\rangle] - [\left\langle{3}\right\rangle] + 8[\left\langle{7}\right\rangle] \in \operatorname{Div}(\operatorname{Spec}{\mathbf{Z}})\).
\item
  \([V(y^2-x^3)] + 2[V(y)] \in \operatorname{Div}({\mathbf{A}}^2_{/ {k}} )\)
\item
  \([0] - [\infty] \in \operatorname{Div}({\mathbf{P}}^1_{/ {k}} )\).
\item
  For \(C\) an irreducible reduced curve, any linear combination of
  closed points.
\end{itemize}

Note that \emph{Cartier} divisors are those locally cut out by a single
equation.

\end{example}

\begin{remark}

Since \(X\) is integral, it has a generic point \(\eta\), so define a
\textbf{rational function} as a nonzero element of
\({\mathcal{O}}_{X, \eta}\). Equivalently, if
\(\operatorname{Spec}R \subseteq X\) is an affine chart, an element of
\(k^{\times}\) where \(k = \operatorname{ff}(R)\). Note that this is
independent of further localizing \(R\)! Any rational function \(\phi\)
on \(X\) gives an element
\(\phi \in \operatorname{ff}{\mathcal{O}}_{X, x} \cong k\) for any point
\(x\in X\). In particular, the standing assumptions (specifically being
regular in codimension 1) implies that \({\mathcal{O}}_{X, x}\) is a DVR
when \(x\) is the generic point of a prime divisor. Let
\(Y \subseteq X\) be a prime divisor, then define \(v_Y(\phi)\) to be
the valuation \(v(\phi)\) in \({\mathcal{O}}_{X, Y}\).

\end{remark}

\begin{example}[?]

The element \(4/7\) is a rational function on
\(\operatorname{Spec}{\mathbf{Z}}\), which is exactly
\({\mathbf{Q}}^{\times}\). Moreover
\(4/7\in {\mathcal{O}}_{\operatorname{Spec}{\mathbf{Z}}, 2}\) and
\({\operatorname{val}}_{\mathfrak{m}}(4/7) = 2\) for
\({\mathfrak{m}}= \left\langle{2}\right\rangle\).

\end{example}

\begin{definition}[Divisors of functions, principal divisors, class groups]

\begin{align*}
\operatorname{Div}(\phi) \coloneqq\sum_{Y \subseteq X \text{prime}} v_Y(\phi) [Y]
.\end{align*}
These are called \textbf{principal divisors}, and form a group
\({ \operatorname{cl}} (X)\) the \textbf{class group}.

\end{definition}

\begin{example}[?]

\(\operatorname{Div}(4/7) = 2[2] - 1[7]\).

\end{example}

\hypertarget{monday-november-29}{%
\section{Monday, November 29}\label{monday-november-29}}

\begin{remark}

Standing assumption: \(X\in {\mathsf{Sch}}\) is

\begin{itemize}
\tightlist
\item
  Integral: covered by \(\operatorname{Spec}R\) for \(R\) integral
  domains
\item
  Noetherian: covered by Noetherian rings.
\item
  Separated: \(\Delta\) is closed.
\item
  Regular in codimension 1:
  \(\dim {\mathcal{O}}_{X, x} = 1\implies {\mathcal{O}}_{X, x}\) is
  regular and thus a DVR.
\end{itemize}

\end{remark}

\begin{definition}[Prime and Weil divisors]

A \textbf{prime divisor} \(Y \subseteq X\) is a closed integral
subscheme of codimension 1, and a \textbf{Weil divisor} is a formal
\({\mathbf{Z}}{\hbox{-}}\)linear combination
\(\sum_{1\leq i \leq k} n_i Y_i\). The divisor is \textbf{effective} if
\(n_i \geq 0\) for all \(i\).

\end{definition}

\begin{example}[?]

If \(\eta\in Y\) is the generic point,
\({\mathcal{O}}_{X, \eta} \coloneqq{\mathcal{O}}_{X, Y}\) is a local
ring of dimension 1, and thus a DVR. This yields a valuation:
\begin{align*}
v_Y: \operatorname{ff}({\mathcal{O}}_{X, Y})^{\times}= k^{\times}\to {\mathbf{Z}}
,\end{align*}
where \(k\) is the residue field of the generic point of \(X\), also
called the \emph{rational functions on \(X\)}.

\end{example}

\begin{definition}[Principal divisors]

If \(f\in k^{\times}\) then there is an associated divisor:
\begin{align*}
\operatorname{Div}(f) \coloneqq\sum_{Y \text{prime divisors}} v_Y(f) [Y]
.\end{align*}
Any divisor of a rational function is \textbf{principal}.

\end{definition}

\begin{example}[?]

For \(X\coloneqq\operatorname{Spec}{\mathbf{Z}}\), the generic point is
\(\left\langle{0}\right\rangle\) and the prime divisors are prime ideals
of height 1, so here just prime ideals \(\left\langle{p}\right\rangle\).
So the integral closed codimension subschemes correspond to primes
\(p\in {\mathbf{Z}}\), and there are valuations
\(v_p: {\mathbf{Q}}^{\times}\to {\mathbf{Z}}\). Write
\(k = \operatorname{ff}({\mathcal{O}}_{X, \left\langle{0}\right\rangle}) = {\mathbf{Q}}\),
then e.g.~\(\operatorname{Div}(4/7) = 2[2] - 1[7]\).

