\input{"preamble.tex"}

\addbibresource{AlgebraicGeometry.bib}

\let\Begin\begin
\let\End\end
\newcommand\wrapenv[1]{#1}

\makeatletter
\def\ScaleWidthIfNeeded{%
 \ifdim\Gin@nat@width>\linewidth
    \linewidth
  \else
    \Gin@nat@width
  \fi
}
\def\ScaleHeightIfNeeded{%
  \ifdim\Gin@nat@height>0.9\textheight
    0.9\textheight
  \else
    \Gin@nat@width
  \fi
}
\makeatother

\setkeys{Gin}{width=\ScaleWidthIfNeeded,height=\ScaleHeightIfNeeded,keepaspectratio}%

\title{
\rule{\linewidth}{1pt} \\
\textbf{
    Algebraic Geometry
  }
    \\ {\normalsize University of Georgia, Fall 2020} \\
  \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 Algebraic Geometry taught by Philip Engel at
the University of Georgia in Fall 2020. As such, any errors or
inaccuracies are almost certainly my own. } } \\
    \end{flushleft}
    \end{minipage}
    \hfill
    \begin{minipage}{.65\linewidth}
    \end{minipage}
  }







\begin{document}

\date{}
\author{D. Zack Garza}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
  \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2021-01-06 }
\end{flushleft}


\newpage

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

\hypertarget{prologue}{%
\section*{Prologue}\label{prologue}}
\addcontentsline{toc}{section}{Prologue}

\hypertarget{references}{%
\subsection{References}\label{references}}

\begin{itemize}
\tightlist
\item
  Gathmann's Algebraic Geometry notes\autocite{AndreasGathmann515}.
\end{itemize}

\hypertarget{notation}{%
\subsection{Notation}\label{notation}}

\begin{itemize}
\tightlist
\item
  If a property \(P\) is said to hold \textbf{locally}, this means that
  for every point \(p\) there is a neighborhood \(U_p \ni p\) such that
  \(P\) holds on \(U_p\).
\end{itemize}

\begin{longtable}[]{@{}ll@{}}
\toprule
\begin{minipage}[b]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
Notation\strut
\end{minipage} &
\begin{minipage}[b]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Definition\strut
\end{minipage}\tabularnewline
\midrule
\endhead
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(k[\mathbf{x}] = k[x_1, \cdots, x_n]\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Polynomial ring in \(n\) indeterminates\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(k(\mathbf{x}) = k(x_1, \cdots, x_n)\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Rational function field in \(n\) indeterminates\\
\(k(\mathbf{x}) = \left\{{f(\mathbf{x}) = p(\mathbf{x})/q(\mathbf{x}), {~\mathrel{\Big|}~}p,q, \in k[x_1, \cdots, x_{n}]}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(\mathcal{U} \rightrightarrows X\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
An open cover\\
\(\mathcal{U} = \left\{{U_j {~\mathrel{\Big|}~}j\in J}\right\}, X = \bigcup_{j\in J}U_j\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(\Delta_X\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
The diagonal\\
\(\Delta_X \coloneqq\left\{{(x, x) {~\mathrel{\Big|}~}x\in X}\right\} \subseteq X\times X\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\({\mathbb{A}}^n_{/k}\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Affine \(n{\hbox{-}}\)space\\
\({\mathbb{A}}^n_{/k} \coloneqq\left\{{\mathbf{a} = {\left[ {a_1, \cdots, a_n} \right]} {~\mathrel{\Big|}~}a_j \in k}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\({\mathbb{P}}^n_{/k}\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Projective \(n{\hbox{-}}\)space\\
\({\mathbb{P}}^n_{/k} \coloneqq\qty{k^n\setminus\left\{{0}\right\}}/x\sim \lambda x\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(V(J), V_a(J)\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Variety associated to an ideal
\(J {~\trianglelefteq~}k[x_1, \cdots, x_{n}]\)\\
\(V_a(J) \coloneqq\left\{{\mathbf{x}\in {\mathbb{A}}^n {~\mathrel{\Big|}~}f(\mathbf{x}) = 0,\, \forall f\in J}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(I(S), I_a(S)\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Ideal associated to a subset \(S \subseteq {\mathbb{A}}^n_{k}\)\\
\(I_a(S) \coloneqq\left\{{f\in k[x_1, \cdots, x_{n}] {~\mathrel{\Big|}~}f(\mathbf{x}) = 0\, \forall \mathbf{x}\in X}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(A(X)\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Coordinate ring of a variety\\
\(A(X) \coloneqq k[x_1, \cdots, x_{n}]/I(X)\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(V_p(J)\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Projective variety of an ideal\\
\(V_p(J) \coloneqq\left\{{\mathbf{x} \in {\mathbb{P}}^n_{/k} {~\mathrel{\Big|}~}f(\mathbf{x}) = 0,\, \forall f\in J}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(I_p(S)\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Projective ideal (?)\\
\(I_p(S) \coloneqq\left\{{f\in k[x_1, \cdots, x_{n}] {~\mathrel{\Big|}~}f \text{ is homogeneous and } f(x) = 0\, \forall x\in S}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(S(X)\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Projective coordinate ring\\
\(S(X) \coloneqq k[x_1, \cdots, x_{n}]/ I_p(X)\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(f^h\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Homogenization\\
\(f^h \coloneqq x_0^{\deg f} f\qty{{x_1 \over x_0}, \cdots, {x_n \over x_0}}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(f^i\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Dehomogenization\\
\(f^i \coloneqq f(1, x_1, \cdots, x_n)\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(J^h\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Homogenization of an ideal\\
\(J^h \coloneqq\left\{{f^j {~\mathrel{\Big|}~}f\in J}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Projective closure of a subset\\
\(\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu \coloneqq V_p(J^h) \coloneqq\left\{{\mathbf{x} \in {\mathbb{P}}^n {~\mathrel{\Big|}~}f^h(\mathbf{x}) = 0\, \forall f\in X}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(D(f)\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Distinguished open set\\
\(D(f) \coloneqq V(f)^c = \left\{{x\in {\mathbb{A}}^n {~\mathrel{\Big|}~}f(x) \neq 0}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(\mathcal{F}\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Presheaf or a sheaf\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(f\in \mathcal{F}(U)\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Section of a presheaf or sheaf\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(\underline{\mathbf{S}}\) where \(S\) is a set\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Locally constant functions valued in \(S\)\\
~\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(\mathcal{F}_p\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Stalk of a sheaf\\
\(\mathcal{F}_p \coloneqq\left\{{(U, \phi) {~\mathrel{\Big|}~}p\in U \text{ open },\, \phi \in \mathcal{F}(U)}\right\}/\sim\)\\
where
\((U, \phi) \sim (U', \phi') \iff \exists p\in W \subset U\cap U' \text{ s.t. } { \left.{{\phi}} \right|_{{W}} } = { \left.{{\phi'}} \right|_{{W}} }\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\(f\in \mathcal{F}_p\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Germs at \(p\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\({\mathcal{O}}_X\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Structure sheaf\\
\({\mathcal{O}}_X \coloneqq\left\{{f:U\to k {~\mathrel{\Big|}~}U \subseteq X \text{ is open}, f \in k(\mathbf{x}) \text{ locally}}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\({\mathcal{O}}_X(U)\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Regular functions on \(U\)\\
\({\mathcal{O}}_X(U) \coloneqq\left\{{f:U\to k {~\mathrel{\Big|}~}f \in k(\mathbf{x}) \text{ locally}}\right\}\)\\
\strut
\end{minipage}\tabularnewline
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.25}}\raggedright
\({\mathcal{O}}_{X, p}\)\strut
\end{minipage} &
\begin{minipage}[t]{(\columnwidth - 1\tabcolsep) * \real{0.74}}\raggedright
Germs of Regular functions?\\
\strut
\end{minipage}\tabularnewline
\bottomrule
\end{longtable}

\hypertarget{summary-of-important-concepts}{%
\subsection{Summary of Important
Concepts}\label{summary-of-important-concepts}}

\begin{itemize}
\tightlist
\item
  What is an affine variety?
\item
  What is the coordinate ring of an affine variety?
\item
  What are the constructions \(V({\,\cdot\,})\) and \(I({\,\cdot\,})\)?
\item
  What is the Nullstellensatz?
\item
  What are the definitions and some examples of:

  \begin{itemize}
  \tightlist
  \item
    The Zariski topology?
  \item
    Irreducibility?
  \item
    Connectedness?
  \item
    Dimension?
  \end{itemize}
\item
  What is the definition of a presheaf?

  \begin{itemize}
  \tightlist
  \item
    What are some examples and counterexamples?
  \end{itemize}
\item
  What is the definition of sheaf?

  \begin{itemize}
  \tightlist
  \item
    What are some examples?
  \item
    What are some presheaves that are not sheaves?
  \end{itemize}
\item
  What is the definition of \({\mathcal{O}}_X\), the sheaf of regular
  functions?

  \begin{itemize}
  \tightlist
  \item
    How does one compute \({\mathcal{O}}_X\) for \(X = D(f)\) a
    distinguished open?
  \end{itemize}
\item
  What is a morphism between two affine varieties?
\item
  What is the definition of separatedness?

  \begin{itemize}
  \tightlist
  \item
    What are some examples of spaces that are and are not separated?
  \end{itemize}
\item
  What is a projective space?
\item
  What is a projective variety?
\item
  What is the projective coordinate ring?
\item
  How does one take the closure of an affine variety \(X\) in projective
  space?
\item
  What is completeness?

  \begin{itemize}
  \tightlist
  \item
    What are some examples and counterexamples of complete spaces?
  \end{itemize}
\end{itemize}

\newpage

\hypertarget{useful-examples}{%
\subsection{Useful Examples}\label{useful-examples}}

\hypertarget{varieties}{%
\subsubsection{Varieties}\label{varieties}}

\begin{itemize}
\tightlist
\item
  \(V(x-p)\) a point.
\item
  \(V(x)\) a coordinate axis
\item
  \(V(xy) \subseteq {\mathbb{A}}^2\) the coordinate axes
\item
  \(V(xy-1) \subseteq {\mathbb{A}}^2\) a hyperbola
\item
  \(V(x_1^2 - x_2^2 - 1) \subseteq {\mathbb{A}}^2_{/{\mathbb{C}}}\)
\item
  \({\mathbb{A}}^2\setminus\left\{{0}\right\}\) is \textbf{not} an
  affine variety or a distinguished open
\end{itemize}

\hypertarget{presheaves-sheaves}{%
\subsubsection{Presheaves / Sheaves}\label{presheaves-sheaves}}

\begin{itemize}
\tightlist
\item
  \(C^\infty({\,\cdot\,}, {\mathbb{R}})\), a sheaf of smooth functions
\item
  \(C^0({\,\cdot\,}, {\mathbb{R}})\), a sheaf of continuous functions
\item
  \(\underline{{\mathbb{R}}}({\,\cdot\,})\), the constant sheaf
  associated to \({\mathbb{R}}\) (locally constant real-valued
  functions)
\item
  \(\operatorname{Hol}({\,\cdot\,}, {\mathbb{C}})\), a sheaf of
  holomorphic functions
\item
  \(K_p\) the skyscraper sheaf:
  \begin{align*}  
  K_p(U) \coloneqq
  \begin{cases}
  k & p\in U \\
  0 & \text{else}.
  \end{cases}
  \end{align*}
\item
  \({\mathcal{O}}_X({\,\cdot\,})\), the sheaf of regular functions on
  \(X\)
\end{itemize}

\hypertarget{the-algebra-geometry-dictionary}{%
\subsection{The Algebra-Geometry
Dictionary}\label{the-algebra-geometry-dictionary}}

Let \(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\),
we're setting up correspondences:

\begin{longtable}[]{@{}ll@{}}
\toprule
Algebra & Geometry\tabularnewline
\midrule
\endhead
\(k[x_1, \cdots, x_{n}]\) & \({\mathbb{A}}^n_{/k}\)\tabularnewline
Maximal ideals \(\mathfrak{m}={x_1 - a_1, \cdots, x_n - a_n}\) & Points
\(\mathbf{a} \coloneqq{\left[ {a_1, \cdots, a_n} \right]} \in {\mathbb{A}}^n\)\tabularnewline
Radical ideals \(J = \sqrt{J} {~\trianglelefteq~}k[x_1, \cdots, x_{n}]\)
& \(V(J)\) the zero locus\tabularnewline
Prime ideals
\(\mathfrak{p}\in \operatorname{Spec}(k[x_1, \cdots, x_{n}])\) &
Irreducible closed subsets\tabularnewline
Minimal prime ideals of \(A(X)\) & Irreducible components of
\(X\)\tabularnewline
\(I(S)\) the ideal of a set & \(S \subseteq {\mathbb{A}}^n\) a
subset\tabularnewline
\(I + J\) & \(V(I) \cap V(J)\)\tabularnewline
\(\sqrt{I(V) + I(W)}\) & \(V\cap W\)\tabularnewline
\(I \cap J, IJ\) & \(V(I) \cup V(J)\)\tabularnewline
\(I(V) \cap I(W), \sqrt{I(V)I(W)}\) & \(V \cup W\)\tabularnewline
\(I(V) : I(W)\) &
\(\mkern 1.5mu\overline{\mkern-1.5muV\setminus W\mkern-1.5mu}\mkern 1.5mu\)\tabularnewline
\(k[x_1, \cdots, x_{n}]/I(X)\) & \(A(X)\) (Functions on
\(X\))\tabularnewline
\(A(X)\) a domain & \(X\) is irreducible\tabularnewline
\(A(X)\) indecomposable & \(X\) is connected\tabularnewline
\(k{\hbox{-}}\)algebra morphism \(A(X)\to A(Y)\) & Morphisms of
varieties \(X\to Y\)\tabularnewline
Krull dimension \(n\) (chaints of primes) & Topological dimension \(n\)
(chains of irreducibles)\tabularnewline
Integral domains \(S(X)\) & Irreducible projective varieties
\(X\)\tabularnewline
\bottomrule
\end{longtable}

\hypertarget{intro-and-motivation-friday-august-21}{%
\section{Intro and Motivation (Friday, August
21)}\label{intro-and-motivation-friday-august-21}}

\hypertarget{coordinate-rings}{%
\subsection{Coordinate Rings}\label{coordinate-rings}}

General idea: functions in a \emph{coordinate ring}
\(R[x_1, \cdots, x_n]/I\) will correspond to the geometry of the
\emph{variety} cut out by \(I\).

\begin{example}

\envlist

\begin{itemize}
\item
  \(x^2 + y^2 - 1\) defines a circle, say, over \({\mathbb{R}}\)
\item
  \(y^2 = x^3-x\) gives an elliptic curve:
\end{itemize}

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-08-21-01-04-22.png}
\caption{An elliptic curve.}
\end{figure}

\begin{itemize}
\item
  \(x^n+y^n-1\): does it even contain a \({\mathbb{Q}}{\hbox{-}}\)point?
  (Fermat's Last Theorem)
\item
  \(x^2 + 1\), which has no \({\mathbb{R}}{\hbox{-}}\)points.
\item
  \(x^2 + y^2 + 1/{\mathbb{R}}\) vanishes nowhere, so its ring of
  functions is not
  \({\mathbb{R}}[x, y] / \left\langle{x^2 + y^2 + 1}\right\rangle\). The
  problem: \({\mathbb{R}}\) is not algebraically closed.
\item
  \(x^2 - y^2 = 0\) over \({\mathbb{C}}\) is not a manifold (no chart at
  the origin):
\end{itemize}

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-08-21-01-23-32.png}
\caption{A non-manifold curve.}
\end{figure}

\begin{itemize}
\item
  \(x+y+1/{\mathbb{F}}_3\), which has 3 points over
  \({\mathbb{F}}_3^2\), but \(f(x, y) = (x^3 - x)(y^3-y)\) vanishes at
  every point

  \begin{itemize}
  \item
    Not possible when algebraically closed. For example, is there a
    nonzero polynomial that vanishes on every point in \({\mathbb{C}}\)?
  \item
    \(V(f) = {\mathbb{F}}_3^2\), so the coordinate ring is zero instead
    of \({\mathbb{F}}_3[x, y]/\left\langle{f}\right\rangle\) This is
    addressed by scheme theory.
  \end{itemize}
\end{itemize}

\end{example}

\hypertarget{harnack-curve-theorem}{%
\subsection{Harnack Curve Theorem}\label{harnack-curve-theorem}}

\begin{theorem}[Harnack Curve Theorem]

If \(f \in {\mathbb{R}}[x, y]\) is of degree \(d\), then\footnote{Actual
  statement: the number of connected components is bounded above by this
  quantity.}
\begin{align*}  
\pi_1 V(f) \subseteq {\mathbb{R}}^2 \leq 1 + {(d-1)(d-2) \over 2}
\end{align*}

\end{theorem}

\begin{example}

Take the curve
\begin{align*}  
X = \left\{{(x, y, z) = (t^3, t^4, t^5) \in {\mathbb{C}}^3 {~\mathrel{\Big|}~}t\in {\mathbb{C}}}\right\}
.\end{align*}

Then \(X\) is cut out by three equations:

\begin{itemize}
\tightlist
\item
  \(y^2 = xz\)
\item
  \(x^2 = yz\)
\item
  \(z^2 = x^2 y\)
\end{itemize}

\end{example}

\begin{exercise}

Show that the vanishing locus of the first two equations above is
\(X\cup L\) where \(L\) is a line.

\end{exercise}

Compare to linear algebra: codimension \(d\) iff cut out by exactly
\(d\) equations.

\hypertarget{connection-to-riemann-surfaces}{%
\subsection{Connection to Riemann
Surfaces}\label{connection-to-riemann-surfaces}}

\begin{example}

Given the Riemann surface
\begin{align*}  
y^2 = (x-1)(x-2)\cdots(x-2n)
,\end{align*}
how does one visualize its solution set?

\end{example}

\begin{fact}

On \({\mathbb{C}}\) with some slits, you can consistently choose a
square root of the RHS.

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-08-21-01-31-47.png}
\caption{Choosing a square root of a polynomial.}
\end{figure}

Away from \(x=1, \cdots, 2n\), there are two solutions for \(y\) given
\(x\).

After gluing along strips, obtain:

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-08-21-01-32-48.png}
\caption{Glusing along strips to obtain a Riemann surface.}
\end{figure}

\end{fact}

\hypertarget{the-nullstellensatz-tuesday-august-25}{%
\section{The Nullstellensatz (Tuesday, August
25)}\label{the-nullstellensatz-tuesday-august-25}}

\hypertarget{radicals-degrees-and-affine-varieties}{%
\subsection{Radicals, Degrees, and Affine
Varieties}\label{radicals-degrees-and-affine-varieties}}

Given \(f\in k[x_1, \cdots, x_n]\), we'll denote by \(f(a)\) the value
of \(f\) at the point \((a_1, \cdots, a_n)\). Let
\(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\) and
\(R\) a ring containing ideals \(I, J\). Recall the definition of the
\emph{radical} of an ideal:

\begin{definition}[Radical]

The \emph{radical} of an ideal \(I {~\trianglelefteq~}R\) is defined as
\begin{align*}  
\sqrt{I} = \left\{{r\in R {~\mathrel{\Big|}~}r^k\in I \text{ for some } k\in {\mathbb{N}}}\right\}
.\end{align*}

\end{definition}

\begin{example}

Let
\begin{align*}
I &= (x_1, x_2^2) \subset {\mathbb{C}}[x_1, x_2] \\
  &= \left\{{ f_1 x_1 + f_2 x_2 {~\mathrel{\Big|}~}f_1, f_2 \in {\mathbb{C}}[x_1, x_2]}\right\}
\end{align*}

Then \(\sqrt{I} = (x_1, x_2)\), since
\(x_2^2 \in I \implies x_2 \in \sqrt{I}\).