\end{example}

\begin{example}[?]

Set \(X\coloneqq{\mathbf{A}}^2_{/ {{\mathbf{C}}}}\) and
\(f(x, y) = x/y\). What is \(v_{[V(x)]}(f)\)? Then
\(\operatorname{Div}(f) = [V(x)] - [V(y)]\), and
\begin{align*}
v_{ [V(x)] }: {\mathbf{C}}[x,y] \left[ { \scriptstyle { {\left\langle{x}\right\rangle}^{-1}}  } \right]  \to ?
.\end{align*}
So the answer is 1.

\end{example}

\begin{proposition}[?]

\(\operatorname{Div}(f)\) is well-defined, i.e.~\(v_f(Y) = 0\) for all
but finitely many \(Y\).

\end{proposition}

\begin{proof}[?]

Let \(f\in k^{\times}= \operatorname{ff}(A)\) for
\(\operatorname{Spec}A \subseteq X\) an affine open. Write \(f=a/b\) for
some \(b\in A\setminus\left\{{0}\right\}\), noting that \(A\) is a
domain since we assumed \(X\) integral. Passing to \(D(b)\), we can
assume \(f\) is a regular function on some affine open \(U\subseteq X\).
Since \(X\setminus U\) is a proper closed subset and \(X\) is
Noetherian, it contains only finitely many prime divisors -- each
irreducible component has \(\operatorname{codim}_X \geq 1\), and
conversely any prime divisor must be an irreducible component, and
Noetherian spaces have finitely many irreducible components. So it
suffices to show \(\operatorname{Div}(f)\) is well-defined for
\(f\in A\) when \(X=\operatorname{Spec}A\). Just use that
\(V(f) \subseteq \operatorname{Spec}A\) is a proper closed subset, the
same argument shows \(V(f)\) contains finitely many prime divisors.
Since \(f\in A\), we have \(f\in {\mathcal{O}}_{X, Y}\) and thus
\(v_Y(f) \geq 0\). Moreover if \(v_Y(f) = 0\) then \(Y \subseteq V(f)\)
-- use that \(f\in pA \left[ { \scriptstyle { {p}^{-1}} } \right]\) and
\(p A \left[ { \scriptstyle { {p}^{-1}} } \right] \cap A = p\) to get
\(f\in p\).

\end{proof}

\begin{definition}[Divisor class groups]

The \textbf{divisor class group} of \(X\) is defined as
\begin{align*}
 \operatorname{Cl} (X) = \operatorname{Div}(X)/\mathop{\mathrm{Prin}}\operatorname{Div}(X)
,\end{align*}
where
\(\mathop{\mathrm{Prin}}\operatorname{Div}(X) = \left\{{\operatorname{Div}f {~\mathrel{\Big\vert}~}f\in k^{\times}}\right\}\).
Since \(v_Y(fg) = v_Y(f) = v_Y(g)\),
\(\mathop{\mathrm{Prin}}\operatorname{Div}(X)\) forms a subgroup of
\(\operatorname{Div}(X)\).

\end{definition}

\begin{example}[?]

Consider \(X\coloneqq\operatorname{Spec}{\mathbf{Z}}\), then
\begin{align*}
\operatorname{Div}(X) = \bigoplus_{p \text{ prime}} {\mathbf{Z}}[p]
.\end{align*}

Then
\(\mathop{\mathrm{Prin}}\operatorname{Div}(\operatorname{Spec}{\mathbf{Z}}) = \operatorname{Div}(X)\)
by sending \(\sum n_p [p] \to \operatorname{Div}(\prod_p p^{n_p})\), so
\(\operatorname{Cl} (\operatorname{Spec}{\mathbf{Z}}) = 0\).

\end{example}

\begin{example}[?]

For \(K \in \mathsf{Number}\mathsf{Field}\) and \({\mathcal{O}}_K\) its
ring of integers, we can consider
\(\operatorname{Cl} ({\mathcal{O}}_K)\). For example,
\(\operatorname{Cl} (\operatorname{Spec}{\mathbf{Z}}[\sqrt{-5}] ) = C_2 = \left\langle{2, 1+\sqrt{-5}}\right\rangle\),
using the Dedekind domains admit unique factorization into prime ideals.

\end{example}

\begin{proposition}[?]

Let \(A\) be a Noetherian domain, then \(A\) is a UFD iff
\(\operatorname{Cl} (\operatorname{Spec}A) = 1\) is trivial.

\end{proposition}

\begin{proof}[?]

Use the lemma that \(A\) is a UFD \(\iff\) every height 1 prime ideal is
principal. Note that \(\left\langle{2, 1 + \sqrt{-5}}\right\rangle\) is
height 1 but not principal!

\(\implies\): Let \(Y \subseteq X = \operatorname{Spec}A\) be a prime
divisor, so \(Y = V(p)\) for \(p\) a height 1 prime ideal, so we can
write \(V(p) = V(f)\) for some \(f\). Then
\(\operatorname{Div}(f) = [Y]\), and any prime divisor is principal, and
now just use that \([Y]\) generate \(\operatorname{Div}(X)\).