\end{example}

\begin{definition}[Degree of an element of {$k[x_1, x_2, \cdots, x_n]$}]

Define \(\deg(f)\) as the largest value of \(i_1 + \cdots + i_n\) such
that the coefficient of \(\prod x_j ^{i_j}\) is nonzero.

\end{definition}

\begin{example}

\(\deg(x_1 + x_2^2 + x_1 x_2^3) = 4\)

\end{example}

\begin{definition}[Affine Variety]

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Affine \(n{\hbox{-}}\)space\footnote{Not \(k^n\), since we won't
    necessarily use the vector space structure (e.g.~adding points).}
  \({\mathbb{A}}^n = {\mathbb{A}}_k^n\) is defined as
  \begin{align*}
  {\mathbb{A}}^n \coloneqq\left\{{(a_1, \cdots, a_n) {~\mathrel{\Big|}~}a_i \in k}\right\}
  \end{align*}
\item
  Let \(S\subset k[x_1, \cdots, x_n]\) be a \textbf{set} of
  polynomials.\footnote{We don't necessarily require \(S\) to be an
    ideal in this definition. We will shortly show that taking the ideal
    it generates yields the same variety.} Then define the
  \textbf{affine variety} of \(S\) as
  \begin{align*}
    V(S) \coloneqq\left\{{x\in {\mathbb{A}}^n {~\mathrel{\Big|}~}f(x) = 0}\right\} \subset {\mathbb{A}}^n
    \end{align*}
\end{enumerate}

\end{definition}

\begin{example}[Examples of affine varieties]

\envlist

\begin{itemize}
\tightlist
\item
  Let \(f(x) = 0\), then \({\mathbb{A}}^n = V\qty{\left\{{f}\right\}}\)
  is an affine variety.
\item
  Any point \((a_1, \cdots, a_n)\in {\mathbb{A}}^n\) is an affine
  variety, uniquely determined by
  \begin{align*}
  V(x_1 - a_1, \cdots, x_n - a_n) = \left\{{a_1, \cdots, a_n}\right\}
  \end{align*}
\item
  For any finite set \(r_1, \cdots, r_k \in {\mathbb{A}}^1\), there
  exists a polynomial \(f\in k[x_1]\) whose roots are \(r_i\).
\end{itemize}

\end{example}

\begin{remark}

We may as well assume \(S\) is an ideal by taking the ideal it
generates,
\begin{align*}
S\subseteq \left\langle{S}\right\rangle = \left\{{\sum g_i f_i {~\mathrel{\Big|}~}g_i \in k[x_1, \cdots, x_n],\, f_i\in S}\right\}
.\end{align*}

\begin{claim}

\begin{align*}  
V(S) = V\qty{\left\langle{S}\right\rangle}
.\end{align*}

\end{claim}

It's clear that \(V(\left\langle{S}\right\rangle) \subset V(S)\).
Conversely, if \(f_1, f_2\) vanish at \(x\in {\mathbb{A}}^n\), then
\(f_1 + f_2\) and \(gf_1\) also vanish at \(x\) for all
\(g\in k[x_1, \cdots, x_n]\). Thus
\(V(S) \subset V(\left\langle{S}\right\rangle)\).

\end{remark}

\hypertarget{ideals-and-properties-of-vcdot}{%
\subsection{\texorpdfstring{Ideals and Properties of
\(V({\,\cdot\,})\)}{Ideals and Properties of V(\{\textbackslash,\textbackslash cdot\textbackslash,\})}}\label{ideals-and-properties-of-vcdot}}

See \cref{useful-algebra-facts} for a review of properties of ideals.

\begin{proposition}[Properties of $V$]

\begin{align*}  
S_1 \subseteq S_2 &\implies V(S_1) \supseteq V(S_2) \tag{1}\\
V(S_1) \cup V(S_2) &= V(S_1 S_2) = V(S_1 \cap S_2) \tag{2} \\
\bigcap V(S_i) &= V\qty{\bigcup S_i} \tag{3}
.\end{align*}

\end{proposition}

We thus have a map
\begin{align*}  
V: \left\{{\text{Ideals in } k[x_1, \cdots, x_n]}\right\} \to \left\{{\text{Affine varieties in } {\mathbb{A}}^n_{/k}}\right\}
.\end{align*}

\begin{definition}[The Ideal of a Set]

Let \(X\subset {\mathbb{A}}^n\) be any set, then \emph{the ideal of
\(X\)} is defined as
\begin{align*}  
I(X) \coloneqq\left\{{f\in k[x_1, \cdots, x_n] {~\mathrel{\Big|}~}f(x) = 0\, \forall x\in X}\right\}
.\end{align*}

\end{definition}

\begin{example}

Let \(X\) be the union of the \(x_1\) and \(x_2\) axes in
\({\mathbb{A}}^2\), then
\begin{align*}
I(X) = \left\langle{x_1 x_2}\right\rangle = \left\{{g x_1 x_2 {~\mathrel{\Big|}~}g\in k[x_1, x_2]}\right\}
.\end{align*}

\end{example}

\begin{proposition}[$I$ is inclusion-reversing]

\begin{align*}  
X_1 \subset X_2 \implies I(X_1) \supset I(X_2)
.\end{align*}

\end{proposition}

\begin{proof}[?]

If \(f\in I(X_2)\), then \(f(x) = 0\) for all \(x\in X_2\). Since
\(X_1 \subset X_2\), we have \(f(x) = 0\) for all \(x\in X_1\), so
\(f\in I(X_2)\).

\end{proof}

\begin{proposition}[The Image of $I$ is Radical]

\begin{align*}  
I(X) = \sqrt{I(X)}
,\end{align*}
i.e.~\(I(X)\) is a radical ideal.

\end{proposition}

\begin{proof}[?]

It's clear that
\begin{align*}
I(X) \subset \sqrt{I(X)} \coloneqq\left\{{f\in k[x_1, \cdots, x_{n}] {~\mathrel{\Big|}~}f^k \in I(X)}\right\}
\end{align*}
since we can simply take \(k=1\) in this definition.

\begin{claim}

For a fixed \(f\in k[x_1, \cdots, x_{n}]\) and any
\(k\in {\mathbb{N}}\),
\begin{align*}  
f(x)^k = 0\,\, \forall x\in X \implies f(x) = 0\,\, \forall x\in X
.\end{align*}

\end{claim}

Granting this claim, if \(f\in \sqrt{I(X)}\) then \(f^k \in I(X)\) and
thus \(f\in I(X)\), completing the proof.

\begin{proof}[?]

Without loss of generality, we can take \(n = 1\) and consider
\(k[x_1, \cdots, x_{n}]\) Toward a contradiction, fix a \(k\) suppose
\(f(x) \neq 0\) but \(f(x)^k = 0\). Then writing
\(f(x) = \sum_{j=1}^d \alpha_j x^j\) where \(d \coloneqq\deg(f)\), we
have \(\alpha_d \neq 0\) and
\begin{align*}  
f(x)^k = \alpha_d^k x^{dk} + \cdots
.\end{align*}
Equating coefficients, we have \(\alpha_d^k = 0\) in the base field. But
fields have no nonzero nilpotents, so we arrive at a contradiction.

\end{proof}

\end{proof}

These maps thus yield correspondences
\begin{align*}  
\left\{{\substack{\text{Ideals in } k[x_1, \cdots, x_n]}}\right\} &\xrightarrow{V} \left\{{\substack{\text{Affine Varieties}}}\right\} \\
\left\{{\substack{\text{Radical Ideals}}}\right\} &\xleftarrow{I} \left\{{\substack{\text{Affine Varieties}}}\right\}
.\end{align*}

We'll find that if we restrict to radical ideals, this will yield a
bijective correspondence.

\hypertarget{the-nullstellensatz-statement-and-proof}{%
\subsection{The Nullstellensatz: Statement and
Proof}\label{the-nullstellensatz-statement-and-proof}}

\begin{theorem}[Hilbert Nullstellensatz (Zero Locus Theorem)]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  For any affine variety \(X\),
  \begin{align*}
  V(I(X)) = X
  .\end{align*}
\item
  For any ideal \(J \subset k[x_1, \cdots, x_n]\),
  \begin{align*}
  I(V(J)) = \sqrt{J}
  .\end{align*}
\end{enumerate}

Thus there is a bijection between radical ideals and affine varieties.

\end{theorem}

Recall the Hilbert Basis Theorem (\cref{thm:hilbert_basis}): any ideal
in a finitely generated polynomial ring over a field is again finitely
generated. We need to show 4 inclusions, 3 of which are easy.

\begin{proof}[$X \subset V(I(X))$ (a)]

If \(x\in X\) then \(f(x) = 0\) for all \(f\in I(X)\). So
\(x\in V(I(X))\), since every \(f\in I(X)\) vanishes at \(x\).

\end{proof}

\begin{proof}[$\sqrt{J} \subset I(V(J))$ (b)]

If \(f\in \sqrt{J}\) then \(f^k \in J\) for some \(k\). Then
\(f^k(x) = 0\) for all \(x\in V(J)\). So \(f(x) = 0\) for all
\(x\in V(J)\). Thus \(f\in I(V(J))\).

\end{proof}

\begin{proof}[$V(I(X)) \subset X$ (c)]

Need to now use that \(X\) is an affine variety. Counterexample:
\(X = {\mathbb{Z}}^2 \subset {\mathbb{C}}^2\), then \(I(X) = 0\). But
\(V(I(X)) = {\mathbb{C}}^2 \not\subset {\mathbb{Z}}^2\). By (b),
\(I(V(J)) \supset \sqrt{J} \supset J\). Since \(V({\,\cdot\,})\) is
order-reversing, taking \(V\) of both sides reverses the containment. So
\(V(I(V(J))) \subset V(J)\), i.e.~\(V(I(X)) \subset X\).

\end{proof}

Thus the hard direction that remains is (d),

\begin{align*}
I(V(J)) \subset \sqrt{J}
\end{align*}

\begin{warnings}[Hard theorem from commutative algebra.]

We'll need Noether Normalization (\cref{thm:noether_normalization}),
which is perhaps more important than the Nullstellensatz!

\end{warnings}

\begin{theorem}[1st Version of Nullstellensatz]\label{thm:nullstellensatz_one}

Suppose \(k\) is algebraically closed and uncountable\footnote{Still
  true in countable case by a different proof.}. Then the maximal ideals
in \(k[x_1, \cdots, x_n]\) are given by
\begin{align*}  
\operatorname{mSpec}\qty{k[x_1, \cdots, x_{n}]} = \left\{{ \left\langle{x_1 - a_1, \cdots, x_n - a_n}\right\rangle {~\mathrel{\Big|}~}a_j\in k}\right\}
.\end{align*}

\end{theorem}

\begin{proof}

Let \({\mathfrak{m}}\) be a maximal ideal, then by the Hilbert Basis
Theorem (\cref{thm:hilbert_basis}),
\({\mathfrak{m}}= \left\langle{f_1, \cdots, f_r}\right\rangle\) is
finitely generated. Let \(L = {\mathbb{Q}}[\left\{{c_i}\right\}]\) where
the \(c_i\) are all of the coefficients of the \(f_i\) if
\(\operatorname{ch}(K) = 0\), \textbf{or}
\({\mathbb{F}}_p[\left\{{c_i}\right\}]\) if
\(\operatorname{ch}(k) = p\). Then \(L\subset k\). Define
\begin{align*}
{\mathfrak{m}}_0 = {\mathfrak{m}}\cap L[x_1, \cdots, x_n]
\end{align*}
Note that by construction, \(f_i \in {\mathfrak{m}}_0\) for all \(i\),
and we can write
\({\mathfrak{m}}= {\mathfrak{m}}_0 \cdot k[x_1, \cdots, x_n]\).

\begin{claim}

\({\mathfrak{m}}_0\) is a maximal ideal.

\end{claim}

If it were the case that
\begin{align*}  
{\mathfrak{m}}_0 \subsetneq {\mathfrak{m}}_0' \subsetneq L[x_1, \cdots, x_n]
,\end{align*}
then
\begin{align*}  
{\mathfrak{m}}_0\cdot k[x_1, \cdots, x_n] \subsetneq {\mathfrak{m}}_0'\cdot k[x_1, \cdots, x_n]  \subsetneq k[x_1, \cdots, x_n]
.\end{align*}

So far, we've constructed a smaller polynomial ring and a maximal ideal
in it. Thus \(L[x_1, \cdots, x_n]/{\mathfrak{m}}_0\) is a field that is
finitely generated over either \({\mathbb{Q}}\) or \({\mathbb{F}}_p\).
So \(L[x_1, \cdots, x_n]/{\mathfrak{m}}_0\) is finite over some
\({\mathbb{Q}}(t_1, \cdots, t_n)\), and since \(k\) is uncountable,
there exists an embedding
\({\mathbb{Q}}(t_1, \cdots, t_n) \hookrightarrow k\).\footnote{Here we
  use the fact that there are only countably many polynomials over a
  countable field.}\\

This extends to an embedding of
\(\phi: L[x_1, \cdots, x_n]/{\mathfrak{m}}_0 \hookrightarrow k\) since
\(k\) is algebraically closed. Letting \(a_i\) be the image of \(x_i\)
under \(\phi\), then \(f(a_1, \cdots, a_n) = 0\) by construction,
\(f_i \in (x_i - a_i)\) implies that \({\mathfrak{m}}= (x_i - a_i)\) by
maximality.

\end{proof}

\hypertarget{more-nullstellensatz-thursday-august-27}{%
\section{More Nullstellensatz (Thursday, August
27)}\label{more-nullstellensatz-thursday-august-27}}

\hypertarget{consequence-of-the-nullstellensatz}{%
\subsection{Consequence of the
Nullstellensatz}\label{consequence-of-the-nullstellensatz}}

Recall Hilbert's Nullstellensatz:

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  For any affine variety, \(V(I(X)) = X\).
\item
  For any ideal \(J{~\trianglelefteq~}k[x_1, \cdots, x_n]\),
  \(I(V(J)) = \sqrt{J}\).
\end{enumerate}

So there's an order-reversing bijection
\begin{align*}  
\left\{{\substack{\text{Radical ideals } k[x_1, \cdots, x_n]}}\right\} 
\mathrel{\operatorname*{\rightleftharpoons}_{V({\,\cdot\,})}^{I({\,\cdot\,})}}
\left\{{\substack{\text{Affine varieties in } {\mathbb{A}}^n}}\right\}
.\end{align*}

In proving \(I(V(J)) \subseteq \sqrt{J}\), we needed Noether
Normalization and an important theorem (\cref{thm:nullstellensatz_one}):
the maximal ideals of \(k[x_1, \cdots, x_n]\) are of the form
\(\left\langle{x-a_1, \cdots, x-a_n}\right\rangle\).

\begin{corollary}[?]

If \(V(I)\) is empty, then \(I = \left\langle{1}\right\rangle\).

\end{corollary}

\begin{slogan}

The only ideals that vanish nowhere are trivial.

\end{slogan}

\begin{proof}

This is because no common vanishing locus \(\implies\) trivial ideal, so
there's a linear combination that equals 1. By contrapositive, suppose
\(I\neq \left\langle{1}\right\rangle\). By Zorn's Lemma, these exists a
maximal ideals \({\mathfrak{m}}\) such that
\(I \subset {\mathfrak{m}}\). By the order-reversing property of
\(V({\,\cdot\,})\), \(V({\mathfrak{m}}) \subseteq V(I)\). By the
classification of maximal ideals,
\({\mathfrak{m}}= \left\langle{x-a_1, \cdots, x-a_n}\right\rangle\), so
\(V({\mathfrak{m}}) = \left\{{a_1, \cdots, a_n}\right\}\) is nonempty.

\end{proof}

\hypertarget{proof-of-remaining-part-of-nullstellensatz}{%
\subsection{Proof of Remaining Part of
Nullstellensatz}\label{proof-of-remaining-part-of-nullstellensatz}}

We now return to the remaining hard part of the proof of the
Nullstellensatz:

\begin{align*}
I(V(J)) \subseteq \sqrt{J}
\end{align*}

\begin{proof}[?]

Let \(f\in V(I(J))\), we want to show \(f\in \sqrt{J}\). Consider the
ideal
\begin{align*}
\tilde J \coloneqq J + \left\langle{ft - 1}\right\rangle \subseteq k[x_1, \cdots, x_n, t]
\end{align*}

\begin{observation}

\(f = 0\) on all of \(V(J)\) by the definition of \(I(V(J))\).

\end{observation}

However, if \(f=0\), then \(ft-1 \neq 0\), so
\begin{align*}
V(\tilde J) = V(G) \cap V(ft-1) = \emptyset
\end{align*}

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-08-27-09-56-33.png}
\caption{Effect, a hyperbolic tube around \(V(J)\), so both can't
vanish}
\end{figure}

Applying the corollary \(\tilde J = (1)\), so
\begin{align*}
1 = \left\langle{ft-1}\right\rangle g_0(x_1, \cdots, x_n, t) + \sum f_i g_i(x_1, \cdots, x_n, t)
\end{align*}
with \(f_i \in J\). Let \(t^N\) be the largest power of \(t\) in any
\(g_i\). Thus for some polynomials \(G_i\), we have
\begin{align*}  
f^N \coloneqq(ft-1) G_0(x_1, \cdots, x_n, ft) + \sum f_i G_i(x_1, \cdots, x_n, ft)
\end{align*}
noting that \(f\) does not depend on \(t\). Now take
\(k[x_1, \cdots, x_n, t]/\left\langle{ft-1}\right\rangle\), so \(ft=1\)
in this ring. This kills the first term above, yielding
\begin{align*}  
f^N = \sum f_i G_i(x_1, \cdots, x_n, 1) \in k[x_1, \cdots, x_n, t]/\left\langle{ft-1}\right\rangle
.\end{align*}

\begin{claim}

There is an inclusion
\begin{align*}  
k[x_1, \cdots, x_n] \hookrightarrow
k[x_1, \cdots, x_n, t]/\left\langle{ft-1}\right\rangle
.\end{align*}

\end{claim}

Since this is injective, this identity also holds in
\(k[x_1, \cdots, x_n]\). But \(f_i\in J\), so \(f\in \sqrt{J}\).

\end{proof}

\begin{exercise}[?]

Why is the claim above true?

\end{exercise}

\begin{example}

Consider \(k[x]\). If \(J\subset k[x]\) is an ideal, it is principal, so
\(J = \left\langle{f}\right\rangle\). We can factor
\(f(x) = \prod_{i=1}^k (x-a_i)^{n_i}\) and
\(V(f) = \left\{{a_1, \cdots, a_k}\right\}\). Then
\begin{align*}
I(V(f)) = \left\langle{(x-a_1)(x-a_2)\cdots(x-a_k)}\right\rangle = \sqrt{J} \subsetneq J
,\end{align*}
so this loses information.

\end{example}

\begin{example}

Let \(J = \left\langle{x-a_1, \cdots, x-a_n}\right\rangle\), then
\(I(V(J)) = \sqrt{J} = J\) with \(J\) maximal. Thus there is a
correspondence
\begin{align*}  
\left\{{\substack{\text{Points of } {\mathbb{A}}^n}}\right\} \iff 
\left\{{\substack{\text{Maximal ideals of }k[x_1, \cdots, x_n]}}\right\}
.\end{align*}

\end{example}

\begin{theorem}[Properties of $I$]

\envlist

\begin{align*}  
I(X_1 \cup X_2)       &= I(X_1) \cap I(X_2) \tag{a} \\
I(X_1) \cap I(X_2) &= \sqrt{I(X_1) + I(X_2)} \tag{b}
.\end{align*}

\end{theorem}

\begin{proof}

We proved (a) on the variety side. For (b), by the Nullstellensatz we
have \(X_i = V(I(X_i))\), so
\begin{align*}  
I(X_1\cap X_2) 
&=
I\qty{ VI(X_1) \cap VI(X_2)} \\
&=
IV\qty{I(X_1) + I(X_2)} \\
&= \sqrt{I(X_1) + I(X_2)}
.\end{align*}

\end{proof}

\begin{example}

Example of property (b):

Take \(X_1 = V(y-x^2)\) and \(X_2 = V(y)\), a parabola and the
\(x{\hbox{-}}\)axis.

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-08-27-10-26-45.png}
\caption{Intersecting \(V(y-x^2)\) and \(V(y)\)}
\end{figure}

Then \(X_1 \cap X_2 = \left\{{(0, 0)}\right\}\), and
\(I(X_1) + I(X_2) = \left\langle{y-x^2, y}\right\rangle = \left\langle{x^2, y}\right\rangle\),
but
\begin{align*}
I(X_1 \cap X_2) = \left\langle{x, y}\right\rangle = \sqrt{\left\langle{x^2, y}\right\rangle}
\end{align*}

\end{example}

\begin{proposition}[?]

If \(f, g\in k[x_1, \cdots, x_n]\), and suppose \(f(x) = g(x)\) for all
\(x\in {\mathbb{A}}^n\). Then \(f = g\).

\end{proposition}

\begin{proof}

Since \(f-g\) vanishes everywhere,
\begin{align*}
f-g \in I({\mathbb{A}}^n) = I(V(0)) = \sqrt{0} = 0
\end{align*}

\end{proof}

More generally suppose \(f(x) = g(x)\) for all \(x\in X\), where \(X\)
is some affine variety. Then by definition, \(f-g \in I(X)\), so a
``natural'' space of functions on \(X\) is \(k[x_1,\cdots, x_n]/I(X)\).

\begin{definition}[Coordinate Ring]

For an affine variety \(X\), the \textbf{coordinate ring of \(X\)} is
\begin{align*}  
A(X) \coloneqq k[x_1, \cdots, x_n]/ I(X)
.\end{align*}

Elements \(f\in A(X)\) are called \emph{polynomial} or \textbf{regular}
functions on \(X\).

\end{definition}

\begin{observation}

The constructions \(V({\,\cdot\,}), I({\,\cdot\,})\) work just as well
with \(A(X)\) instead of \(k[x_1, \cdots, x_{n}]\) and \(X\) instead of
\({\mathbb{A}}^n\).

Given any \(S\subset A(Y)\) for \(Y\) an affine variety,
\begin{align*}  
V(S) = V_Y(S) \coloneqq\left\{{x\in Y {~\mathrel{\Big|}~}f(x) = 0\,\,\forall f\in S}\right\}
.\end{align*}

Given \(X\subset Y\) a subset,
\begin{align*}  
I(X) = I_Y(X) \coloneqq\left\{{f\in A(Y) {~\mathrel{\Big|}~}f(x) = 0\,\,\forall x\in X}\right\} \subseteq A(Y)
.\end{align*}

\end{observation}

\begin{example}

For \(X\subset Y \subset {\mathbb{A}}^n\), we have
\(I(X) \supset I(Y) \supset I({\mathbb{A}}^n)\), so we have maps

\begin{center}
\begin{tikzcd}
A({\mathbb{A}}^n) \ar[rrrr, twoheadrightarrow, "{\,\cdot\,}/I(X)", bend left] \ar[rr, "{\,\cdot\,}/I(Y)"', twoheadrightarrow] 
&& A(Y)\ar[rr, twoheadrightarrow, "{\,\cdot\,}/I(X)"'] 
&& A(X) \\
\end{tikzcd}
\end{center}

\end{example}

\begin{theorem}[Relative Nullstellensatz]

Let \(X\subset Y\) be an affine subvariety, then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(A(X) = A(Y) / I_Y(X)\)
\item
  There is a correspondence
  \begin{align*}  
  \left\{{\substack{\text{Affine subvarieties of }Y}}\right\} 
  &\iff \left\{{\substack{\text{Radical ideals in }A(Y)}}\right\} \\
  X &\mapsto I_Y(X) \\
  V_Y(J) &\mapsfrom J
  .\end{align*}
\end{enumerate}

\end{theorem}

\begin{proof}

Properties are inherited from the case of \({\mathbb{A}}^n\), see
exercise in Gathmann.

\end{proof}

\begin{example}

Let \(Y = V(y-x^2) \subset {\mathbb{A}}^2/{\mathbb{C}}\) and
\(X = \left\{{(1, 1)}\right\} = V(x-1, y-1)\subset {\mathbb{A}}^2/{\mathbb{C}}\).

Then there is an inclusion
\(\left\langle{y-x^2}\right\rangle \subset \left\langle{x-1, y-1}\right\rangle\),
e.g.~by Taylor expanding about the point \((1, 1)\). and thus there is a
map

\begin{center}
\begin{tikzcd}
A({\mathbb{A}}^n)\ar[r]\ar[d, equal] & A(Y) \ar[r]\ar[d, equal]   & A(X)\ar[d, equal] \\
k[x, y]\ar[r]               & k[x, y]/\left\langle{y-x^2}\right\rangle\ar[r, dotted, "\exists"] & k[x, y]/\left\langle{x-1, y-1}\right\rangle
\end{tikzcd}
\end{center}

\end{example}

\hypertarget{zariski-topology-tuesday-september-01}{%
\section{Zariski Topology (Tuesday, September
01)}\label{zariski-topology-tuesday-september-01}}

\hypertarget{the-zariski-topology}{%
\subsection{The Zariski Topology}\label{the-zariski-topology}}

Last time:
\begin{align*}  
V(I) &= \left\{{x\in {\mathbb{A}}^n {~\mathrel{\Big|}~}f(x) = 0 \, \forall x\in I}\right\} \\
I(X) &= \left\{{f\in k[x_1, \cdots, x_n] {~\mathrel{\Big|}~}f(x) = 0\, \forall x\in X}\right\}
.\end{align*}

We proved the Nullstellensatz \(I(V(J)) = \sqrt{J}\), defined the
coordinate ring of an affine variety \(X\) as
\begin{align*}
A(X) \coloneqq k[x_1, \cdots, x_n] / I(X)
\end{align*}
the ring of \emph{regular} (polynomial) functions on \(X\). Recall that
a \emph{topology} on \(X\) can be defined as a collection of closed
subsets of \(X\) that are closed under arbitrary intersections and
finite unions. A subset \(Y\subset X\) inherits a subspace topology with
closed sets of the form \(Z\cap Y\) for \(Z\subset X\) closed in \(X\).

\begin{definition}[Zariski Topology]

Let \(X\) be an affine variety. The closed sets are affine subvarieties
\(Y\subset X\).

\end{definition}

\begin{proposition}[The Zariski topology is a topology]

This satisfies the axioms for a topological space.

\end{proposition}

\begin{proof}[?]

\envlist

\begin{itemize}
\item
  We have \(\emptyset, X\) closed, since

  \begin{enumerate}
  \def\labelenumi{\arabic{enumi}.}
  \tightlist
  \item
    \(V_X(1) = \emptyset\),
  \item
    \(V_X(0) = X\)
  \end{enumerate}
\item
  Closure under finite unions:

  Let \(V_X(I), V_X(J)\) be closed in \(X\) with \(I, J \subset A(X)\)
  ideals. Then \(V_X(IJ) = V_X(I) \cup V_X(J)\).
\item
  Closure under intersections:

  We have
  \(\bigcap_{i\in \sigma} V_X(J) = V_X\qty{ \sum_{i\in \sigma} J_i}\).
\end{itemize}

\end{proof}

\begin{remark}

There are few closed sets, so this is a ``weak'' topology.

\end{remark}

\begin{example}[Closedness differs in the analytic topology]

Compare the classical topology on \({\mathbb{A}}^1_{/{\mathbb{C}}}\) to
the Zariski topology. Consider the set
\begin{align*}
A\coloneqq\left\{{x\in {\mathbb{A}}^1_{/{\mathbb{C}}} {~\mathrel{\Big|}~}{\left\lVert {x} \right\rVert} \leq 1}\right\}
\end{align*}
which is closed in the classical topology. However, \(A\) is \emph{not}
closed in the Zariski topology, since the closed subsets are finite sets
or the whole space. In fact, the topology here is the cofinite topology.

\end{example}

\begin{example}

Let \(f: {\mathbb{A}}^1_{/k} \to {\mathbb{A}}^1_{/k}\) be any injective
map. Then \(f\) is necessarily continuous wrt the Zariski topology. Thus
the notion of continuity is too weak in this situation.

\end{example}

\begin{warnings}[The topology on the product is not the product topology]

Consider \(X\times Y\) a product of affine varieties. Then there is a
product topology where open sets are of the form
\(\bigcup_{i=1}^n U_i \times V_i\) with \(U_i, V_i\) open in \(X, Y\)
respectively. This is the wrong topology! On
\({\mathbb{A}}^1 \times{\mathbb{A}}^1 = {\mathbb{A}}^2\), the diagonal
\(\Delta \coloneqq V(x-y)\) is closed in the Zariski topology on
\({\mathbb{A}}^2\) but not in the product topology.

\end{warnings}

\hypertarget{irreducibility-and-connectedness}{%
\subsection{Irreducibility and
Connectedness}\label{irreducibility-and-connectedness}}

\begin{example}

Consider \({\mathbb{A}}^2_{/{\mathbb{C}}}\), so the closed sets are
curves and points. Observation:
\(V(x_1 x_2 ) \subset {\mathbb{A}}^2_{/{\mathbb{C}}}\) decomposed into
the union of the coordinate axes \(X_1 \coloneqq V(x_1)\) and
\(X_2 \coloneqq V(x_2)\). The Zariski topology can detect these
decompositions.

\end{example}

\begin{definition}[Irreducibility and Connectedness]

Let \(X\) be a topological space.

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(X\) is \textbf{reducible} iff there exist nonempty proper closed
  subsets \(X_1 ,X_2 \subset X\) such that \(X = X_1 \cup X_2\).
  Otherwise, \(X\) is said to be \emph{irreducible}.
\item
  \(X\) is \textbf{disconnected} if there exist \(X_1, X_2 \subset X\)
  such that \(X = X_1 {\coprod}X_2\). Otherwise, \(X\) is said to be
  \emph{connected}.
\end{enumerate}

\end{definition}

\begin{example}

\(V(x_1 x_2)\) is reducible but connected.

\end{example}

\begin{example}

\({\mathbb{A}}^1_{/{\mathbb{C}}}\) is \emph{not} irreducible, since we
can write
\begin{align*}
{\mathbb{A}}^1_{/{\mathbb{C}}} = \left\{{{\left\lVert {x} \right\rVert} \leq 1}\right\} \cup\left\{{{\left\lVert {x} \right\rVert} \geq 1}\right\}
\end{align*}

\end{example}

\begin{proposition}[?]

Let \(X\) be a disconnected affine variety with
\(X = X_1 {\coprod}X_2\). Then \(A(X) \cong A(X_1) \times A(X_2)\).

\end{proposition}

\begin{proof}

We have
\begin{align*}
X_1 \cup X_2 = X \implies I(X_1) \cap I(X_2) = I(X) = (0) \in A(X)
,\end{align*}
recalling that the coordinate ring \(A(X)\) is a quotient by \(I(X)\).
Since \(X_1 \cap X_1 = \emptyset\), we have
\begin{align*}  
I(X_1 \cap X_2) = \sqrt{I(X_1) + I(X_2) } = I(\emptyset) = \left\langle{1}\right\rangle
.\end{align*}

Thus \(I(X_1) + I(X_2) = \left\langle{1}\right\rangle\), and by the
Chinese Remainder Theorem, the following map is an isomorphism:
\begin{align*}  
A(X) \to A(X)/I(X_1) \times A(X) / I(X_2)
.\end{align*}

However, the codomain is precisely \(A(X_1) \times A(X_2)\).

\end{proof}

\hypertarget{irreducibility-on-the-algebra-side}{%
\subsubsection{Irreducibility on the Algebra
Side}\label{irreducibility-on-the-algebra-side}}

\begin{proposition}[?]

An affine variety \(X\) is irreducible \(\iff\) \(A(X)\) is an integral
domain.

\end{proposition}

\begin{proof}

\(\implies\): By contrapositive, suppose \(f_1, f_2 \in A(X)\) are
nonzero with \(f_1 f_2 = 0\). Let \(X_i \coloneqq V(f_i)\), then
\begin{align*}
X= V(0) = V(f_1 f_2) = X_1 \cup X_2
\end{align*}
which are closed and proper since \(f_i \neq 0\).

\hfill\break

\(\impliedby\): Suppose \(X\) is reducible with \(X = X_1 \cup X_2\)
with \(X_i\) proper and closed. Define \(J_i \coloneqq I(X_i)\), then by
part (a) of the Nullstellensatz.
\begin{align*}  
V(J_i) = V(I(X_i)) = X_i \implies J_i \neq 0
.\end{align*}
So there exists a nonzero \(f_i \in J_i = I(X_i)\), so \(f_i\) vanishes
on \(X_i\). But then
\begin{align*}
V(f_1) \cup V(f_2) \supset X_1 \cup X_2 = X
,\end{align*}
so \(X= V(f_1 f_2)\) and
\(f_1 f_2 \in I(X) = \left\langle{0}\right\rangle\) and \(f_1 f_2 = 0\),
and \(A(X)\) is thus not a domain.

\end{proof}

\begin{example}

Let \(X = \left\{{\mathbf{p}^1, \cdots, \mathbf{p}^d}\right\}\) be a
finite set in \({\mathbb{A}}^n\). The Zariski topology on \(X\) is the
discrete topology, and
\(X = {\coprod}_{i=1}^d \left\{{\mathbf{p}^i}\right\}\). So
\begin{align*}  
A(X) = A\qty{\displaystyle\coprod_{i=1}^d \left\{{\mathbf{p}^i}\right\}} = \prod_{i=1}^d A\qty{\left\{{\mathbf{p}^i}\right\}} = \prod_{i=1}^d \frac{ k[x_1, \cdots, x_n]} {\left\langle{x_1 - p^i_1, \cdots, x_n - p^i_n}\right\rangle }
\end{align*}
where \(p_j^i\) is the \(j\)th component of \(\mathbf{p}^i\).

\end{example}

\begin{example}

Set \(V(x_1 x_2) = X\), then
\(A(X) = k[x_1, x_2]/ \left\langle{x_1 x_2}\right\rangle\). This not
being a domain (since \(x_1 x_2 = 0\)) corresponds to
\(X = V(x_1) \cup V(x_2)\) not being irreducible.

\end{example}

\begin{example}

Let \(X_1\) be the \(xy{\hbox{-}}\)plane and \(X_2\) be the line
parallel to the \(y{\hbox{-}}\)axis through
\({\left[ {0,0,1} \right]}\), and let \(X= X_1 {\coprod}X_2\):

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-09-01-10-43-00.png}
\caption{Union of a plane and a parallel line.}
\end{figure}

Then \(X_1 = V(z)\) and \(X_2 = V(x, z-1)\), and
\begin{align*}
I(X) = \left\langle{z}\right\rangle \cdot \left\langle{x, z-1}\right\rangle=  \left\langle{xz, z^2 - z}\right\rangle
\end{align*}
The coordinate ring is then given by
\begin{align*}
A(X) = { {\mathbb{C}}[x, y, z] \over \left\langle{xz, z^2 - z}\right\rangle } = { {\mathbb{C}}[x, y, z] \over  \left\langle{z}\right\rangle } \oplus { {\mathbb{C}}[x, y,z] \over \left\langle{x, z-1}\right\rangle }
\end{align*}

\end{example}

\hypertarget{irreducibility-thursday-september-03}{%
\section{Irreducibility (Thursday, September
03)}\label{irreducibility-thursday-september-03}}

\hypertarget{irreducibility-and-prime-ideals}{%
\subsection{Irreducibility and Prime
Ideals}\label{irreducibility-and-prime-ideals}}

Recall that the Zariski topology is defined on an affine variety
\(X = V(J)\) with \(J {~\trianglelefteq~}k[x_1, \cdots, x_n]\) by
describing the closed sets.

\begin{proposition}[?]

\(X\) is irreducible if its coordinate ring \(A(X)\) is a domain.

\end{proposition}

\begin{proposition}[?]

There is a 1-to-1 correspondence
\begin{align*}  
\left\{{\substack{\text{Irreducible subvarieties} \\ \text{of }X}}\right\}
\iff
\left\{{\substack{\text{Prime ideals} \\ \text{in }A(X)}}\right\}
.\end{align*}

\end{proposition}

\begin{proof}

Suppose \(Y\subset X\) is an affine subvariety. Then
\begin{align*}  
A(X) / I_X(Y) = A(Y)
.\end{align*}

By the Nullstellensatz, there is a bijection between subvarieties of
\(X\) and radical ideals of \(A(X)\) where \(Y\mapsto I_X(Y)\). A
quotient is a domain iff quotienting by a prime ideal, so \(A(Y)\) is a
domain iff \(I_X(Y)\) is prime.

\end{proof}

Recall that \({\mathfrak{p}}{~\trianglelefteq~}R\) is prime when
\(fg\in {\mathfrak{p}}\iff f\in {\mathfrak{p}}\) or
\(g\in {\mathfrak{p}}\). Thus
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu \mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu = 0\)
in \(R/{\mathfrak{p}}\) implies
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu = 0\) or
\(\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu = 0\) in
\(R/{\mathfrak{p}}\), i.e.~\(R/{\mathfrak{p}}\) is a domain. Finally,
note that prime ideals are radical (easy proof).

\begin{example}

Consider \({\mathbb{A}}^2/{\mathbb{C}}\) and some subvarieties \(C_i\):

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-09-03-09-47-09.png}
\caption{Subvarieties}
\end{figure}

Then irreducible subvarieties correspond to prime ideals in
\({\mathbb{C}}[x, y]\). Here \(C_1, C_3\) correspond to \(V(f), V(g)\)
for \(f,g\) irreducible polynomials, whereas \(C_2\) corresponds to a
maximal ideal, i.e.~\(V(x_1 - a_1, x_2 - a_2)\). Note that
\(I(C_1 \cup C_2 \cup C_3)\) is not a prime ideal, since the variety is
reducible as the union of 3 closed subsets.