\(\impliedby\): Suppose \(\operatorname{Cl} (X) = 0\) and let
\(Y \subseteq X\) be a prime divisor with \(\operatorname{Div}(f) = Y\)
for some \(f\in k^{\times}\). We want to show \(V(f) = Y\). If
\(\operatorname{Div}(f) = [Y]\), then for all \(Y' \subseteq X\) prime
divisors we have \(v_{Y'}(f) \geq 0\). By ring theory, \(f\in A\). If
\(V(f) = 1\) then
\(f\in pA \left[ { \scriptstyle { {p}^{-1}} } \right]\) for \(p=V(y)\),
so \(f\in pA \left[ { \scriptstyle { {p}^{-1}} } \right] \cap A\) and
thus \(f\in p\). The claim is that \(p = \left\langle{f}\right\rangle\)
-- suppose \(g\in p\), then \(v_{Y'}(g) \geq 0\) and \(v_Y(g) \geq 1\)
implies \(v_{Y'}(g/f) \geq 0\) for all \(Y'\). So \(g/f\in A\), making
\(g\in \left\langle{f}\right\rangle\).

\end{proof}

\begin{remark}

Use that valuations are non-negative on prime divisors and that the
valuations are either 0 or 1.

\end{remark}

\hypertarget{wednesday-december-01}{%
\section{Wednesday, December 01}\label{wednesday-december-01}}

\begin{remark}

Recall:

\begin{itemize}
\item
  \(X\in{\mathsf{Sch}}\) is Noetherian, integral, separated, regular in
  codimension 1,
\item
  \(Y \subseteq X\) a prime divisor is an integral closed codimension 1
  subscheme,
\item
  \(\operatorname{Div}(X) = {\mathbf{Z}} { \left[ \scriptstyle { \left\{{\text{Prime divisors}}\right\} } \right] }\),
\item
  \(\mathop{\mathrm{Prin}}\operatorname{Div}(X) = \left\{{\operatorname{Div}(f) {~\mathrel{\Big\vert}~}f\in k^{\times}, \text{ the rational functions on } X}\right\}\)
  where \(\operatorname{Div}(f) \coloneqq\sum_{Y} v_Y(f) [Y]\),
\item
  \(\operatorname{Cl} (X) \coloneqq\operatorname{Div}(X) / \mathop{\mathrm{Prin}}\operatorname{Div}(X)\).
\end{itemize}

We proved that if \(X=\operatorname{Spec}A\),
\begin{align*}
 \operatorname{Cl} (X) = 1 \iff A \text{ is a UFD}
.\end{align*}
This shows that any height 1 prime ideal contained in \(A\) is
principal. The key commutative algebra fact was
\begin{align*}
\bigcap_{\operatorname{ht}(p) = 1} A \left[ { \scriptstyle { {p}^{-1}}  } \right]  = A
.\end{align*}
.

\end{remark}

\begin{remark}

A quick review of why \(\operatorname{Cl} (X) = 1 \implies\) every
\(p\in \operatorname{Spec}A\) with \(\operatorname{ht}(p) = 1\) is
principal. Let \(Y = V(p)\), then \([Y]\in \operatorname{Cl} (X) = 1\)
means that \([Y] = \operatorname{Div}(\phi)\) for some
\(\phi \in k^{\times}\). Since \(v_{Y'} \geq 0\) for all \(Y'\) (since
they're just zero for \(Y\neq Y'\)) implies that \(f\in A\), and the
claim is that \(\left\langle{\phi}\right\rangle = p\). Taking
\(f\in p\), then \(\operatorname{Div}(f) = [Y] + {\varepsilon}\) where
\({\varepsilon}\geq 0\) is effective. Then
\(\operatorname{Div}(f/\phi)\geq 0\) is effective,
i.e.~\(f/\phi \in A \left[ { \scriptstyle { {p'}^{-1}} } \right]\) for
all \(p'\). But then
\(f/\phi\in \bigcap_{\operatorname{ht}(p) = 1} A \left[ { \scriptstyle { {p}^{-1}} } \right] = A\),
so \(f\in \left\langle{\phi}\right\rangle\). So
\(p = \left\langle{\phi}\right\rangle\).

\end{remark}

\begin{proof}[of the commutative algebra fact]

To show
\(\bigcap_{\operatorname{ht}(p)=1} A \left[ { \scriptstyle { {p}^{-1}} } \right] = A\),
let \(a/b\in k^{\times}\), then
\(\dim A/\left\langle{b}\right\rangle = \dim A - 1\). Why? Let
\(\mkern 1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu_0 \subset \cdots \subset A/\left\langle{b}\right\rangle\),
then
\(q^{-1}(\mkern 1.5mu\overline{\mkern-1.5mup\mkern-1.5mu}\mkern 1.5mu_0) = p \ni b\).

\end{proof}

\begin{remark}

The geometric analog: if \(X\coloneqq\operatorname{Spec}A\) and \(V(p)\)
is a variety, intersecting with a hyperplane yields a codimension 1
locus (in nice cases).

\end{remark}

\begin{example}[Affine space]

For \(k\in \mathsf{Field}\) not necessarily algebraically closed,
\begin{align*}
 \operatorname{Cl} ({\mathbf{A}}^n_{/ {k}} ) = 1
.\end{align*}

Proof: \(k\) is a UFD, so \(k[x_1, \cdots, x_{n}]\) is a UFD, so apply
the proposition.