\end{example}

\begin{example}

A finite set is irreducible iff it contains only one point.

\end{example}

\begin{example}

Any irreducible topological space is connected, since irreducible
requires a union but connectedness requires a \emph{disjoint} union.

\end{example}

\begin{example}

\({\mathbb{A}}^n/k\) is irreducible: by prop 2.8, its irreducible iff
the coordinate ring is a domain. However
\(A({\mathbb{A}}^n) = k[x_1, \cdots, x_n]\), which is a domain.

\end{example}

\begin{example}

\(V(x_1 x_2)\) is not irreducible, since it's equal to
\(V(x_1) \cup V(x_2)\).

\end{example}

\begin{definition}[Noetherian Space]

A \emph{Noetherian} topological space \(X\) is a space with no infinite
strictly decreasing sequence of closed subsets.

\end{definition}

\begin{proposition}[?]

An affine variety \(X\) with the Zariski topology is a Noetherian space.

\end{proposition}

\begin{proof}

Let \(X_0 \supsetneq X_1 \supsetneq \cdots\) be a decreasing sequence of
closed subspaces. Then \(I(X_0) \subsetneq I(X_1) \subsetneq\). Note
that these containments are strict, otherwise we could use
\(V(I(X_1)) = X_1\) to get an equality in the original chain.

Recall that a ring \(R\) is Noetherian iff every ascending chain of
ideals terminates. Thus it suffices to show that \(A(X)\) is Noetherian.

We have \(A(X) = k[x_1, \cdots, x_{n}] / I(X)\), and if this had an
infinite chain \(I_1 \subsetneq I_2 \subsetneq \cdots\) lifts to a chain
in \(k[x_1, \cdots, x_{n}]\), which is Noetherian. A useful fact: \(R\)
Noetherian implies that \(R[x]\) is Noetherian, and fields are always
Noetherian.

\end{proof}

\begin{remark}

Any subspace \(A\subset X\) of a Noetherian space is Noetherian. To see
why, suppose we have a chain of closed sets in the subspace topology,
\begin{align*}  
A\cap X_0 \supsetneq A\cap X_1 \supsetneq \cdots
.\end{align*}

Then \(X_0 \supsetneq X_1 \supsetneq \cdots\) is a strictly decreasing
chain of closed sets in \(X\). Why strictly decreasing:
\begin{align*}
\cap^n X_i = \cap^{n+1} X_i \implies A\cap^n X_i = A\cap^{n+1} X_i
,\end{align*}
yielding a contradiction.

\end{remark}

\begin{proposition}[Important: Noetherian spaces are finite unions of closed irreducibles]

Every Noetherian space \(X\) is a finite union of irreducible closed
subsets, i.e.~\(X = \bigcup_{i=1}^k X_i\). If we further assume
\(X_i \not\subset X_j\) for all \(i, j\), then the \(X_i\) are unique up
to permutation.

\end{proposition}

\begin{remark}

The \(X_i\) are the \textbf{components} of \(X\). In the previous
example \(C_1 \cup C_2 \cup C_3\) has three components.

\end{remark}

\begin{proof}

\begin{claim}

Such a finite decomposition exists.

\end{claim}

If \(X\) is irreducible, then \(X=X\) and this holds. Otherwise, write
\(X = X_1 \cup X_2\) with \(X_i\) proper closed subsets. If \(X_1\) and
\(X_1'\) are irreducible, we're done, so otherwise suppose wlog \(X_1'\)
is not irreducible. Then we can express
\(X = X_1 \cup\qty{X_2 \cup X_2'}\) with \(X_2, X_2' \subset X_1'\)
closed and proper. Thus we can obtain a tree whose leaves are proper
closed subsets:

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-09-03-10-15-53.png}
\caption{Image}
\end{figure}

This tree terminates because \(X\) is Noetherian: if it did not, this
would generate an infinite decreasing chain of subspaces.

\begin{claim}

This decomposition is unique if no two components are contained in the
other.

\end{claim}

Suppose
\begin{align*}  
X= \bigcup_{i=1}^k X_i = \bigcup_{j=1}^\ell X_j'
.\end{align*}

Note that \(X_i \subset X\) implies that
\(X_i = \bigcup_{j=1}^\ell X_i \cap X_j'\). But \(X_i\) is irreducible
and this would express \(X_i\) as a union of proper closed subsets, so
some \(X_i \cap X_j'\) is \emph{not} a proper closed subset. Thus
\(X_i = X_i \cap X_j'\) for some \(j\), which forces
\(X_i \subset X_j'\). Applying the same argument to \(X_j'\) to obtain
\(X_j' \subset X_k\) for some \(k\). Then
\(X_i \subset X_j' \subset X_k\), but \(X_ i \not\subset X_j\) when
\(j\neq i\). So \(X_i = X_j' = X_k\), forcing the \(X_i\) to be unique
up to permutation.

\end{proof}

Recall from ring theory: for \(I\subset R\) and \(R\) Noetherian, \(I\)
has a \emph{primary decomposition} \(I = \bigcap_{i=1}^k Q_i\) with
\(\sqrt{Q_i}\) prime. Assuming the \(Q_i\) are minimal in the sense that
\(\sqrt{Q_i} \not\subset \sqrt{Q_j}\) for any \(i, j\), this
decomposition is unique.

Applying this to \(I(X) {~\trianglelefteq~}k[x_1, \cdots, x_{n}] = R\)
yields
\begin{align*}  
I(X) = \bigcap_{i=1}^k Q_i 
\implies
X  = V(I(X)) = \bigcup_{i=1}^k V(Q_i)
.\end{align*}

Letting \(P_i = \sqrt{Q_i}\), noting that the \(P_i\) are prime and thus
radical, we have \(V(Q_i) = V(P_i)\). Writing \(X = \bigcup V(P_i)\), we
have \(I(V(P_i)) = P_i\) and thus \(A(V(P_i)) = R/P_i\) is a domain,
meaning \(V(P_i)\) are irreducible affine varieties. Conversely, if we
express \(X = \bigcup X_i\), we have
\(I = I\qty{\bigcup X_i} = \bigcap I(X_i) = \bigcap P_i\) which are
irreducible since they are prime.

\begin{remark}

There is a correspondence
\begin{align*}  
\left\{{\substack{\text{Irreducible components} \\ \text{of } X}}\right\} 
\iff
\left\{{\substack{\text{Minimal prime ideals} \\ \text{in } A(X)}}\right\}
,\end{align*}
where here \emph{minimal} is the condition that no pair of ideals
satisfies a subset containment.

\end{remark}

In what follows, let \(X\) be an irreducible topological space.

\begin{proposition}[1]

The intersection of nonempty two open sets is \emph{never} empty.

\end{proposition}

\begin{proof}

Let \(U, U'\) be open and \(X\setminus U, X\setminus U'\) closed. Then
\(U\cap U' = \emptyset \iff (X\setminus U) \cup(X\setminus U') = X\),
but this is not possible since \(X\) is irreducible.\footnote{Irreducible
  iff any two nonempty open sets intersect.}

\end{proof}

\begin{proposition}[?]

Any nonempty open set is dense, i.e.~if \(U\subset X\) is open then its
closure \(\operatorname{cl}_X(U)\) is dense in \(X\).

\end{proposition}

\begin{proof}

Write \(X = \operatorname{cl}_X(U) \cup(X\setminus U)\). Since
\(X\setminus U \neq X\) and \(X\) is irreducible, we have
\(\operatorname{cl}_X(U) = X\).

\end{proof}

\hypertarget{dimension-tuesday-september-08}{%
\section{Dimension (Tuesday, September
08)}\label{dimension-tuesday-september-08}}

Review: we discussed irreducible components. Recall that the
\emph{Zariski topology} on an affine variety \(X\) has affine
subvarieties as closed sets, and a \emph{Noetherian space} has no
infinitely decreasing chains of closed subspaces. We showed that any
Noetherian space has a decomposition into irreducible components
\(X = \cup X_i\) with \(X_i\) closed, irreducible, and unique such that
no two are subsets of each other. Applying this to affine varieties, a
descending chain of subspaces \(X_0 \supsetneq X_1 \cdots\) in \(X\)
corresponds to an increasing chain of ideals
\(I(X_0) \subsetneq I(X_1) \cdots\) in \(A(X)\). Since
\(k[x_1, \cdots, x_{n}]\) is a Noetherian ring, this chain terminates,
so affine varieties are Noetherian.

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

\begin{definition}[Dimensions]

Let \(X\) be a topological space.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  The \textbf{dimension}
  \(\dim X \in {\mathbb{N}}\cup\left\{{\infty}\right\}\) is either
  \(\infty\) or the length \(n\) of the longest chain of
  \emph{irreducible} closed subsets
  \(\emptyset \neq Y_0 \subsetneq \cdots \subsetneq Y_n \subset X\)
  where \(Y_n\) need not be equal to \(X\).\footnote{Note that we count
    the number of nontrivial strict subset containments in this chain.}
\item
  The \textbf{codimension} of \(Y\) in \(X\),
  \(\operatorname{codim}_X(Y)\), for an irreducible subset
  \(Y\subseteq X\) is the length of the longest chain
  \(Y\subset Y_0 \subsetneq Y_1 \cdots \subset X\).
\end{enumerate}

\end{definition}

\begin{example}

Consider \({\mathbb{A}}^1/k\), what are the closed subsets? The finite
sets, the empty set, and the entire space.

What are the irreducible closed subsets? Every point is a closed subset,
so sets with more than one point are reducible. So the only irreducible
closed subsets are \(\left\{{a}\right\}, {\mathbb{A}}^1/k\), since an
affine variety is irreducible iff its coordinate ring is a domain and
\(A({\mathbb{A}}^1/k) = k[x]\). We can check

\begin{center}
\begin{tikzcd}
    {\emptyset} & {\left\{{a}\right\}} & {{\mathbb{A}}^1_k} \\
    {Y_0} & {Y_1} & {Y_2}
    \arrow[from=1-1, to=2-1, equal]
    \arrow[from=1-2, to=2-2, equal]
    \arrow[from=1-3, to=2-3, equal]
    \arrow[from=1-1, to=1-2, hook]
    \arrow[from=1-2, to=1-3, hook]
    \arrow[from=2-1, to=2-2, hook]
    \arrow[from=2-2, to=2-3, hook]
\end{tikzcd}
\end{center}

which is of length \(1\), since there is one nontrivial containment
\(Y_1 \subsetneq Y_2\), and so \(\dim({\mathbb{A}}^1/k) = 1\).

\end{example}

\begin{example}

Consider \(V(x_1 x_2) \subset {\mathbb{A}}^2/k\), the union of the
\(x_i\) axes. Then the closed subsets are \(V(x_1), V(x_2)\), along with
finite sets and their unions. What is the longest chain of irreducible
closed subsets?

Note that \(k[x_1, x_2] / \left\langle{x_1}\right\rangle \cong k[x_2]\)
is a domain, so \(V(x_i)\) are irreducible. So we can have a chain
\begin{align*}  
\emptyset \subsetneq \left\{{a}\right\} \subsetneq V(x_1) \subset X
,\end{align*}
where \(a\) is any point on the \(x_2{\hbox{-}}\)axis, so
\(\dim(X) = 1\).

The only closed sets containing \(V(x_1)\) are \(V(x_1)\cup S\) for
\(S\) some finite set, which can not be irreducible.

\end{example}

\begin{remark}

You may be tempted to think that if \(X\) is Noetherian then the
dimension is finite. However, finite dimension requires a bounded length
on descending/ascending chains, whereas Noetherian only requires
``termination'', which may not happen in a bounded number of steps. So
this is \textbf{false}!

\end{remark}

\begin{example}

Take \(X = {\mathbb{N}}\) and define a topology by setting closed
subsets be the sets \(\left\{{0, \cdots, n}\right\}\) as \(n\) ranges
over \({\mathbb{N}}\), along with \({\mathbb{N}}\) itself. Is \(X\)
Noetherian? Check descending chains of closed sets:

\begin{align*}  
{\mathbb{N}}\supsetneq \left\{{0, \cdots, N}\right\} \supsetneq \left\{{0, \cdots, N-1}\right\} \cdots
,\end{align*}

which has length at most \(N\), so it terminates and \(X\) is
Noetherian. But note that all of these closed subsets
\(X_N \coloneqq\left\{{0, \cdots, N}\right\}\) are irreducible. Why? If
\(X_n = X_i \cup X_j\) then one of \(i, j\) is equal to \(N\), i.e
\(X_i, X_j = X_N\). So for every \(N\), there exists a chain of
irreducible closed subsets of length \(N\), implying that
\(\dim({\mathbb{N}}) = \infty\).

\end{example}

\begin{remark}

Let \(X\) be an affine variety. There is a correspondence
\begin{align*}  
\left\{{\substack{\text{Chains of irreducible closed subsets} \\ Y_0 \subsetneq \cdots \subsetneq Y_n \text{ in } X}}\right\}
\left\{{\substack{\text{Chains of prime ideals} \\ P_0\supsetneq \cdots \supsetneq P_n \text{ in } A(X)}}\right\}
.\end{align*}
Why? We have a correspondence between closed subsets and radical ideals.
If we specialize to irreducible, we saw that these correspond to radical
ideals \(I\subset A(X)\) such that \(A(Y) \coloneqq A(X) / I\) is a
domain, which precisely correspond to prime ideal in \(A(X)\).

\end{remark}

We thus make the following definition:

\begin{definition}[Krull Dimension]

The \emph{Krull dimension} of a ring \(R\) is the length \(n\) of the
longest chain of prime ideals
\begin{align*}  
P_0 \supsetneq P_1 \supsetneq \cdots \supsetneq P_n
.\end{align*}

\end{definition}

\begin{remark}

This uses the key fact from commutative algebra: a finitely generated
\(k{\hbox{-}}\)algebra \(M\) satisfies

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(M\) has finite \(k{\hbox{-}}\)dimension
\item
  If \(M\) is a domain, every maximal chain has the same length.
\end{enumerate}

\end{remark}

\begin{remark}

From scheme theory: for any ring \(R\), there is an associated
topological space \(\operatorname{Spec}R\) given by the set of prime
ideals in \(R\), where the closed sets are given by
\begin{align*}  
V(I) = \left\{{\text{Prime ideals } {\mathfrak{p}}{~\trianglelefteq~}R {~\mathrel{\Big|}~}I\subseteq {\mathfrak{p}}}\right\}
.\end{align*}

If \(R\) is a Noetherian ring, then \(\operatorname{Spec}(R)\) is a
Noetherian space.

\end{remark}

\begin{example}

Using the fact above, let's compute \(\dim {\mathbb{A}}^n/k\). We can
take the following chain of prime ideals in \(k[x_1, \cdots, x_{n}]\):
\begin{align*}  
0 \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
.\end{align*}

By applying \(V({\,\cdot\,})\) we obtain
\begin{align*}  
{\mathbb{A}}^n/k \supsetneq {\mathbb{A}}^{n-1}/k \cdots \supsetneq {\mathbb{A}}^0/k = \left\{{0}\right\} \supsetneq \emptyset
,\end{align*}
where we know each is irreducible and closed, and it's easy to check
that these are maximal:

If there were an ideal
\(\left\langle{x_1, x_2}\right\rangle \subset P \subset \left\langle{x_1, x_2, x_3}\right\rangle\),
then take
\(P\cap k[x_1, x_2, x_3] / \left\langle{x_1, x_2}\right\rangle\) which
would yield a polynomial ring in \(k[x_1]\). But we know the only
irreducible sets in \({\mathbb{A}}^1/k\) are a point and the entire
space.

So this is a chain of maximal length, implying
\(\dim {\mathbb{A}}^n/k = n\).

\end{example}

\hypertarget{dimension-thursday-september-10}{%
\section{Dimension (Thursday, September
10)}\label{dimension-thursday-september-10}}

Recall that the dimension of a ring \(R\) is the length of the longest
chain of prime ideals. Similarly, for an affine variety \(X\), we
defined \(\dim X\) to be the length of the longest chain of irreducible
closed subsets.

These notions of dimension of the same when taking \(R = A(X)\),
i.e.~\(\dim {\mathbb{A}}^n/k = n\).

\begin{proposition}[Dimensions]

Let
\(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\).

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  The dimension of \(k[x_1, \cdots, x_n]\) is \(n\).
\item
  All maximal chains of prime ideals have length \(n\).
\end{enumerate}

\end{proposition}

\hypertarget{proof-of-dimension-proposition}{%
\subsection{Proof of Dimension
Proposition}\label{proof-of-dimension-proposition}}

The case for \(n=0\) is trivial, just take
\(P_0 = \left\langle{0}\right\rangle\). For \(n=1\), easy to see since
the only prime ideals in \(k[x]\) are \(\left\langle{0}\right\rangle\)
and \(\left\langle{x-a}\right\rangle\), since any polynomial factors
into linear factors.\\

Let \(P_0 \subsetneq \cdots \subsetneq P_m\) be a maximal chain of prime
ideals in \(k[x_1, \cdots, x_{n}]\); we then want to show that \(m=n\).
Assume \(P_0 = \left\langle{0}\right\rangle\), since we can always
extend our chain to make this true (using maximality). Then \(P_1\) is a
minimal prime and \(P_m\) is a maximal ideal (and maximals are prime).

\begin{claim}

\(P_1\) is principle, i.e.~\(P_1 = \left\langle{f}\right\rangle\) for
some irreducible \(f\).

\end{claim}

\hypertarget{proof-that-p_1-is-principle}{%
\subsubsection{\texorpdfstring{Proof That \(P_1\) is
Principle}{Proof That P\_1 is Principle}}\label{proof-that-p_1-is-principle}}

\begin{claim}

\(k[x_1, \cdots, x_{n}]\) is a unique factorization domain. This follows
since \(k\) is a UFD since it's a field, and \(R\) a UFD
\(\implies R[x]\) is a UFD for any \(R\).

\begin{quote}
See Gauss' lemma.
\end{quote}

\end{claim}

\begin{claim}

In a UFD, minimal primes are principal. Let \(r \in P\), and write
\(r = u \prod p_i^{n_i}\) with \(p_i\) irreducible and \(u\) a unit. So
some \(p_i\in P\), and \(p_i\) irreducible implies
\(\left\langle{p_i}\right\rangle\) is prime. Since
\(0 \subsetneq \left\langle{p_i}\right\rangle \subset P\), but \(P\) was
prime and assumed minimal, so \(\left\langle{p_i}\right\rangle = P\).

\end{claim}

The idea is to now transfer the chain
\(P_0 \subsetneq \cdots \subsetneq P_m\) to a maximal chain in
\(k[x_1, \cdots, x_{n-1}]\). The first step is to make a linear change
of coordinates so that \(f\) is monic in the variable \(x_n\).

\begin{example}

Take \(f=x_1x_2 + x_3^2 x_4\) and map \(x_3 \mapsto x_3 + x_4\).

\end{example}

So write
\begin{align*}  
f(x_1, \cdots, x_n) = x_n^d + f_1(x_1, \cdots, x_{n-1}) x_n^{d-1} + \cdots + f_d(x_1, \cdots, x_{n-1})
.\end{align*}

We can then descend to \(k[x_1, \cdots, x_{n}]\) to
\(k[x_1, \cdots, x_{n}]/\left\langle{f}\right\rangle\):

\begin{center}
\begin{tikzcd}
P_0 \ar[r] & P_1 \ar[r]\ar[d]                   & \cdots \ar[r]\ar[d] & P_m\ar[d] \\
           & P_1/P_1 \ar[r]\ar[d]               & \cdots \ar[r]\ar[d] & P_m/P_1\ar[d] \\
           & P_1/P_1 \cap k[x_1, \cdots, x_{n-1}] \ar[r] & \cdots \ar[r]       & (P_m / P_1) \cap k[x_1, \cdots, x_{n-1}]
\end{tikzcd}
\end{center}

The first set of downward arrows denote taking the quotient, and the
upward is taking inverse images, and this preserves strict inequalities.

\begin{definition}[Integral Extension]

An \emph{integral} ring extension \(R\hookrightarrow R'\) of \(R\) is
one such that all \(r' \in R'\) satisfying a monic polynomial with
coefficients in \(R\), where \(R'\) is finitely generated.

\begin{quote}
In this case, also implies that \(R'\) is a finitely-generated \(R\)
module.
\end{quote}

\end{definition}

In this case,
\(k[x_1, \cdots, x_{n-1}] \hookrightarrow k[x_1, \cdots, x_{n}] /\left\langle{f}\right\rangle\)
is an integral extension. We want to show that the intersection step
above also preserves strictness of inclusions, since it preserves
primality.

\begin{lemma}

Suppose \(P', Q' \subset R'\) are distinct prime ideals with
\(R\hookrightarrow R'\) an integral extension. Then if
\(P'\cap R = Q'\cap R\), neither contains the other,
i.e.~\(P'\not\subset Q'\) and \(Q'\not\subset P'\).

\end{lemma}

\begin{proof}

Toward a contradiction, suppose \(P' \subset Q'\), we then want to show
that \(Q'\supset P'\). Let \(a\in Q'\setminus P'\) (again toward a
contradiction), then
\begin{align*}  
R/\qty{P'\cap R} \hookrightarrow R'/P'
\end{align*}
is integral.

Then
\(\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu \neq 0\)
in \(R'/P'\), and there exists a monic polynomial of minimal degree that
\(\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu\)
satisfies,
\(p(x) = x^n + \sum_{i=2}^n \mkern 1.5mu\overline{\mkern-1.5muc\mkern-1.5mu}\mkern 1.5mu_i x^{n-i}\).
This implies
\(\mkern 1.5mu\overline{\mkern-1.5muc\mkern-1.5mu}\mkern 1.5mu_n \in Q'/P'\)
(which will contradict \(c_n \in P'\)), since if
\(\mkern 1.5mu\overline{\mkern-1.5muc\mkern-1.5mu}\mkern 1.5mu_n = 0\)
then factoring out \(x\) yields a lower degree polynomial that
\(\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu\)
satisfies.

But then
\(\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu_n \in Q'\cap R\),
so ???

\end{proof}

Question: Given \(R\hookrightarrow R'\) is an integral extension, can we
lift chains of prime ideals?

Answer: Yes, by the ``Going Up'' Theorem: given \(P\subset R\) prime,
there exists \(P'\subset R'\) prime such that \(P'\cap R = P\).
Furthermore, we can lift \(P_1 \subset P_2\) to \(P_1' \subset P_2'\),
as well as ``lifting sandwiches'':

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-09-10-10-18-40.png}
\caption{Image}
\end{figure}

In this process, the length of the chain decreased since
\(\left\langle{0}\right\rangle\) was deleted, but otherwise the chains
are in bijective correspondence. So the inductive hypothesis applies.
\(\hfill\blacksquare\)

\hypertarget{using-dimension-theory}{%
\subsection{Using Dimension Theory}\label{using-dimension-theory}}

Key fact used: the dimension doesn't change under integral extensions,
i.e.~if \(R\hookrightarrow R'\) is integral then \(\dim R = \dim R'\).

\begin{proposition}

Any affine variety has finite dimension.

\end{proposition}

\begin{proof}

We have \(\dim X = \dim A(X)\), where
\(A(X) \coloneqq k[x_1, \cdots, x_{n}] I\) for some
\(I(X)=\sqrt{I(X)}\).

The noether normalization lemma (used in proof of nullstellensatz) shows
that a finitely generated \(k{\hbox{-}}\)algebra is an integral
extension of some polynomial ring \(k[y_1, \cdots, y_d]\). I.e., the
following extension is integral:
\begin{align*}  
k[y_1, \cdots, y_d] \hookrightarrow k[x_1, \cdots, x_n]/I
.\end{align*}

We can conclude that \(\dim A(X) = d < \infty\).

\end{proof}

\begin{proposition}[?]

Let \(X, Y\) be irreducible affine varieties. Then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(\dim X\times Y = \dim X + \dim Y\).
\item
  \(Y\subset X \implies \dim X = \dim Y + \operatorname{codim}_X Y\).
\item
  If \(f\in A(X)\) is nonzero, then any component of \(V(f)\) has
  codimension 1.
\end{enumerate}

\end{proposition}

\begin{proof}

\begin{remark}

Why is \(X\times Y\) again an affine variety? If
\(X\subset {\mathbb{A}}^n/k\), \(Y\subset {\mathbb{A}}^m/k\) with
\(X = V(I), Y = V(J)\), then
\(X\times Y \subset {\mathbb{A}}^n/k \times{\mathbb{A}}^m/k = {\mathbb{A}}^{n+m}/k\)
can be given by taking
\(I+J {~\trianglelefteq~}k[x_1, \cdots, x_n, y_1, \cdots, y_m]\) using
the natural inclusions of \(k[x_1, \cdots, x_{\ell}]\).

Note that we can write
\begin{align*}  
k[x_1, \cdots, x_n, y_1, \cdots, y_m] = k[x_1, \cdots, x_{n}] \otimes_k k[y_1, \cdots, y_n]
\end{align*}
where we think of \(x_i = x_i \otimes 1, y_j = 1 \otimes y_j\). We thus
map \(I, J\) to \(I\otimes 1 + 1\otimes J\) and obtain
\(V(I\otimes 1 + 1\otimes J) = X\times Y\) and
\(A(X\times Y) = A(X)\otimes_k A(Y)\).

In general, for \(k{\hbox{-}}\)algebras \(R,S\),
\begin{align*}  
R/I \otimes_k S/J \cong R\otimes_k S / \left\langle{I\otimes 1 + 1\otimes J}\right\rangle
.\end{align*}

\end{remark}

\begin{remark}

For \(R,S\) finitely generated \(k{\hbox{-}}\)algebras,
\(\dim R\otimes_k S = \dim R + \dim S\).

\end{remark}

Part (a) is proved by the above remarks.

For part (b), the statement is equivalent to \(P\subset A(X)\) with
\(I(Y) \subset P\) is a member of some maximal chain, along with the
statement that all maximal chains are the same length.

\end{proof}

\hypertarget{codimension-and-hypersurfaces-tuesday-september-15}{%
\section{Codimension and Hypersurfaces (Tuesday, September
15)}\label{codimension-and-hypersurfaces-tuesday-september-15}}

Recall the dictionary in \cref{the-algebra-geometry-dictionary}, and the
fact that \(A(X) \coloneqq k[x_1, \cdots, x_{n}]\) contains no nilpotent
elements. We also had some results about dimension

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\dim X<\infty\) and \(\dim {\mathbb{A}}^n = n\).
\item
  \(\dim Y + \operatorname{codim}_X Y = \dim X\) when \(Y\subset X\) is
  irreducible.
\item
  Only over
  \(\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu = k\),
  \(\operatorname{codim}_X V(f) = 1\).
\end{enumerate}

\begin{example}

Take \(V(x^2+y^2) \subset {\mathbb{A}}^2_{/{\mathbb{R}}}\).

\end{example}

\begin{definition}[Curves, Surfaces, and Hypersurfaces]

An affine variety \(Y\) of

\begin{itemize}
\tightlist
\item
  \(\dim Y = 1\) is a \textbf{curve},
\item
  \(\dim Y = 2\) is a \textbf{surface},
\item
  \(\operatorname{codim}_X Y = 1\) is a \textbf{hypersurface in \(X\)}
\end{itemize}

\end{definition}

Question: Is every hypersurface the vanishing locus of a \emph{single}
polynomials \(f\in A(X)\)?

Answer: This is true iff \(A(X)\) is a UFD.

\hypertarget{codimension}{%
\subsection{Codimension}\label{codimension}}

\begin{definition}[Codimension in a Ring]

\(\operatorname{codim}_R {\mathfrak{p}}\) is the length of the longest
chain
\begin{align*}P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_n = {\mathfrak{p}}.\end{align*}

\end{definition}

Recall that \(f\) is irreducible if
\(f = f_1 f_2 \implies f_i \in R^{\times}\) for one \(i\), and \(f\) is
prime iff \(\left\langle{f}\right\rangle\) is a prime ideal, or
equivalently \(f\divides ab \implies f\divides a\) or \(f\divides b\).

Note that prime implies irreducible, since \(f\) divides itself.

\begin{proposition}[?]

Let \(R\) be a Noetherian domain, then TFAE

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  All prime ideals of codimension 1 are principal.
\item
  \(R\) is a UFD.
\end{enumerate}

\end{proposition}

\begin{proof}

\(a\implies b\):

Let \(f\) be a nonzero non-unit, we'll show it admits a prime
factorization. If \(f\) is not irreducible, then \(f = f_1 f_1'\), both
non-units. If \(f_1'\) is not irreducible, we can repeat this, to get a
chain
\begin{align*}  
\left\langle{f}\right\rangle \subsetneq \left\langle{f_1'}\right\rangle \subsetneq \left\langle{f_2'}\right\rangle \subsetneq \cdots
,\end{align*}
which must terminate.

This yields a factorization \(f = \prod f_i\) with \(f_i\) irreducible.
To show that \(R\) is a UFD, it thus suffices to show that the \(f_i\)
are prime. Choose a minimal prime ideal containing \(f\). We'll use
Krull's Principal Ideal Theorem: if you have a minimal prime ideal
\({\mathfrak{p}}\) containing \(f\), its codimension
\(\operatorname{codim}_R {\mathfrak{p}}\) is one. By assumption, this
implies that \({\mathfrak{p}}= \left\langle{g}\right\rangle\) is
principal. But \(g\divides f\) with \(f\) irreducible, so \(f,g\) differ
by a unit, forcing \({\mathfrak{p}}= \left\langle{f}\right\rangle\). So
\(\left\langle{f}\right\rangle\) is a prime ideal.\\

\(b\implies a\):

Let \({\mathfrak{p}}\) be a prime ideal of codimension 1. If
\({\mathfrak{p}}= \left\langle{0}\right\rangle\), it is principal, so
assume not. Then there exists some nonzero non-unit
\(f\in {\mathfrak{p}}\), which by assumption has a prime factorization
since \(R\) is assumed a UFD. So \(f=\prod f_i\).

Since \({\mathfrak{p}}\) is a prime ideal and \(f\in{\mathfrak{p}}\),
some \(f_i\in {\mathfrak{p}}\). Then
\(\left\langle{f_i}\right\rangle \subset {\mathfrak{p}}\) and
\({\mathfrak{p}}\) minimal implies
\(\left\langle{f_i}\right\rangle = {\mathfrak{p}}\), so
\({\mathfrak{p}}\) is principal.

\end{proof}

\begin{corollary}[?]

Every hypersurface \(Y\subset X\) is cut out by a single polynomial, so
\(Y=V(f)\), iff \(A(X)\) is a UFD.

\end{corollary}

\hypertarget{hypersurfaces-and-prime-ideals}{%
\subsection{Hypersurfaces and Prime
Ideals}\label{hypersurfaces-and-prime-ideals}}

\begin{example}

Apply this to \(R=A(X)\), we find that there is a bijection
\begin{align*}  
\operatorname{codim}1 \text{ prime ideals}
\iff 
\operatorname{codim}1 \text{ closed irreducible subsets }Y\subset X,\text{ i.e. hypersurfaces}
.\end{align*}

Taking
\(A(X) = {\mathbb{C}}[x,y,z]/\left\langle{x^2+y^2-z^2}\right\rangle\),
whose real points form a cone:

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-09-15-10-26-24.png}
\caption{Image}
\end{figure}

Note that \(x^2 + y^2 = (x-iy)(x+iy) = z^2\) in this quotient, so this
is not a UFD.

Then taking a line through its surface is a codimension 1 subvariety not
cut out by a single polynomial. Such a line might be given by
\(V(x + iy, z)\), which is 2 polynomials, so why not codimension 2?

Note that \(V(z)\) is the union of the lines

\begin{itemize}
\tightlist
\item
  \(z = 0, x + iy= 0\),
\item
  \(z=0, x - iy = 0\).
\end{itemize}

Note that it suffices to show that this ring has an irreducible that is
not prime. Supposing \(z = f_1 f_2\), some \(f_i\) is a unit, then \(z\)
is not prime because \(z\divides xy\) but divides neither of \(x,y\).

\end{example}

\begin{example}

Note that \(k[x_1, \cdots, x_{n}]\) is a UFD since \(k\) is a UFD.
Applying the corollary, every hypersurface in \({\mathbb{A}}^n\) is cut
out by a single irreducible polynomial.

\end{example}

\begin{definition}[Pure Dimension of Affine Varieties]

An affine variety \(X\) is of \textbf{pure dimension \(d\)} iff every
irreducible component \(X_i\) is of dimension \(d\).

\end{definition}

\begin{remark}

Note that \(X\) is a Noetherian space, so has a unique decomposition
\(X = \cup X_i\). Given \(X\subset {\mathbb{A}}^n/k\) of pure dimension
\(n-1\), \(X = \cup X_i\) with \(X_i\) hypersurfaces with
\(I(X_j) = \left\langle{f_j}\right\rangle\),
\(I(X) = \left\langle{f}\right\rangle\) where \(f = \prod f_i\).

\end{remark}

\begin{definition}[Degree of a Hypersurface]

Given such an \(X\), define the \textbf{degree of a hypersurface} as the
degree of \(f\) where \(I(X) = \left\langle{f}\right\rangle\).

\end{definition}

\hypertarget{regular-functions-thursday-september-17}{%
\section{Regular Functions (Thursday, September
17)}\label{regular-functions-thursday-september-17}}

\begin{quote}
See chapter 3 in the notes.
\end{quote}

We'll next want to attach certain rings of functions to a space.

\begin{example}[Some examples of regular functions]

\envlist

\begin{itemize}
\tightlist
\item
  \(X\) a manifold or an open set in \({\mathbb{R}}^n\) has a ring of
  \(C^\infty\) functions.
\item
  \(X \subset {\mathbb{C}}\) has a ring of holomorphic functions.
\item
  \(X\subset {\mathbb{R}}\) has a ring of real analytic functions
\end{itemize}

These all share a common feature: it suffices to check if a function is
a member on an arbitrary open set about a point, i.e.~they are
\textbf{local}.

\end{example}

\hypertarget{defining-regular-functions}{%
\subsection{Defining Regular
Functions}\label{defining-regular-functions}}

\begin{definition}[Regular Functions]

Let \(X\) be an affine variety and \(U\subseteq X\) open. A
\textbf{regular function} on \(U\) is a function \(\phi: U\to k\) such
that \(\phi\) is ``locally a fraction'', i.e.~a ratio of polynomial
functions. More formally, for all \(p\in U\) there exists a \(U_p\) with
\(p\in U_p \subseteq U\) such that \(\phi(x) = g(x)/ f(x)\) for all
\(x\in U_p\) with \(f, g\in A(X)\).

\end{definition}

\begin{example}

For \(X\) an affine variety and \(f\in A(X)\), consider the open set
\(U\coloneqq V(f)^c\). Then \({1\over f}\) is a regular function on
\(U\), so for \(p\in U\) we can take \(U_p\) to be all of \(U\).

\end{example}

\begin{example}

For \(X = {\mathbb{A}}^1\), take \(f=x-1\). Then \({x\over x-1}\) is a
regular function on \({\mathbb{A}}^1 \setminus\left\{{1}\right\}\).

\end{example}

\begin{example}

Let \(X + V(x_1 x_4 - x_2 x_3)\) and
\begin{align*}  
U \coloneqq X\setminus V(x_2, x_4) = \left\{{{\left[ {x_1, x_2, x_3, x_4} \right]} {~\mathrel{\Big|}~}x_1 x_4 = x_2 x_3, x_2\neq 0 \text{ or } x_4\neq 0 }\right\}
.\end{align*}

Define
\begin{align*}  
\phi: U &\to K \\
{\left[ {x_1, x_2, x_3, x_4} \right]} &\mapsto
\begin{cases}
{x_1\over x_2} & \text{if } x_2 \neq 0 \\
{x_3\over x_4} & \text{if } x_4 \neq 0
\end{cases}
.\end{align*}

This is well-defined on
\(\left\{{x_2\neq 0}\right\} \cap\left\{{x_4 \neq 0}\right\}\), since
\({x_1\over x_2} = {x_3 \over x_4}\). Note that this doesn't define an
element of \(k\) at \({\left[ {0,0,0,1} \right]}\in U\). So this is not
globally a fraction.

\end{example}

Notation: we'll let \({\mathcal{O}}_X(U)\) is the ring of regular
function on \(U\).

\begin{proposition}[?]

Let \(U\subset X\) be an affine variety and
\(\phi \in {\mathcal{O}}_X(U)\). Then
\(V(\phi) \coloneqq\left\{{x\in U {~\mathrel{\Big|}~}\phi(x) = 0}\right\}\)
is closed in the subspace topology on \(U\).

\end{proposition}

\begin{proof}

For all \(a\in U\) there exists \(U_a\subset U\) such that
\(\phi = g_a/f_a\) on \(U_a\) with \(f_a, g_a \in A(X)\) with
\(f_a \neq 0\) on \(U_a\). Then
\begin{align*}  
\left\{{x\in U_a {~\mathrel{\Big|}~}\phi(x) \neq 0}\right\} = U_a \setminus V(g_a)\cap U_a
\end{align*}
is an open subset of \(U_a\), so taking the union over \(a\) again
yields an open set. But this is precisely \(V(\phi)^c\).

\end{proof}

\begin{proposition}

Let \(U\subset V\) be open in \(X\) an \emph{irreducible} affine
variety. If \(\phi_1, \phi_2 \in {\mathcal{O}}_X(V)\) agree on \(U\),
then they are equal.