\end{example}

\begin{example}[Number fields]

For \(K\in \mathsf{Number}\mathsf{Field}\) and \({\mathcal{O}}_K\) its
ring of integers,
\begin{align*}
 \operatorname{Cl} (\operatorname{Spec}{\mathcal{O}}_K) = 1 \iff {\mathcal{O}}_K \text{ is a UFD}
,\end{align*}
and this coincides with \(\operatorname{Cl} ({\mathcal{O}}_K)\) from
number theory.

\end{example}

\begin{example}[A geometric non-example]

If \(X\coloneqq V(xy-z^2) \subseteq {\mathbf{A}}^3_{/ {k}}\), then
\(\operatorname{Cl} (X) \neq 1\) since \(xy = zz\) in \(A\), exhibiting
failure of unique factorization. How to find an irreducible subscheme:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{36pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-12-01_12-06.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

We use that
\(\left\langle{x, z}\right\rangle \in k[x,y,z] / \left\langle{xy-z^2}\right\rangle\)
is not principal. Note that if \(Y=V(p)\) then \(2[Y] = 0\) in
\(\operatorname{Cl} (X)\): show that
\(x\in A \left[ { \scriptstyle { {p}^{-1}} } \right]\), then
\(v_p(x) = 2\).

\end{example}

\hypertarget{friday-december-03}{%
\section{Friday, December 03}\label{friday-december-03}}

\begin{theorem}[?]

For \(X\coloneqq{\mathbf{P}}^n_{/ {k}}\) and
\(D\in \operatorname{Div}(X)\), define
\begin{align*}
\deg D \coloneqq\sum n_i \deg Y_i && \text{where } D = \sum n_i [Y_i]
.\end{align*}
Let \(H \coloneqq\left\{{x_0 = 0}\right\}\) by a hyperplane, then

\begin{itemize}
\tightlist
\item
  \(D\sim \deg(D) H\)
\item
  \(f\in k^{\times}\implies \deg(f) = 0\),
\item
  \(\deg: \operatorname{Cl} (X)\to {\mathbf{Z}}\) is an isomorphism.
\end{itemize}

\end{theorem}

\begin{proof}[?]

Missed, see Hartshorne.

\end{proof}

\begin{proposition}[?]

For \(Z \subseteq X\) proper and closed with
\(U \coloneqq X\setminus Z\), if
\(\operatorname{codim}Z \mathop{\mathrm{=}}2\), then
\(\operatorname{Cl} (X) \cong \operatorname{Cl} (U)\). If \(Z\) is
irreducible and \(\operatorname{codim}Z = 1\), there is an exact
sequence
\({\mathbf{Z}}\xrightarrow{f} \operatorname{Cl} (X) \to \operatorname{Cl} (U) \to 0\)
where \(f(1) = [Z]\).

\end{proposition}

\begin{remark}

Note that this \({\mathbf{Z}}\to \operatorname{Cl} (X)\) isn't injective
in general: take \(X\coloneqq{\mathbf{A}}^n\) so
\(\operatorname{Cl} (X) = 1\).

\end{remark}

\begin{proof}[?]

Define a map
\begin{align*}
\phi: \operatorname{Div}(X) &\to \operatorname{Div}(U) \\
Y &\mapsto
\begin{cases}
Y \cap U &  Y \cap U\neq \emptyset
\\
0 & Y \cap U = \emptyset.
\end{cases}
\end{align*}
Then \(\phi\) is \({\mathbf{Z}}{\hbox{-}}\)linear, and
\(k^{\times}(X) \cong k^{\times}(U)\), and descends to a map
\(\operatorname{Cl} (X) \to \operatorname{Cl} (U)\). Moreover
\(\ker \phi\) is generated by prime divisors contained in \(Z\), so if
\(\operatorname{codim}Z\geq 2\) this is empty and we have an
isomorphism. Otherwise if \(\operatorname{codim}Z = 1\) with \(Z\)
irreducible, then the only prime divisors in \(Z\) is \(Z\) itself, so
\(Z\) generates \(\ker \phi\).

\end{proof}

\begin{example}[?]

Let \(Z \subseteq {\mathbf{P}}^2\) be an irreducible degree \(d\) curve
and let \(U = {\mathbf{P}}^2_{/ {k}} \setminus\left\{{0}\right\}\). Then
\(\operatorname{Cl} (U) = C_d\) is cyclic of order \(d\). We have
\begin{align*}
{\mathbf{Z}}&\to  \operatorname{Cl} ({\mathbf{P}}^2_{/ {k}} ) \cong {\mathbf{Z}}\to  \operatorname{Cl} (U) \to 0 \\
1 &\mapsto [Z] \cong \deg d
.\end{align*}

\end{example}

\hypertarget{divisors-on-curves}{%
\subsection{Divisors on Curves}\label{divisors-on-curves}}

\begin{definition}[Curve]

Let \(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\),
then a \textbf{curve} \(X\in {\mathsf{Sch}}_{/ {k}}\) is an integral (so
reduced) separated of finite type of dimension 1. We say \(X\) is
\textbf{complete} if \(X\to \operatorname{Spec}k\) is proper, and
\textbf{smooth} if \(X\) is regular (equivalently regular in codimension
1).

\end{definition}

\begin{example}[?]

\(X\coloneqq V(f) \subseteq {\mathbf{P}}^2_{/ {k}}\) where \(f\) is
irreducible. This is complete since it is closed in
\({\mathbf{P}}^2_{/ {k}}\), which is proper.

\end{example}

\begin{proposition}[?]