\end{proposition}

\begin{proof}

\(V(\phi_1 - \phi_2)\) contains \(U\) and is closed in \(V\). It
contains
\(\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu\cap V\),
by an earlier lemma, \(X\) irreducible implies that
\(\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu = X\) and
so \(V(\phi_1 - \phi_2) =V\).

\end{proof}

\begin{question}

Let \(U\subset V \subset {\mathbb{R}}^n\) be open. If
\(\phi_1, \phi_2 \in C^\infty(V)\) such that \(\phi_1, \phi_2\) are
equal when restricted \(U\subset V\). Does this imply
\(\phi_1 = \phi_2\)?

\end{question}

\begin{answer}

For \({\mathbb{R}}^n\), no, there exist smooth bump functions. You can
make a bump function on \(V\setminus U\) and extend by zero to \(U\).
For \({\mathbb{C}}\) and holomorphic functions, the answer is yes, by
the uniqueness of analytic continuation.

\end{answer}

\hypertarget{distinguished-open-sets}{%
\subsection{Distinguished Open Sets}\label{distinguished-open-sets}}

\begin{definition}[(Important) Distinguished Open Sets]

A \textbf{distinguished open set} in an affine variety is one of the
form
\begin{align*}  
D(f) \coloneqq X\setminus V(f) = \left\{{x\in X {~\mathrel{\Big|}~}f(x) \neq 0}\right\}
.\end{align*}

\end{definition}

\begin{proposition}

The distinguished open sets form a base of the Zariski topology.

\end{proposition}

\begin{proof}

Given \(f, g\in A(X)\), we can check:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Closed under finite intersections: \(D(f) \cap D(g) = D(fg)\).
\item

  \begin{align*}U = X\setminus V(f_1, \cdots, f_k) = V\setminus\bigcap V(f_i) = \bigcup D(f_i),\end{align*}
  and any open set is a \emph{finite} union of distinguished opens by
  the Hilbert basis theorem.
\end{enumerate}

\end{proof}

\begin{proposition}[?]

The regular functions on \(D(f)\) are given by
\begin{align*}  
{\mathcal{O}}_X(D(f)) = \left\{{{ g \over f^n} {~\mathrel{\Big|}~}g\in A(X), n\in {\mathbb{N}}}\right\} = A(X)_{\left\langle{f}\right\rangle}
,\end{align*}
the localization of \(A(X)\) at \(\left\langle{f}\right\rangle\).

\end{proposition}

Note that if \(f=1\), then \({\mathcal{O}}_X(X) = A(X)\).

\begin{proposition}[?]

Note that \({g\over f^n} \in {\mathcal{O}}_X(D(f))\) since \(f^n\neq 0\)
on \(D(f)\). Let \(\phi: D(f) \to k\) be a regular function. By
definition, for all \(a\in D(f)\) there exists a local representation as
a fraction \(\phi = g_a/f_a\) on \(U_a\ni a\). Note that \(U_a\) can be
covered by distinguished opens, one of which contains \(a\). Shrink
\(U_a\) if necessary to assume it is a distinguished open set
\(U_a = D(h_a)\).\\

Now replace
\begin{align*}  
\phi = {g_a \over f_a} = {g_a h_a \over f_a h_a}
,\end{align*}
which makes sense because \(h_a\neq 0\) on \(U_a\). We can assume wlog
that \(h_a = f_a\). Why? We have \(\phi = {g_a \over f_a}\) on
\(D(f_a)\). Since \(f_a\) doesn't vanish on \(U_a\), we have
\(V(f_a h_a) = V(h_a)\) since \(V(f_a) \subset D(h_a)^c = V(h_a)\).
Consider \(U_a = D(f_a)\) and \(U_b = D(f_b)\), on which
\(\phi = {g_a\over f_a}\) and \(\phi = {g_b \over f_b}\) respectively.
On \(U_a\cap U_b = D(f_a f_b)\), these are equal,
i.e.~\(f_b g_a = f_a g_b\) in the coordinate ring \(A(X)\).\\

Then \(D(f) = \bigcup_a D(f_a)\), so take the component
\(V(f) = \cap V(f_a)\) by the Nullstellensatz
\(f\in I(V(f_a)) = I(V(g_a, a\in D_f)) = \sqrt{f_a {~\mathrel{\Big|}~}a\in D_f}\).
Then there exists an expression \(f^n = \sum k_a f_a\) as a finite sum,
so set \(g - \sum g_a k_a\).

\begin{claim}

\(\phi = g/f^n\) on \(D(f)\).

\end{claim}

This follows because on \(D(f_b)\), we have \(\phi = {g_b \over f_b}\),
and so \(gf_b = \sum k_a g_a f_b\).

\begin{quote}
Finish next class
\end{quote}

\end{proposition}

\hypertarget{distinguished-opens-tuesday-september-22}{%
\section{Distinguished Opens (Tuesday, September
22)}\label{distinguished-opens-tuesday-september-22}}

\hypertarget{computing-the-regular-functions}{%
\subsection{Computing the Regular
Functions}\label{computing-the-regular-functions}}

Given an affine variety \(X\) and \(U\subseteq X\) open, a \emph{regular
function} \(\phi: U\to k\) is one locally (wrt the Zariski topology) a
fraction. We write the set of regular functions as \({\mathcal{O}}_X\).

\begin{example}

\(X = V(x_1 x_4 - x_2 x_3)\) on \(U = V(x_2, x_4)^c\), the following
function is regular:
\begin{align*}  
\phi: U &\to k \\
x &\mapsto 
\begin{cases}
{x_1\over x_2} & x_2 \neq 0 \\ \\
{x_3 \over x_4} & x_4 \neq 0
\end{cases}
.\end{align*}
Note that this is not globally a fraction.

\end{example}

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

\begin{definition}[Distinguished Open Sets]

A \emph{distinguished open set} \(D(f) \subseteq X\) for some
\(f\in A(X)\) is
\(V(f)^c \coloneqq\left\{{x\in X {~\mathrel{\Big|}~}f(x) \neq 0}\right\}\).

\end{definition}

These are useful because the \(D(f)\) form a base for the Zariski
topology.

\begin{proposition}[?]

For \(X\) an affine variety, \(f\in A(X)\), we have
\begin{align*} 
{\mathcal{O}}_X(D(f)) = \left\{{ {g\over f^n} {~\mathrel{\Big|}~}g\in A(X), n\in {\mathbb{N}}}\right\}
.\end{align*}

\end{proposition}

\begin{proof}

The first reduction we made was that \(\phi \in {\mathcal{O}}_X(D(f))\)
is expressible as \(g_a / f_a\) on distinguished opens \(D(f_a)\)
covering \(D(f)\). We also noted that
\begin{align*}
{g_a \over f_a} = {g_b \over f_b} \text{ on } D(f_a) \cap D(f_b) \implies f_b g_a = f_a g_b \text{ in } A(X)
.\end{align*}
\\

The second step was writing \(D(f) = \cup D(f_a)\), and so
\(V(f) = \cap_a V(f_a)\) implies that
\(f\in I\qty{ V\qty{ \left\{{f_a {~\mathrel{\Big|}~}a\in U}\right\} }}\).
By the Nullstellensatz,
\(f\in \sqrt{\left\langle{f_a {~\mathrel{\Big|}~}a\in U}\right\rangle}\),
so \(f^N = \sum k_a f_a\) for some \(N\). So construct
\(g = \sum k_a g_a\), then compute
\begin{align*}  
gf_b = \sum_a k_a g_a f_b = \sum_a k_a g_b f_a = g_b \sum k_a f_a = g_b f^N
.\end{align*}
Thus \(g/f^N = g_b / f_b\) for all \(b\), and we can thus conclude
\begin{align*}  
\phi \coloneqq\left\{{{g_b \over f_b} \text{ on } D(f_b)}\right\} = g/f^N
.\end{align*}

\end{proof}

\begin{corollary}[?]

For \(X\) an affine variety, \({\mathcal{O}}_X(X) = A(X)\).

\end{corollary}

\begin{warnings}[Things go wrong when $k\neq \bar k$]

For \(k\) not algebraically closed, the proposition and corollary are
both false. Take \(X = {\mathbb{A}}^1/{\mathbb{R}}\), then
\({1\over x^2+1} \in {\mathbb{R}}(x)\), but
\({\mathcal{O}}_X(X) \neq A(X) = {\mathbb{R}}[x]\).

\end{warnings}

\hypertarget{structure-sheaf-of-distinguished-opens}{%
\subsection{Structure Sheaf of Distinguished
Opens}\label{structure-sheaf-of-distinguished-opens}}

\begin{definition}[Localization]

Let \(R\) be a ring and \(S\) a set closed under multiplication, then
the localization at \(S\) is defined by
\begin{align*}  
R_S \coloneqq\left\{{r/s {~\mathrel{\Big|}~}r\in R, s\in S}\right\} / \sim
.\end{align*}
where \(r_1/s_1 \sim r_2/s_2 \iff s_3(s_2 r_1 - s_1 r_2) = 0\) for some
\(s_3 \in S\).

\end{definition}

\begin{example}

Let \(f\in R\) and take
\(S = \left\{{f^n {~\mathrel{\Big|}~}n\geq 1}\right\}\), then
\(R_f \coloneqq R_S\).

\end{example}

\begin{corollary}[?]

\({\mathcal{O}}_X(D(f)) = A(X)_f\) is the localization of the coordinate
ring.

\end{corollary}

These requires some proof, since the LHS literally consists of functions
on the topological space \(D(f)\) while the RHS consists of formal
symbols.

\begin{proof}

Consider the map
\begin{align*}  
A(X)_f &\to {\mathcal{O}}_X(D(f)) \\
``g/f^n" &\mapsto g/f^n: D(f) \to k
.\end{align*}

By definition, there exists a \(k\geq 0\) such that
\begin{align*}  
f^k(f^m g - f^n g') = 0 
\implies
f^k(f^m g - f^n g') = 0 \text{ as a function on } D(f)
.\end{align*}
Since \(f^k \neq 0\) on \(D(f)\), we have \(f^m g = f^n g'\) as a
function on \(D(f)\), so \(g/f^n = g'/g^m\) as functions on \(D(f)\).\\

\textbf{Surjectivity}: By the proposition, we have surjectivity,
i.e.~any element of \(|OO_x(D(f))\) can be represented by some
\(g/f^n\).\\

\textbf{Injectivity}: Suppose \(g/f^n\) defines the zero function on
\(D(f)\), then \(g = 0\) on \(D(f)\) implies that \(fg=0\) on \(X\)
(i.e.~\(fg= 0 \in A(X)\)), and we can write
\(f(g\cdot 1 - f^n\cdot 0) = 0\). Then \(g/f^n\sim 0/1 \in A(X)_f\),
which forces \(g/f^n = 0\in A(X)_f\).

\end{proof}

\hypertarget{presheaves-and-sheaves}{%
\subsection{Presheaves and Sheaves}\label{presheaves-and-sheaves}}

Idea: spaces on functions on topological spaces.

\begin{definition}[Presheaf]

A \emph{presheaf} (of rings) \(\mathcal{F}\) on a topological space is

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  For every open set \(U\subset X\) a ring \(\mathcal{F}(U)\).
\item
  For any inclusion \(U\subset V\) a restriction map
  \(\operatorname{Res}_{VU}: \mathcal{F}(V) \to \mathcal{F}(U)\)
  satisfying
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(\mathcal{F}(\emptyset) = 0\).
\item
  \(\operatorname{Res}_{UU} = \operatorname{id}_{\mathcal{F}(U)}\).
\item
  \(\operatorname{Res}_{VW} \circ \operatorname{Res}_{UV} = \operatorname{Res}_{UW}\).
\end{enumerate}

\end{definition}

\begin{example}

The smooth functions on \({\mathbb{R}}\) with the standard topology,
\(\mathcal{F} = C^\infty\) where \(C^\infty(U)\) is the set of smooth
functions \(U\to {\mathbb{R}}\). It suffices to check the restriction
condition, but the restriction of a smooth function is smooth: if \(f\)
is smooth on \(U\), it is smooth at every point in \(U\), i.e.~all
derivatives exist at all points of \(U\). So if \(V\subset U\), all
derivatives of \(f\) will exist at points \(x \in V\), so \(f\) will be
smooth on \(V\).

Note that this also works with continuous functions.

\end{example}

\begin{definition}[Sheaf]

A \emph{sheaf} is a presheaf satisfying an additional gluing property:
given \(\phi_i \in \mathcal{F}(U_i)\) such that
\({\left.{{\phi_i}} \right|_{{U_i\cap U_j}} } = {\left.{{\phi_j}} \right|_{{U_i \cap U_j}} }\),
then there exists a unique \(\phi\in \mathcal{F}(\cup_i U_i)\) such that
\({\left.{{\phi}} \right|_{{U_i}} } = \phi_i\).

\end{definition}

\hypertarget{categories-and-presheaves-thursday-september-24}{%
\section{Categories and Presheaves (Thursday, September
24)}\label{categories-and-presheaves-thursday-september-24}}

\hypertarget{regular-functions-vs-holomorphic-functions}{%
\subsection{Regular Functions vs Holomorphic
Functions}\label{regular-functions-vs-holomorphic-functions}}

Recall that we defined the \emph{regular functions}
\({\mathcal{O}}_X(U)\) on an open set \(U\subset X\) an affine variety
as the set of functions \(\phi: U\to k\) such that \(\phi\) is locally a
fraction, i.e.~for all \(p\in U\) there exists a neighborhood of \(p\),
say \(U_p \subset U\), such that \(\phi\) restricted to \(U_p\) is given
by \(g_p \over f_p\) for some \(f_p, g_p \in A(X)\).

We proved that on a distinguished open set \(D(f) = V(f)^c\), we have
\({\mathcal{O}}_X(D(f)) = A(X)_f\). An important example was that
\({\mathcal{O}}_X(X) = A(X)\).

\begin{question}

If \(X\) is a variety over \({\mathbb{C}}\), does
\(A(X) = \operatorname{Hol}(X)\)?

\end{question}

\begin{answer}

The answer is no, since taking
\({\mathbb{A}}^1/{\mathbb{C}}\cong {\mathbb{C}}= X\) we obtain
\(A(X) = {\mathbb{C}}[x]\) but for example
\(e^z \in \operatorname{Hol}(X)\). On the other hand, if you require
that \(f\in \operatorname{Hol}(X)\) is meromorphic at \(\infty\),
i.e.~\(f({1\over z})\) is meromorphic at zero, then you do get
\({\mathbb{C}}[z]\). This is an example of GAGA!

\end{answer}

\begin{quote}
Review: what is a category?
\end{quote}

\begin{quote}
Review: what is a presheaf?
\end{quote}

\hypertarget{sheafifying-tuesday-september-29}{%
\section{Sheafifying (Tuesday, September
29)}\label{sheafifying-tuesday-september-29}}

\hypertarget{presheaves-that-are-not-sheaves}{%
\subsection{Presheaves That Are Not
Sheaves}\label{presheaves-that-are-not-sheaves}}

Recall the definition of a presheaf: a sheaf of rings on a space is a
contravariant functor from its category of open sets to ring, such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(F(\emptyset) = 0\)
\item
  The restriction from \(U\) to itself is the identity,
\item
  Restrictions compose.
\end{enumerate}

\begin{example}[?]

\envlist

\begin{itemize}
\tightlist
\item
  Smooth functions on \({\mathbb{R}}^n\)
\item
  Holomorphic functions on \({\mathbb{C}}\)
\end{itemize}

\end{example}

Recall the definition of sheaf: a presheaf satisfying \emph{unique}
gluing: given \(f_i \in \mathcal{F}(U_i)\), such that
\({\left.{{f_i}} \right|_{{U_i \cap U_j}} } = {\left.{{f_j}} \right|_{{U_i\cap U_j}} }\)
implies that there exists a unique \(f\in \mathcal{F}(\cup U_i)\) such
that \({\left.{{f}} \right|_{{U_i}} } = f_i\).

\begin{question}

Are the constant functions on \({\mathbb{R}}\) a presheaf and/or a
sheaf?

\end{question}

\begin{answer}

This is a presheaf but not a sheaf. Set
\(\mathcal{F}(U) = \left\{{f: U\to {\mathbb{R}}{~\mathrel{\Big|}~}f(x) = c}\right\} \cong {\mathbb{R}}\)
with \(\mathcal{F}(\emptyset) = 0\). Can check that restrictions of
constant functions are constant, the composition of restrictions is the
overall restriction, and restriction from \(U\) to itself gives the
function back.

Given constant functions \(f_i \in \mathcal{F}(U_i)\), does there exist
a unique constant function \(\mathcal{F}(\cup U_i)\) restricting to
them? No: take \(f_1 = 1\) on \((0, 1)\) and \(f_2 = 2\) on \((2, 3)\).
Can check that they both restrict to the zero function on the
intersection, since these sets are disjoint.

\end{answer}

\hypertarget{locally-constant-sheaves}{%
\subsection{Locally Constant Sheaves}\label{locally-constant-sheaves}}

How can we make this into a sheaf? One way: weaken the topology. Another
way: define another presheaf \(\mathcal{G}\) on \({\mathbb{R}}\) given
by \emph{locally} constant function,
i.e.~\(\left\{{f: U\to {\mathbb{R}}{~\mathrel{\Big|}~}\forall p\in U, \exists U_p\ni p,\, { \left.{{f}} \right|_{{U_p}} } \text{ is constant}}\right\}\).
Reminiscent of definition of regular functions in terms of local
properties.

\begin{example}

Let \(X = \left\{{p, q}\right\}\) be a two-point space with the discrete
topology, i.e.~every subset is open. Then define a sheaf by
\begin{align*}  
\emptyset &\mapsto 0 \\
\left\{{p}\right\} &\mapsto R \\
\left\{{q}\right\} &\mapsto S \\
\implies \left\{{p, q}\right\} &\mapsto R\times S
,\end{align*}
where the sheaf condition forces the assignment of the whole space to be
the product. Note that the first 3 assignments are automatically
compatible, which means that we need a unique \(f\in \mathcal{F}(X)\)
restricting to \(R\) and \(S\). In other words, \(\mathcal{F}(X)\) needs
to be unique and have maps to \(R, S\), but this is exactly the
universal property of the product.

\end{example}

\begin{example}

Consider the presheaf on \(X\) given by
\(\mathcal{F}(X) = R\times S \times T\). Taking
\(T = {\mathbb{Z}}/2{\mathbb{Z}}\), we can force uniqueness to fail: by
projecting to \(R, S\), there are two elements in the fiber, namely
\((r,s,0)\mapsto r,s\) and \((r,s,1)\mapsto r,s\).

\end{example}

\begin{example}

Let \(X = \left\{{a, b, c}\right\}\) and
\(\tau = \left\{{\emptyset, \left\{{a}\right\}, \left\{{a, b}\right\}, \left\{{a, c}\right\}}\right\}\).
Can check that it's closed under finite intersections and arbitrary
unions, so this forms a topology. Now make the assignments
\begin{align*}
\left\{{a}\right\}    & \mapsto A \\
\left\{{b}\right\}    & \mapsto B \\
\left\{{a, b}\right\} & \mapsto C \\
X         & \mapsto ?
.\end{align*}

We have a situation like the following:

\begin{center}
\begin{tikzcd}
               & \mathcal{F}(X)\ar[ld]\ar[rd] & \\
B\ar[rd]       &                              & C\ar[ld] \\
               & A\ar[d]                      & \\
               & \emptyset                    &
\end{tikzcd}
\end{center}

Unique gluing says that given \(r\in B, s\in C\) such that
\(\phi_B(r) = \phi_C(s)\), there should exist a unique
\(t\in \mathcal{F}(X)\) such that
\({ \left.{{t}} \right|_{{\left\{{a, b}\right\}}} } = r\) and
\({ \left.{{t}} \right|_{{\left\{{a, c}\right\}}} } = s\). This recovers
exactly the fiber product.
\begin{align*}  
B \times_A C \coloneqq\left\{{(r, s) \in B\times C {~\mathrel{\Big|}~}\phi_B(r) = \phi_C(s) \in A}\right\}
.\end{align*}

\end{example}

\hypertarget{the-structure-sheaf-is-a-sheaf}{%
\subsection{The Structure Sheaf is a
Sheaf}\label{the-structure-sheaf-is-a-sheaf}}

\begin{example}

Let \(X\) be an affine variety with the Zariski topology and let
\(\mathcal{F} \coloneqq{\mathcal{O}}_X\) be the sheaf of regular
functions:
\begin{align*}  
{\mathcal{O}}_X(U) \coloneqq\left\{{f: U\to k {~\mathrel{\Big|}~}\forall p\in U,\, \exists U_p \ni p,\,\, { \left.{{f}} \right|_{{U_p}} } ={g_p \over h_p} }\right\}
.\end{align*}

Is this a presheaf? We can check that there are restriction maps:
\begin{align*}  
{\mathcal{O}}_X(U) &\to {\mathcal{O}}_X(V) \\
\left\{{f: U\to K}\right\} &\mapsto \left\{{{ \left.{{f}} \right|_{{V}} }(x) \coloneqq f(x) \text{ for } x \in V }\right\}
.\end{align*}
This makes sense because if \(V\subset U\), any \(x\in V\) is in the
domain of \(f\). Given that \(f\) is locally a fraction, say
\(\rho = g_p / h_p\) on \(U_p \ni p\), is
\({ \left.{{\phi}} \right|_{{V}} }\) locally a fraction? Yes: for all
\(p\in V\subset U\), \(\phi = g_p / f_p\) on \(U_p\) and this remains
true on \(U_p \cap V\).

To check that \({\mathcal{O}}_X\) is a sheaf, given a set of regular
functions \(\left\{{\phi_i: U_i \to k}\right\}\) agreeing on
intersections, define
\begin{align*}  
\phi: \cup U_i &\to k\\
\phi(x) &\coloneqq\phi_i(x) \text{ if }x\in U_i
.\end{align*}

This is well-defined, since if \(x\in U_i \cap U_j\),
\(\phi_i(x) = \phi_j(x)\) since both restrict to the same function on
\(U_i \cap U_j\) by assumption.

Why is \(\phi\) locally a fraction? We need to check that for all
\(p\in U \coloneqq\cup U_i\) there exists a \(U_p \ni p\) with
\({ \left.{{\phi}} \right|_{{U_p}} } = g_p/h_p\). But any
\(p\in \cup U_i\) implies \(p\in U_i\) for some \(i\). Then there exists
an open set \(U_{i, p} \ni p\) in \(U_i\) such that
\({ \left.{{\phi}} \right|_{{U_{i, p}}} } = g_p / h_p\) by definition of
a regular function. So take \(U_p = U_{i, p}\) and use the fact that
\({ \left.{{\phi}} \right|_{{U_i}} } = \phi_i\) along with compatibility
of restriction.

\end{example}

\begin{remark}

General observation: any presheaf of functions is a sheaf when the
functions are defined by a local property, i..e any property that can be
checked at \(p\) by considering an open set \(U_p \ni p\).

As in the examples of smooth or holomorphic functions, these were local
properties. E.g. checking that a function is smooth involves checking on
an open set around each point. On the other hand, being a constant
function is not a local property.

\end{remark}

\hypertarget{restriction-stalks-sections}{%
\subsection{Restriction, Stalks,
Sections}\label{restriction-stalks-sections}}

\begin{definition}[Restriction of a (Pre)sheaf]

Given a sheaf \(\mathcal{F}\) on \(X\) and an open set \(U\subset X\),
we can define a sheaf \({ \left.{{\mathcal{F}}} \right|_{{U}} }\) on
\(U\) (with the subspace topology) by defining
\({ \left.{{\mathcal{F}}} \right|_{{U}} }(V) \coloneqq\mathcal{F}(V)\)
for \(U\subseteq V\).

\end{definition}

\begin{definition}[Stalks]

Let \(\mathcal{F}\) be a sheaf on \(X\) and \(p\in X\) a point. The
\emph{stalk} of \(\mathcal{F}\) at \(p\), denoted \(\mathcal{F}_p\) for
\(p\in U\), is defined by
\begin{align*}  
\mathcal{F}_p \coloneqq\left\{{(U, \phi) {~\mathrel{\Big|}~}\phi \in \mathcal{F}(U) }\right\} / \sim
\end{align*}
where \((U, \phi) \sim (V, \phi')\) iff there exists a
\(W\subset U\cap V\) and \(p\in W\) such that
\({ \left.{{\phi}} \right|_{{W}} } = { \left.{{\phi}} \right|_{{W}} }'\).

\end{definition}

\begin{example}

What is the stalk of \(\operatorname{Hol}({\mathbb{C}})\) at \(p=0\)?

Examples of equivalent elements in this stalk:

\begin{figure}
\centering
\includegraphics{figures/image_2020-09-29-10-38-22.png}
\caption{O}
\end{figure}

In this case
\begin{align*}  
\operatorname{Hol}({\mathbb{C}})_0 = \left\{{\phi = \sum_{i>0}c_i z^i {~\mathrel{\Big|}~}\phi \text{ has a positive radius of convergence}}\right\}
.\end{align*}

\end{example}

\begin{definition}[Sections and Germs]

An element \(f\in \mathcal{F}(U)\) is called a \emph{section} over
\(U\), and elements of the stalk \(f\in \mathcal{F}_p\) are called
\emph{germs} at \(p\).

\end{definition}

\hypertarget{stalks-and-localizations-thursday-october-01}{%
\section{Stalks and Localizations (Thursday, October
01)}\label{stalks-and-localizations-thursday-october-01}}

\hypertarget{review}{%
\subsection{Review}\label{review}}

Recall that a sheaf of rings on a topological space \(X\) is a ring
\(\mathcal{F}(U)\) for all open sets \(U\subset X\) satisfying four
properties:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Initial to terminal: the empty set is mapped to zero.\footnote{The
    initial object in the category of Sets is the empty set, and the
    terminal object in the category of Rings is the zero ring. This
    ``swap'' comes from the fact that presheaves should be
    contravariant.}
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\item
  Identities: the morphism \(\mathcal{F}(U)\to \mathcal{F}(U)\) is the
  identity.
\item
  Composition: given \(W\subset V\subset U\) we have a commutative
  diagram

  \begin{center}
  \begin{tikzcd}
  \mathcal{F}(U) \ar[rr, "\operatorname{Res}_{UV}"] \ar[rd, "\operatorname{Res}_{UW}"'] &                & \mathcal{F}(V) \ar[ld, "\operatorname{Res}_{VW}"] \\
                      & \mathcal{F}(W) &
  \end{tikzcd}
  \end{center}
\item
  Gluing: given sections \(s_i \in\mathcal{F}(U_i)\) which agree on
  overlaps (restrict to the same function on \(U_i\cap U_j\)), there is
  a unique \(s\in \mathcal{F}(\cup U_i)\).
\end{enumerate}

\begin{example}

If \(X\) is an affine variety with the Zariski topology,
\(\mathcal{O}_X\) is a sheaf of regular functions, where we recall
\(\mathcal{O}_X(U)\) are the functions \(\phi: U\to k\) that are locally
a fraction.

\end{example}

\hypertarget{computing-stalks}{%
\subsection{Computing Stalks}\label{computing-stalks}}

Recall that the \textbf{stalk} of a sheaf \(\mathcal{F}\) at a point
\(p\in X\), is defined as
\begin{align*}  
\mathcal{F}_p \coloneqq\left\{{(U, \phi) {~\mathrel{\Big|}~}p\in U \text{ open },\, \phi \in \mathcal{F}(U)}\right\}/\sim
.\end{align*}
where \((U, \phi) \sim (U', \phi')\) if there exists a
\(p\in W \subset U\cap U'\) such \(\phi, \phi'\) restricted to \(W\) are
equal.

Recall that a \textbf{local ring} is a ring with a unique maximal ideal
\({\mathfrak{m}}\). Given a prime ideal \({\mathfrak{p}}\in R\), so
\(ab\in {\mathfrak{p}}\implies a,b\in {\mathfrak{p}}\), the complement
\(R\setminus P\) is closed under multiplication. So we can localize to
obtain
\begin{align*}
R_{\mathfrak{p}}&\coloneqq\left\{{a/s {~\mathrel{\Big|}~}s\in R\setminus P, a\in R}\right\}/\sim \\
\text{ where } a'/s' \sim a/s &\iff \exists t\in R\setminus P \text{ such that } t(a's - as') = 0
\end{align*}

\begin{warnings}[Notation for localizations]

Note that \(R_f\) is localizing at the powers of \(f\), whereas
\(R_{\mathfrak{p}}\) is localizing at the \emph{complement} of
\({\mathfrak{p}}\).

\end{warnings}

Since maximal ideals are prime, we can localize any ring \(R\) at a
maximal ideal \(\mathfrak{m} \in \operatorname{mSpec}(R)\) to obtain
\(R_{\mathfrak{m}}\), and this will be a local ring. Why? The ideals in
\(R_{\mathfrak{m}}\) biject with ideals in \(R\) contained in
\({\mathfrak{m}}\). Thus all ideals in \(R_{\mathfrak{m}}\) are
contained in the maximal ideal generated by \({\mathfrak{m}}\),
i.e.~\({\mathfrak{m}}R_{\mathfrak{m}}\).

\begin{lemma}[Stalks are isomorphic to localizations]

Let \(X\) be an affine variety, then
\begin{align*}  
{\mathcal{O}}_{X, p}\coloneqq\qty{{\mathcal{O}}_X}_p \cong A(X)_{\mathfrak{m}_p}
.\end{align*}
In words: the stalk of the sheaf of regular functions
\({\mathcal{O}}_{X, p} \coloneqq({\mathcal{O}}_X)_p\) is isomorphic to
the localization \(A(X)_{{\mathfrak{m}}_p}\) where
\({\mathfrak{m}}_p \coloneqq I(\left\{{p}\right\})\).

\end{lemma}

\begin{proof}

We can write
\begin{align*}  
A(X)_{{\mathfrak{m}}_p} \coloneqq\left\{{{g\over f} {~\mathrel{\Big|}~}g\in A(X),\, f\in A(X)\setminus{\mathfrak{m}}_p}\right\} / \sim \\
\text{ where } g_1/f_1 \sim g_2/f_2 \iff \exists h(p) \neq 0 \text{ where }0 = h(f_2 g_1 - f_1 g_2)
.\end{align*}
where the \(f\) are regular functions on \(X\) such that \(f(p) = 0\).

We can also write
\begin{align*}  
{\mathcal{O}}_{X, p} \coloneqq\left\{{(U, \phi) {~\mathrel{\Big|}~}p\in U,\, \phi \in {\mathcal{O}}_X(U) }\right\} /\sim 
\\ \text{ where } (U, \phi) \sim (U', \phi') 
\iff \exists p\in W \subset U\cap U' \text{ s.t. } { \left.{{\phi}} \right|_{{W}} } = { \left.{{\phi'}} \right|_{{W}} }
.\end{align*}

So we can define a map
\begin{align*}  
\Phi: A(X)_{{\mathfrak{m}}_p} &\to {\mathcal{O}}_{X, p} \\
{g\over f} &\mapsto \qty{D_f, {g\over f}}
.\end{align*}

\textbf{Step 1:} There are equivalence relations on both sides, so we
need to check that things are well-defined.

We have
\begin{align*}  
g/f \sim g'/f' &\iff \exists g \text{ such that } h(p) \neq 0,\, h(gf' - g'f)=0 \in A(X) \\
&\iff \text{the functions } {g\over f}, {g' \over f'} \text{ agree on } W\coloneqq D(f) \cap D(f') \cap D(h) \\
&\implies (D_f, g/f) \sim (D_{f'}, g'/f')
,\end{align*}
since there exists a \(W\subset D_f \cap D_{f'}\) such that
\(g/f, g'/f'\) are equal.\\

\textbf{Step 2:} Surjectivity, since this is clearly a ring map with
pointwise operations.

Any germ can be represented by \((U, \phi)\) with
\(\phi \in {\mathcal{O}}_X(U)\). Since the sets \(D_f\) form a base for
the topology, there exists a \(D_f\subset U\) containing \(p\). By
definition, \((U, \phi) = (D_f, { \left.{{\phi}} \right|_{{D_f}} })\) in
\({\mathcal{O}}_{X, p}\).

Using the proposition that \({\mathcal{O}}_X(D(f)) = A(X)_f\), this
implies that \({ \left.{{\phi}} \right|_{{D_f}} } = g/f^n\) for some
\(n\) and \(f(p) \neq 0\), so \((U, \phi)\) is in the image of
\(\Phi\).\\

\textbf{Step 3}: Injectivity. We want to show that \(g/f\mapsto 0\)
implies that \(g/f = 0 \in A(X)_{{\mathfrak{m}}_p}\).

Suppose that \((D_f, g/f) = 0 \in {\mathcal{O}}_{X, p}\) and
\((U, \phi) = 0 \in {\mathcal{O}}_{X,p}\), then there exists an open
\(W\subset D_f\) containing \(p\) such that after passing to some
distinguished open \(D_h\ni p\) such that \(\phi = 0\) on \(D_h\). Wlog
we can assume \(\phi = 0\) on \(U\), since we could shrink \(U\)
(staying in the same equivalence class) to make this true otherwise.
Then \(\phi = g/f\) on \(D_h\), using that
\({\mathcal{O}}_X(D_f) = A(X)_f\), so \(g/f = 0\) here. So there exists
a \(k\) such that
\begin{align*}  
f^k(g\cdot 1 - 0\cdot f) = 0 \in A(X) \implies f^k g=0 \in A(X)_{{\mathfrak{m}}_p}
.\end{align*}

We can thus conclude that
\begin{align*}  
{\mathcal{O}}_{X, p} \cong A(X)_{{\mathfrak{m}}_p}
.\end{align*}

\end{proof}

\begin{example}

Let \(X = \left\{{p, q}\right\}\) with the discrete topology with the
sheaf \(\mathcal{F}\) given by
\begin{align*}  
p &\mapsto R \\
q &\mapsto S \\
X &\mapsto R\times S
.\end{align*}
Then \(\mathcal{F}_p = R\), since if \(U\) is open and \(p\in\ U\) then
either \(U= \left\{{p}\right\}\) or \(U = X\). We can check that for
\((r, s)\) a section of \(\mathcal{F}\), we have an equivalence of germs
\begin{align*}
(X, (r, s)) \sim (\left\{{p}\right\}, r) \text{ since } \left\{{p}\right\} \subset X\cap\left\{{p}\right\}
\end{align*}
Here \(X\) plays the role of \(U\), \(\left\{{p}\right\}\) of \(U'\),
and the last \(\left\{{p}\right\}\) the role of \(W \subset U\cap U'\).

\begin{align*}  
{\mathcal{O}}_{X, p} &\to A(X) \\
(\left\{{p}\right\}, r) &\mapsto r \\
\mathcal{F}_p &\cong R
.\end{align*}

\end{example}

\begin{example}

Let \(M\) be a manifold and consider the sheaf \(C^\infty\) of smooth
functions on \(M\). Then the stalk \(C_p^\infty\) at \(p\) is defined as
the set of smooth functions in a neighborhood of \(p\) modulo functions
being equivalent if they agree on a small enough ball
\(B_\varepsilon(p)\). This contains a maximal ideal
\({\mathfrak{m}}_p\), the smooth functions vanishing at \(p\). Then
\({\mathfrak{m}}_p^2\) is again an ideal, and as sets,
\begin{align*}  
{\mathfrak{m}}^2 = \left\{{f {~\mathrel{\Big|}~}{\partial}_i {\partial}_j f\mathrel{\Big|}_p = 0,\, \forall i,j}\right\}
.\end{align*}
Thus
\({\mathfrak{m}}_p/{\mathfrak{m}}_p^2 \cong \left\{{{\partial}_v}\right\}^\vee\),
the \emph{dual} of the set of directional derivatives.

\end{example}

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

Problem: what should a map of affine varieties be? A bad definition
would be just taking the continuous maps: for example, any bijection
\({\mathbb{A}}^1_{_{\mathbb{C}}}\) is a homeomorphism in the Zariski
topology. Why? This coincides with the cofinite topology, and the
preimage of a cofinite set is cofinite.

\begin{question}

How do we fix this?

\end{question}

\begin{answer}

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(f:X\to Y\) should be continuous, i.e.~\(f^{-1} (U)\) is open
  whenever \(U\) is open.
\item
  Given \(U\subset Y\) open and \(\phi \in {\mathcal{O}}_Y(U)\), the
  function \(\phi \circ f: f^{-1}(U) \to k\) should be regular.
\end{enumerate}

We'll take these conditions to be the definition of a morphism
\(X\to Y\).

\end{answer}

\begin{example}

For smooth manifolds, we similarly require that there is a pullback that
preserves smooth functions:
\begin{align*}  
f^*: C^\infty(U) \to C^\infty(f^{-1}(U))
.\end{align*}

\end{example}

\hypertarget{ringed-spaces-ch.-4-tuesday-october-06}{%
\section{Ringed Spaces (Ch. 4, Tuesday, October
06)}\label{ringed-spaces-ch.-4-tuesday-october-06}}

\begin{definition}[Ringed Spaces]

A \textbf{ringed space} is a topological space \(X\) together with a
sheaf \({\mathcal{O}}_X\) of rings.

\end{definition}

\begin{example}

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(X\) an affine variety and \({\mathcal{O}}_X\) its ring of regular
  functions.
\item
  \(X\) a manifold over \({\mathbb{R}}^n\) with \({\mathcal{O}}_X\) a
  ring of smooth or continuous functions on \(X\).
\item
  \(X = \left\{{p, q}\right\}\) with the discrete topology and
  \({\mathcal{O}}_X\) given by \(p\mapsto R, q\mapsto S\).
\item
  Let \(U\subset X\) an open subset of \(X\) an affine variety. Then
  declare \({\mathcal{O}}_U\) to be
  \({ \left.{{{\mathcal{O}}_X}} \right|_{{U}} }\).
\end{enumerate}

\end{example}

Recall that the restriction of a sheaf \(\mathcal{F}\) to an open subset
\(U\subset X\) is defined by
\({ \left.{{\mathcal{F}}} \right|_{{U}} }(V) = \mathcal{F}(V)\).

\begin{example}[Skyscraper Sheaf]

Let \(X\) be a topological space and \(p\in X\) a point. The
\textbf{skyscraper sheaf at \(p\)} is defined by
\begin{align*}  
k_p(U) \coloneqq
\begin{cases}
k & p\in U \\
0 & p\not\in U
\end{cases}
.\end{align*}

\end{example}

\begin{remark}

As a convention, we'll always assume that \({\mathcal{O}}_X\) is a sheaf
of functions, so \({\mathcal{O}}_X(U)\) is a subring of all
\(k{\hbox{-}}\)valued functions on \(U\). Moreover,
\(\operatorname{Res}_{UV}\) is restriction of \(k{\hbox{-}}\)valued
functions.