If \(f:X\to Y\) is a morphism of curves and \(X\) is complete and
nonsingular, then \(\operatorname{im}f\) is either a point or all of
\(Y\). If \(\operatorname{im}= Y\), then \(f\) is finite.

\end{proposition}

\begin{proof}[?]

\(X\) proper implies \(f(X)\) is closed in \(Y\), and \(X\) irreducible
implies \(f(X)\) is irreducible. Since \(Y\) is irreducible, this forces
\(f(X) = {\operatorname{pt}}\) or \(Y\). Let \(V \subseteq Y\) be an
affine open and \(U \coloneqq f^{-1}(V)\), the claim is that \(U\) is
affine and the pullback
\(f^*: {\mathcal{O}}_Y(V) \to {\mathcal{O}}_X(U)\) is a module-finite
extension. We have a map on function fields \(f^*:k(X) \to k(Y)\), and
since \(\dim X, \dim Y =1\), these are fields of transcendence degree 1
over \(k\). Therefore \(f^*\) is a finite extension of fields (use
Noether normalization), and \({\mathcal{O}}_Y(V) \subseteq k(Y)\). We
can write \({\mathcal{O}}_Y(V) = \cap_{p\in V} {\mathcal{O}}_{Y, p}\).
The following gives module-finiteness:

\begin{claim}

The integral closure of \({\mathcal{O}}_Y(V)\) in \(k(X)\) is
\({\mathcal{O}}_X(U)\).

\end{claim}

To see that \(U\) is affine: exercise!

\end{proof}

\hypertarget{curves-and-divisors-ramification-and-degree-monday-december-06}{%
\section{Curves and Divisors: Ramification and Degree (Monday, December
06)}\label{curves-and-divisors-ramification-and-degree-monday-december-06}}

\begin{remark}

Recall that we defined a curve as a 1-dimensional integral separated
scheme of finite type over an algebraically closed field. Here
nonsingular corresponds to regular, and complete corresponds to proper.
We were proving the following:

\end{remark}

\begin{proposition}[?]

If \(f:X\to Y\) is a morphism of curves with \(X\) complete and
nonsingular, then

\begin{itemize}
\tightlist
\item
  \(f(X) = {\operatorname{pt}}\) or all of \(Y\)
\item
  If \(f(X) = Y\), then \(f\) is finite and \(f^*: K(Y) \to K(X)\) is a
  finite extension of fields.
\end{itemize}

\end{proposition}

\begin{remark}

If \(\operatorname{Spec}B \hookrightarrow Y\) is an affine open, then
defining \(A\) as the integral closure of \(B\) in \(K(X)\) we get
\(\operatorname{Spec}A\hookrightarrow X\) and
\(\operatorname{Spec}A = f^{-1}(\operatorname{Spec}B)\). This relies on
\(X\) being complete and nonsingular -- prove this as an exercise.

\end{remark}

\begin{definition}[Degree of a surjective morphism of curves]

Let \(f:X\to Y\) as before and suppose \(f\) is surjective. The
\textbf{degree} of \(f\) is defined as
\begin{align*}
\deg f \coloneqq[K(X): K(Y)]
.\end{align*}

\end{definition}

\begin{remark}

Define a pullback of divisors
\begin{align*}
f^*: \operatorname{Div}(Y) &\to \operatorname{Div}(X)
,\end{align*}
defined on closed points (and extended
\({\mathbf{Z}}{\hbox{-}}\)linearly) as follows: let \(q\in Y\) be a
closed point, then since \(Y\) regular in codimension 1 there exists a
generator \(t\in K(Y)\) such that \(t\in {\mathcal{O}}_{Y, q}\) and
\({\mathfrak{m}}_q = \left\langle{t}\right\rangle\). We'll call \(t\) a
\textbf{local parameter} at \(q\). Take an open \(U\ni q\) where \(t\)
is regular and \(V(t) = \left\{{q}\right\}\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-12-06_11-49.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Write \(f^* t\in {\mathcal{O}}_X(f^{-1}(U))\), then
\begin{align*}
\operatorname{Div}_{f^{-1}(U)} f^*(t) \coloneqq f^*[q] = \sum_{f(p) = q} v_p(f^* t)[p]
.\end{align*}

\end{remark}

\begin{example}[?]

Consider \(C = \left\{{y^2 = x^3+ax+b}\right\}\) and
\(X\coloneqq V(C)\), and take the projection
\begin{align*}
X &\to {\mathbf{A}}^1_{/ {k}} \\
(x, y) &\mapsto x
.\end{align*}

Assume the discriminant \(\Delta(a, b)\neq 0\) so \(X\) is nonsingular
and the roots of \(f(x) \coloneqq x^3 + ax + b\) are distinct:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-12-06_11-56.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Consider \((a, b) = (0, -1)\) so \(y^2 = x^3-1\) and work over
\({\mathbf{C}}\). Check that \((x, y) = (0, \pm i)\) are solutions, and
we can write
\begin{align*}
f^* [0] = \sum_{f(p) = 0} v_p(f^* x) = 1\cdot [(0, i)] + 1\cdot [(0, -i)]
,\end{align*}
since
\({\mathfrak{m}}_p = \left\langle{ x, y\mp i}\right\rangle = \left\langle{x}\right\rangle\).
Similarly,
\begin{align*}
f^*[1] = v_{(1, 0)}\left\langle{x-1}\right\rangle = 2[(1, 0)]
,\end{align*}
so the function \(x\) is not a local coordinate at 1, but \(y\) is.
Consider
\(\qty{ k[x, y] \over \left\langle{y^2-x^3-+1}\right\rangle } \left[ { \scriptstyle { {\left\langle{x-1, y}\right\rangle}^{-1}} } \right]\);
then \({\mathfrak{m}}= \left\langle{x-1, y}\right\rangle\) and we can
factor \(y^2 = x^3-1 = (x-1)(x^2 + x +1)\), and we can invert to write
\(x-1 = {y^2\over x^2+x+1} \in {\mathfrak{m}}^2\setminus{\mathfrak{m}}^3\).

\end{example}

\begin{remark}

The punchline: even though the size of the set-theoretic fibers changed,
in both cases we pulled back degree 1 divisors and got degree 2
divisors, and this is evidently a 2-to-1 cover. Note that this example
wasn't complete, but we can take the projective closure by homogenizing
to get \(V(y^2z = ^3 + axz^2 + bz^3)\), and we can extend our map
\(\pi:X\to {\mathbf{A}}^1\) to
\(\tilde \pi: \tilde X\to {\mathbf{P}}^1\) by mapping the new point to
\(\infty\).

\end{remark}

\begin{definition}[Ramification, branching]

Let \(f:X\to Y\) be a morphism of smooth complete curves with
\(f(X) = Y\). A \textbf{ramification point} of \(f\) is a point
\(p\in X\) where \(e_p(f) \coloneqq v_p(f^* t) >1\) for \(t\) a local
parameter at \(q=f(p)\). Such a point \(q\) is said to be a
\textbf{branch point}. The \textbf{ramification divisor} of \(f\) is
defined as
\begin{align*}
R_f \coloneqq\sum_{p\in X} \qty{e_p(f) - 1}[p]
.\end{align*}

\end{definition}

\begin{remark}

This is a finite sum: show that for all but finitely many points (i.e.~a
Zariski open), the pullback of a local parameter will again be a local
parameter on the cover, potentially after subtracting a constant to
shift the image to 0. More precisely, for any \(f\in K(X)\), \(f-f(p)\)
will be a local parameter at \(p\) for a Zariski open.

\end{remark}

\begin{proposition}[?]

Let \(f:X\to Y\) be a nonconstant morphism of smooth complete curves,
then
\begin{align*}
\deg f^* D = \deg f \cdot \deg D
,\end{align*}
where \(\deg\qty{\sum n_p [p]} = \sum n_p\).

\end{proposition}

\begin{proof}[?]

The 30s version: write \(\operatorname{Spec}V \subset Y\) with \(A\)
defined as the integral closure of \(B\) in \(K(X)\). Then \(B\to A\) is
module-finite of dimension \(\deg f = [K(X) : K(Y)]\). Taking there is
an induced map on the local ring
\begin{align*}
B \left[ { \scriptstyle { {q}^{-1}}  } \right]  &\to \bigoplus _{f(p) = q} A \left[ { \scriptstyle { {p}^{-1}}  } \right]  \\
t &\mapsto \bigoplus f^* t
.\end{align*}
Then
\begin{align*}
\dim\qty{\bigoplus A \left[ { \scriptstyle { {p}^{-1}}  } \right]  / t \bigoplus A \left[ { \scriptstyle { {p}^{-1}}  } \right]  / B \left[ { \scriptstyle { {q}^{-1}}  } \right]  / tB \left[ { \scriptstyle { {q}^{-1}}  } \right] } = 
\dim\qty{ \bigoplus _{f(p) = q} k[t] / t^{e_p(f)} / k} = \deg f
.\end{align*}

Note that this uses CRT:
\(A/tA \cong \bigoplus _{f(p) = q} A \left[ { \scriptstyle { {p}^{-1}} } \right] / tA \left[ { \scriptstyle { {p}^{-1}} } \right]\).

\end{proof}

\hypertarget{tuesday-december-07}{%
\section{Tuesday, December 07}\label{tuesday-december-07}}

\begin{proposition}[?]

For \(f\in K(C)^{\times}\) for \(C\) a smooth complete curve,
\(\deg(\operatorname{Div}f) = 0\).

\end{proposition}

\begin{proof}[?]

If \(f\) is constant this is trivial, so assume not. Define
\begin{align*}
U_1 &\coloneqq\left\{{p\in C {~\mathrel{\Big\vert}~}v_p(F) \geq 0 }\right\} \\
U_2 &\coloneqq\left\{{p\in C {~\mathrel{\Big\vert}~}v_p(F) \leq 0 }\right\}
.\end{align*}
Then \(U_1 \cap U_2\) is precisely the set of closed points of \(C\).
Suppose \(f\) is regular on \(U_1\), so \(1/f\) is regular on \(U_2\).
Define a map \(\widehat{f}: C\to {\mathbf{P}}^1\) by writing
\({\mathbf{P}}^1 = {\mathbf{A}}_1 \cup{\mathbf{A}}_1\) and defining
\({ \left.{{f}} \right|_{{U_i}} }: U_i\to {\mathbf{A}}^1\) to map into
the \(i\)th factor. Note that
\({ \left.{{\widehat{f}}} \right|_{{U_2}} } = 1/f\). Then
\begin{align*}
\operatorname{Div}(f) = \sum_{p\in C} v_p(f) [p] = f^*\qty{ [0] - [\infty]} \coloneqq f^*(D)
.\end{align*}

Then \(\deg(\operatorname{Div}(f)) = \deg f \cdot \deg (D) = 0\) since
\(\deg(D) = 0\), noting that \(\deg f\) is the degree of the
corresponding field extension.

\end{proof}

\begin{corollary}[?]

For a smooth complete curve, the degree map descends to a well-defined
map:
\begin{align*}
\deg:  \operatorname{Cl} (C) &\to {\mathbf{Z}}\\
\sum n_p p &\mapsto \sum n_p
.\end{align*}

\end{corollary}

\begin{definition}[$\Cl_0$]

Define
\begin{align*}
 \operatorname{Cl} _0(C) \coloneqq\ker\qty{ \operatorname{Cl} (C) \xrightarrow{\deg} {\mathbf{Z}}}
.\end{align*}

\end{definition}

\begin{definition}[Elliptic curve]

An \textbf{elliptic curve} \(E\) is a smooth complete genus 1 curve over
\(k\) with a distinguished closed point \(0\in E\) called the origin.
Complex analytically, \(E = {\mathbf{C}}/ \Lambda\) where
\(\Lambda \subset {\mathbf{C}}\) is an integral lattice with
\(\Lambda \otimes_{\mathbf{Z}}{\mathbf{R}}\cong {\mathbf{C}}\), so the
basis vectors are independent.

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-12-07_11-49.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{definition}

\begin{example}[?]

The cubic in \({\mathbf{P}}^3_{/ {k}}\) defined by
\begin{align*}
C \coloneqq V(zy^2 = x^3 + axz^2 + bz^3), \quad 0 \coloneqq[0: 1 : 0]
.\end{align*}

\end{example}

\begin{definition}[Weierstrass $\wp$ function]

Define a complex analytic function
\begin{align*}
\wp: {\mathbf{C}}&\to {\mathbf{P}}^1 \\
z &\mapsto {1\over z^2} + \sum_{\lambda\in \Lambda\setminus\left\{{0}\right\}} {1\over (z-\lambda)^2} - {1\over \lambda^2} = z^{-2} + { \mathsf{O}} (z^2)
.\end{align*}

\end{definition}

\begin{remark}

Note that

\begin{itemize}
\tightlist
\item
  \(\wp(z + \lambda) = \wp(z)\) for all \(\lambda \in \Lambda\)
\item
  \(\wp: E\to {\mathbf{P}}^1\), so \(\wp\in {\mathbf{C}}(E)^{\times}\)
\item
  \(\wp'(z) = \sum_{\lambda\in \Lambda} {-2 \over (z - \lambda)^3}\in {\mathbf{C}}(E)^{\times}= -{2\over z^3} + { \mathsf{O}} (z)\).
\item
  \(\wp'(z)^2 = {4\over z^6} + {c_1\over z^2} + { \mathsf{O}} (1)\)
\item
  \(\wp(z)^3 = z^{-6} + c_2z^{-2} + { \mathsf{O}} (1)\)
\end{itemize}

So there is a relation
\begin{align*}
F: \quad \wp'(z)^2 = 4\wp(z)^3 + G_2(\Lambda)\wp(z) + G_3( \Lambda) + { \mathsf{O}} (z)
.\end{align*}

Note that this cancels the poles at the lattice points, making it a
bounded holomorphic function and thus constant. Since it's
\({ \mathsf{O}} (z)\), this forces it to be zero. So define a map
\begin{align*}
E &\to {\mathbf{P}}^2_{/ {{\mathbf{C}}}} \\
z &\mapsto [\wp(z): \wp'(z): 1]
,\end{align*}
and note that \(0\mapsto [0:1:0]\) so this factors through
\(V(zy^2 = 4x^3 + G_2( \Lambda)xz^2 + G_3(\Lambda) z^3 )\)
biholomorphically, using that \(\deg \wp, \wp' = 2,3\) to get
injectivity. This makes \(E\) an algebraic variety.

\end{remark}

\begin{remark}

An aside: suppose \(f: C_1\to C_2\) is a degree 1 holomorphic map of
compact complex curves. Then \(f'=0\) at only finitely many points, so
\(f\) is invertible on an open set and \(f^{-1}\) extends continuously
to \(C_2\). Now use the Riemann removable singularity theorem: extending
a holomorphic function continuously over a puncture implies that the new
function is holomorphic.

\end{remark}

\begin{remark}

Why is this algebraic structure unique? Use an overpowered theorem:
Serre's GAGA, i.e.~there is a unique variety structure on a compact
complex manifold over \({\mathbf{C}}\). In our case, it suffices to show
\(\wp(z-c), \wp'(z-c)\in {\mathbf{C}}(\wp, \wp')\). In fact, the
rational functions are given by
\(K(X) = \operatorname{ff}\qty{{\mathbf{C}}[\wp, \wp']/\left\langle{\text{a cubic}}\right\rangle}\).

\end{remark}

\begin{remark}

Write \(E\) for the vanishing locus of the cubic \(F\) above, and
consider a map
\begin{align*}
{\mathbf{C}}/ \Lambda  { \, \xrightarrow{\sim}\, }E \subseteq {\mathbf{P}}^2_{/ {{\mathbf{C}}}} 
.\end{align*}
Consider a line \(L \subseteq {\mathbf{P}}^2_{/ {{\mathbf{C}}}}\), then
\(L \cap C = p+q+r = \operatorname{Div}(L)\) generically. We claim
\(p+q+r \equiv 0 \operatorname{mod}\Lambda\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-12-07_12-11.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

To prove this: write \(L_0 \coloneqq V(z)\), so
\(\operatorname{Div}(L_0) = 3[0]\), and consider \(L_0 \cap V(F)\). We
have \(\operatorname{Div}(L/L_0) = [p] + [q] +[r] - 3[0]\).

\begin{claim}

If \(f\in K(E)^{\times}\) and \(\operatorname{Div}f = \sum n_p [p]\)
then \(\sum n_p p \equiv 0 \operatorname{mod}\Lambda\) after taking
these as honest points in \({\mathbf{C}}\).

\end{claim}

To prove this, do some kind of contour integral over the fundamental
domain and use lattice periodicity of \(f\). This yields
\(p+q+r-3\cdot 0 \in \Lambda\), so given any two points we can solve for
the third.

\end{remark}

\begin{remark}

This can be used as a reduction algorithm:
\begin{align*}
[p_1] + [p_2] + 2[p_3] - 4[p_4] 
&= [p_1 + p_2] + [0] + 2[p_3] - 4[p_4] \in \operatorname{Div}^0(E) \\
&= 2[p_3 - p_4] - 2[p_4] - 2[0] \\
\cdots &= [p] - [0]
\implies 
,\end{align*}
so there is an isomorphism
\begin{align*}
E &\to  \operatorname{Cl} ^0(E) \\
p &\mapsto [p] - [0]
.\end{align*}

\end{remark}

\hypertarget{appendix}{%
\section{Appendix}\label{appendix}}

\hypertarget{notation}{%
\subsection{Notation}\label{notation}}

\begin{itemize}
\item
  \(\operatorname{ff}(R)\) denotes the fraction field (or field of
  quotients) of \(R\).
\item
  \(R \left[ { \scriptstyle { {S}^{-1}} } \right]\) is the ring \(R\)
  localized at the multiplicative set \(S \subseteq R\), i.e.~the subset
  of the fraction field \(\operatorname{ff}(R)\) with denominators only
  in \(S\). This differs from the usual notation \(S^{-1}R\).
\item
  \({ {\mathbf{Z}}_{\widehat{p}} }\) is the \(p{\hbox{-}}\)adic
  integers, i.e.~the ring \(R={\mathbf{Z}}\) completed at the ideal
  \(\left\langle{p}\right\rangle\). This differs from the usual notation
  \({\mathbf{Z}}_p\).
\item
  \(R{\llbracket t \rrbracket }\) is the topological ring of formal
  power series in \(t\), i.e.~infinite sums \(\sum_{i\geq 0} r_i t^i\)
  with the \(t{\hbox{-}}\)adic topology.
\item
  \(R{\left(\left( t \right)\right) }\) is the topological ring of
  formal Laurent series, i.e.~half-infinite sums
  \(\sum_{i\geq -N} r_i t^i\).

  \begin{itemize}
  \tightlist
  \item
    Note that
    \(R {\left(\left( t \right)\right) } = R{\llbracket t \rrbracket } \left[ { \scriptstyle { {S}^{-1}} } \right]\)
    where \(S = \left\{{1, x, x^2,\cdots}\right\}\). If \(R\) is a
    field,
    \(R{\left(\left( t \right)\right) } = \operatorname{ff}(R{\llbracket t \rrbracket })\).
  \end{itemize}
\end{itemize}

\hypertarget{facts}{%
\subsection{Facts}\label{facts}}

\begin{remark}

Some useful facts:

\begin{itemize}
\tightlist
\item
  The equalizer diagram for a sheaf \({\mathcal{F}}\):
\end{itemize}

\begin{center}
\begin{tikzcd}
    \mathsf{\emptyset} 
    \ar[r]
    &
    {\mathcal{F}}(U)
    \stackar{3}[r]
    &
    \displaystyle\prod_{i\in I} {\mathcal{F}}(U_{i})
    \stackar{5}[r]
    &   
    \displaystyle\prod_{i < j \in I} {\mathcal{F}}(U_{ij})
    \cdots
\end{tikzcd}
\end{center}

\begin{itemize}
\tightlist
\item
  The inverse image / pushforward (``direct image'') adjunction:
\end{itemize}

\begin{align*}
\adjunction{f_*}{f^{-1}}{{\mathsf{Sh}}(X)}{{\mathsf{Sh}}(Y)} \implies {\mathsf{Sh}}(X)(f^{-1}{\mathcal{G}}, {\mathcal{F}})  { \, \xrightarrow{\sim}\, }{\mathsf{Sh}}(Y)({\mathcal{G}}, f_* {\mathcal{F}})
.\end{align*}

\end{remark}

\addsec{ToDos}
\listoftodos[List of Todos]
\cleardoublepage

% Hook into amsthm environments to list them.
\addsec{Definitions}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={definition}, numwidth=3.5em]
\cleardoublepage

\addsec{Theorems}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={theorem,proposition}, numwidth=3.5em]
\cleardoublepage

\addsec{Exercises}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={exercise}, numwidth=3.5em]
\cleardoublepage

\addsec{Figures}
\listoffigures
\cleardoublepage



\newpage
\printbibliography[title=Bibliography]


\end{document}