\end{remark}

\hypertarget{morphisms-of-ringed-spaces}{%
\subsection{Morphisms of Ringed
Spaces}\label{morphisms-of-ringed-spaces}}

\begin{definition}[Morphisms of Ringed Spaces]

A \textbf{morphism of ringed spaces}
\begin{align*}  
(X, {\mathcal{O}}_X) \xrightarrow{f}  (Y, {\mathcal{O}}_Y)
\end{align*}
is a continuous map \(X\to Y\) such that for all opens \(U \subset Y\)
and any \(\phi \in {\mathcal{O}}_Y(U)\), the pullback satisfies
\(f^* \phi \in {\mathcal{O}}_X(f^{-1}(U))\).

\end{definition}

\begin{slogan}

Pullbacks of regular functions are regular.

\end{slogan}

\begin{remark}

We'll need to use th convention that \({\mathcal{O}}_X\) is a sheaf of
\(K{\hbox{-}}\)valued functions in order to make sense of pullbacks. In
general, for schemes with \(U \subset Y\) and \(f^{-1} (U) \subset X\),
we'll need some analog of
\(f^*: {\mathcal{O}}_Y(U) \to {\mathcal{O}}_X(f^{-1} (U) )\) to make
sense of ``composing'' or ``restricting'' sections. We still need
continuity, however, so that \(f^{-1}(U)\) is open when \(U\) is open
and thus \({\mathcal{O}}_X(f^{-1}(U))\) makes sense.

\end{remark}

\begin{example}

If \((X, {\mathcal{O}}_X)\) is a ringed space associated to an affine
variety, then we assume \({\mathcal{O}}_X(U)\) are literally functions
on \(U\). Morphisms of open subsets is again defined by morphisms of
ringed spaces.

\end{example}

\begin{example}

Let \(X = {\mathbb{A}}^1/k\) and \(U \coloneqq D(x)\), then
\(D(f) = {\mathbb{A}}^1\setminus\left\{{0}\right\}\). Then
\(\iota: U\hookrightarrow X\) is continuous. Given an arbitrary
distinguished open set \(D(f) \subset {\mathbb{A}}^1\), we know from
previous results that
\begin{align*}  
{\mathcal{O}}_X(D(f)) \coloneqq{\mathcal{O}}_{{\mathbb{A}}^1}(D(f)) = A({\mathbb{A}}^1)_{\left\langle{f}\right\rangle} = k[x]_{\left\langle{f}\right\rangle} \coloneqq\left\{{g/f^n {~\mathrel{\Big|}~}g\in k[x]}\right\}
.\end{align*}
We want to show that
\(\iota: (U, {\mathcal{O}}_U) \hookrightarrow(X, {\mathcal{O}}_X)\) is a
morphism of ringed spaces where
\({\mathcal{O}}_U(V) \coloneqq{\mathcal{O}}_X(V)\). Does \(\iota^*\)
pull back regular functions to regular functions? Yes, since
\begin{align*}
\iota^{-1} (D(f)) = D(f) \cup D(x) = D(xf)
\end{align*}
and thus
\begin{align*}
g/f^n \in {\mathcal{O}}_U(\iota^{-1}(D(f))) = {\mathcal{O}}_U(D(xf))
\end{align*}
where we've used that \(f^n \neq 0 \implies xf\neq 0\).

\end{example}

\begin{example}

A non-example: take
\begin{align*}  
h: {\mathbb{A}}^1 &\to {\mathbb{A}}^1 \\
x & \mapsto 
\begin{cases}
x & x \neq \pm 1 \\
-x & x= \pm 1
\end{cases}
.\end{align*}
This is continuous because the Zariski topology on \({\mathbb{A}}^1\) is
the cofinite topology (since the closed sets are finite), so any
injective map is continuous since inverse images of cofinite sets are
again cofinite.

\begin{question}

Does \(h\) define a morphism of ringed spaces? I.e., is the pullback of
a regular function on an open still regular?

\end{question}

\begin{answer}

Take \(U = {\mathbb{A}}^1\) and the regular function
\(x\in {\mathcal{O}}_{{\mathbb{A}}^1}({\mathbb{A}}^1)\). Then
\(h^*x = x\circ h\), so
\begin{align*}  
(x\circ h)(p) =
\begin{cases}
p & p\neq \pm 1 \\
-p & p= \pm 1 
\end{cases}
\not \in k[x]
\end{align*}
since this is clearly not a polynomial: if two polynomials agree on an
infinite set of points, they are equal.

\end{answer}

\end{example}

\begin{example}

Consider
\begin{align*}
\iota: ({\mathbb{R}}^2, C^\infty) \hookrightarrow({\mathbb{R}}^3, C^\infty)
\end{align*}
is the inclusion of a coordinate hyperplane. To say that this is a
morphism of ringed spaces, we need that for all
\(U\subset {\mathbb{R}}^3\) open and \(f:U\to {\mathbb{R}}\) a smooth
function, we want \(i^* f\in C^\infty (\iota^{-1}(U))\). But this is the
same as \(f\circ \iota \in C^\infty({\mathbb{R}}^2\cap U)\), which is
true.

\end{example}

\hypertarget{gluing-morphisms}{%
\subsection{Gluing Morphisms}\label{gluing-morphisms}}

\begin{proposition}[Ringed spaces form a category]

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  They can be composed: if \(\phi \in {\mathcal{O}}_Z(U)\), then
  \(g^* \phi \in {\mathcal{O}}_Y(g^{-1}(U))\) and so
  \(f^* g^* \phi \in {\mathcal{O}}_X(f^{-1} g^{-1} (U))\).
\item
  The identity is a morphism.
\end{enumerate}

Thus ringed spaces form a category, since composition is associative.

\end{proposition}

\begin{lemma}[Gluing for Morphisms]

Let \(f:X\to Y\) be a continuous map between ringed spaces. Assume there
exists an open cover
\(\left\{{U_i}\right\}_{i\in I}\rightrightarrows X\) such that
\({ \left.{{f}} \right|_{{U_i}} }\) is a morphism, then \(f\) is a
morphism.

\end{lemma}

\begin{remark}

Slogan: it suffices to check a morphism on an open cover.

\end{remark}

\begin{proof}[of part (a)]

Part a: Need to check that \(f\) is continuous, can compute
\begin{align*}  
f^{-1}(V) = \bigcup_{i\in I} U_i \cap f^{-1}(V) = \bigcup_{i\in I} { \left.{{f}} \right|_{{U_i}} }^{-1} (V)
.\end{align*}
but the latter is open as a union of open sets, where each constituent
set is open by assumption.

\end{proof}

\hypertarget{morphisms-glue-thursday-october-08}{%
\section{Morphisms Glue (Thursday, October
08)}\label{morphisms-glue-thursday-october-08}}

We continue the proof that morphisms glue.

\begin{proof}[part b]

We want to show that \(f^*\) sends sections of \({\mathcal{O}}_Y\) to
sections of \({\mathcal{O}}_X\) (e.g.~regular functions pullback). Let
\(V\subset Y\) be open and \(\phi \in {\mathcal{O}}_Y(V)\), then
\begin{align*}  
{ \left.{{f^* \phi}} \right|_{{U_i \cap f^{-1} (V)}} }
\qty{ { \left.{{f^* \phi}} \right|_{{U_i \cap f^{-1} (V)}} } }^* \phi \in 
{\mathcal{O}}_X(U_i f^{-1} (V))
.\end{align*}

Since pullback commutes with restriction, \(f^* \phi\) is the unique
\(k{\hbox{-}}\)valued function for which
\begin{align*}  
{ \left.{{f^* \phi}} \right|_{{U_i \cap f^{-1} V}} } =
{ \left.{{f}} \right|_{{U_i\cap f^{-1} V}} }^* \phi
.\end{align*}
and all of the latter functions agree on overlaps \(U_i \cap U_j\). This
by unique gluing, \(f^* \phi \in {\mathcal{O}}_X(f^{-1}(V))\).

\end{proof}

\hypertarget{morphisms-have-regular-components}{%
\subsection{Morphisms Have Regular
Components}\label{morphisms-have-regular-components}}

\begin{proposition}[Morphisms between affine varieties have regular functions as components]

Let \(U\subset X\) be open in an affine variety and let
\(Y\subset {\mathbb{A}}^n\) be another affine variety. Then the
morphisms \(U\to Y\) of ringed spaces are the maps of the form
\(f = {\left[ {f_1, \cdots, f_n} \right]}: U\to {\mathbb{A}}^n\) such
that \(f(U) \subset Y\) and \(f_i \in {\mathcal{O}}_X(U)\) for all
\(i\).

\end{proposition}

\begin{proof}[$\implies$]

Assume that \(f: U\to Y\) is a morphism. Then the coordinate functions
\(Y\xrightarrow{y_i} {\mathbb{A}}_1\) are regular functions, since they
generate \({\mathcal{O}}_Y(Y) = k[y_1, \cdots, y_n]/I(Y)\). Then
\(f^* y_i\) is a regular function, so define \(f_i \coloneqq f^* y_i\).
But then \(f = {\left[ {f_1, \cdots, f_n} \right]}\).

\end{proof}

\begin{proof}[$\impliedby$]

Conversely suppose
\(f \coloneqq{\left[ {f_1, \cdots, f_n} \right]}: U\to Y \subset {\mathbb{A}}^n\)
is a map such that \(f_i \in {\mathcal{O}}_U(U)\). We want to show that
\(f\) is a morphism, i.e.~that the pullback of every regular function is
regular. We thus need to show the following:

\begin{claim}

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(f\) is continuous, and
\item
  \(f^*\) pulls back regular functions.
\end{enumerate}

\end{claim}

\begin{proof}[of (1)]

Suppose \(Z\) is closed, then it suffices to show \(f^{-1} (Z)\) is
closed. Then \(Z = V(g_1, \cdots, g_n)\) for some \(g_i \in A(Y)\). So
we can write
\begin{align*}  
f^{-1} (Z) = \left\{{
x\in U {~\mathrel{\Big|}~}g_i(f_1(x), \cdots, f_n(x)  ) = 0\, \forall i
}\right\}
.\end{align*}
The claim is that the functions \(g_i\) are regular, i.e.~in
\({\mathcal{O}}_U(U)\), because the \(g_i\) are polynomials in regular
functions, which form a ring. This is the common vanishing locus of
\(m\) regular functions on \(U\). By lemma 3.4, the vanishing locus of a
regular function is closed, so \(f^{-1} (Z)\) is closed.

\end{proof}

\begin{proof}[of (2)]

For 2, let \(\phi \in {\mathcal{O}}_Y(W)\) be a regular function on
\(W\subset Y\) open. Then
\begin{align*}  
f^* \phi  = \phi \circ f: f^{-1} (W) &\to K \\
x &\mapsto \phi(f_1(x), \cdots, f_n(x))
.\end{align*}
We want to show that this is a regular function. Since the \(f_i\) are
regular functions, they are locally fractions, so for all
\(x\in f^{-1} (W)\) there is a neighborhood of \(U_x\ni x\) such that
(by intersecting finitely many neighborhoods) all of the \(f_i\) are
fractions \(a_i/b_i\). Then at a point \(p = {\left[ {f_i(x)} \right]}\)
in the image, there exists an open neighborhood \(W_p\) in \(W\) such
that \(\phi = U/V\). But then
\(\phi{{\left[ {a_i /b_i} \right]}} = (U/V)({\left[ {a_i/b_i} \right]})\),
which is evaluation of a fraction of functions on fractions.

\end{proof}

\end{proof}

\begin{example}

Let \(Y \coloneqq V(xy-1) \subset{\mathbb{A}}^2\) and
\(U\coloneqq D(x) = {\mathbb{A}}^1\setminus\left\{{0}\right\}\subset {\mathbb{A}}^1\).
Note that
\begin{align*}  
A(Y)    &= {k[x, y] \over \left\langle{xy-1}\right\rangle} \\
A({\mathbb{A}}^1) &= k[t]
,\end{align*}
and if \(f_1 \coloneqq t, f_2\coloneqq t^{-1}\), then
\(f_1, f_2 \in {\mathcal{O}}_U(U)\). So we can define a map
\begin{align*}  
{\left[ {f_1, f_2} \right]}: U &\to Y  \\
p &\mapsto {\left[ {p, {1\over p} } \right]}
\end{align*}
whose image lies in \(Y\). Conversely, there is a map
\begin{align*}  
V(xy - 1) &\to U = D(0) \subset  {\mathbb{A}}^1 \\
{\left[ {x, y} \right]} &\mapsto x
.\end{align*}
This a morphism from \(V(xy - 1)\) to \({\mathbb{A}}^1\), since the
coordinates are regular functions. Since the image is contained in
\(U\), the definitions imply that this is in fact a morphism of ringed
spaces. We thus have mutually inverse maps
\begin{align*}  
U       &\mathrel{\operatorname*{\rightleftharpoons}_{t\mapsto {\left[ {t, t^{-1} } \right]}}^{x\mapsfrom {\left[ {x, y} \right]}}} V(xy-1) \\ 
,\end{align*}
so \(U\cong V(xy-1)\) as ringed spaces.

\end{example}

\begin{slogan}

Maps of affine varieties (or their open subsets) are given by functions
whose coordinates are regular.

\end{slogan}

\hypertarget{morphisms-of-varieties-on-the-algebra-side}{%
\subsection{Morphisms of Varieties on the Algebra
Side}\label{morphisms-of-varieties-on-the-algebra-side}}

\begin{corollary}[Morphisms of varieties corresponds to $k\dash$algebra morphisms of coordinate rings]

Let \(X, Y\) be affine varieties, then there is a correspondence
\begin{align*}  
\left\{{\substack{\text{Morphisms } X\to Y }}\right\}
&\iff
\left\{{\substack{k{\hbox{-}}\text{algebra morphisms } A(Y) \to A(X)}}\right\} \\
X\to Y &\mapsto A(Y) \to A(X) \\
f &\mapsto f^* {\mathcal{O}}_Y(Y) = {\mathcal{O}}_X(X)
.\end{align*}

Thus there is an equivalence of categories between reduced\footnote{An
  algebra is \textbf{reduced} iff it has no nonzero nilpotent elements.}
\(k{\hbox{-}}\)algebras and affine varieties.

\end{corollary}

\begin{proof}

We have a map in the forward direction. Conversely, given a
\(k{\hbox{-}}\)algebra morphism \(g:A(Y) \to A(X)\), we need to
construct a morphism \(f\) such that \(f^* = g\). Let
\(Y\subset {\mathbb{A}}^n\) with coordinate functions
\(y_1, \cdots, y_n\). Then
\(f_i = g(y_i) \in A(X) = {\mathcal{O}}_X(X)\). Set
\(f = {\left[ {f_1, \cdots, f_n} \right]}\). Then by the proposition,
\(f\) is a morphism to \({\mathbb{A}}^n\). Letting
\(h\in A({\mathbb{A}}^n)\), we have

\begin{align*}
(f^*h)(x) 
&= h(f(x)) \\
&= h({\left[ {f_1(x) , \cdots, f_n(x)} \right]}) \\
&= h(g(y_1), \cdots, g(y_n)) \\ 
&= g(h)(x) \qquad\text{since $g$ is an algebra morphism, $h$ is a polynomial}
\end{align*}
which follows since \(f_i(x) = g(y_i)(x)\), where \(g:A(Y) \to A(X)\).
So \(f^*(h) = g(h)\) for all \(h\in A({\mathbb{A}}^n)\), so the pullback
of \(f\) is \(g\). We now need to check that it's contained in the
image. Let \(h\in I(Y)\), then \(f^*(h) = g(h) = 0\) since
\(h = 0 \in A(Y)\). So \(\operatorname{im}f \subset Y\). Since the
coordinate \(f_i\) are regular, this is a morphism, and we have
\(f^* = g\) as desired.

\end{proof}

\hypertarget{isomorphisms-of-affine-varieties-tuesday-october-13}{%
\section{Isomorphisms of Affine Varieties (Tuesday, October
13)}\label{isomorphisms-of-affine-varieties-tuesday-october-13}}

Last time: proved that if \(X, Y\) are affine varieties then there is a
bijection
\begin{align*}  
\left\{{\substack{\text{Morphisms} \\ f:X\to Y}}\right\}
&\iff
\left\{{\substack{\text{$k{\hbox{-}}$algebra morphisms}\\ A(Y) \to A(X)}}\right\}
\\
f & \mapsto f^*: {\mathcal{O}}_Y(Y) \to {\mathcal{O}}_X(X)
.\end{align*}

\begin{remark}

A morphism \(f:X\to Y\) is by definition a morphism of ringed spaces
where \({\mathcal{O}}_X, {\mathcal{O}}_Y\) are the sheaves of regular
functions. This shows \(X\cong Y\) as ringed spaces iff
\(A(X) \cong A(Y)\) as \(k{\hbox{-}}\)algebras.

\end{remark}

\hypertarget{counterexample-isomorphisms-are-not-just-bijective-morphisms}{%
\subsection{Counterexample: Isomorphisms Are Not Just Bijective
Morphisms}\label{counterexample-isomorphisms-are-not-just-bijective-morphisms}}

\begin{warnings}[Isomorphisms are not necessarily bijective morphisms]

Let \(X = V(y^2 - x^3) \subset {\mathbb{A}}^2\) and define a map
\begin{align*}  
f: {\mathbb{A}}^1 &\to X \\
t &\mapsto {\left[ {t^2, t^3} \right]}
,\end{align*}
This is a morphism by proposition 4.7 in \autocite{AndreasGathmann515},
since the coordinates \(t^2, t^3\) are regular functions. Then \(f\) is
a bijection, since we can define a piecewise inverse
\begin{align*}  
f^{-1}: X &\to {\mathbb{A}}^1 \\
{\left[ {x, y} \right]} &\mapsto 
\begin{cases}
y/x & x\neq 0 \\
0 & \text{else}.
\end{cases}
\end{align*}
However, \(f^{-1}\) is not a morphism, since it is not in \(A(X)\) and
thus not a regular function on \(X\). For instance, pulling back the
function \(g(t) = t\) yields
\begin{align*}  
\qty{ (f^{-1})^* g} \qty{ {\left[ {x, y} \right]} } = 
\begin{cases}
y/x & x\neq 0\\
0 & x=y=0
\end{cases}
\quad 
\not \in A(X)
.\end{align*}
Since \(f\) is a morphism, however, we can still consider the
corresponding map of \(k{\hbox{-}}\)algebras:
\begin{align*}  
f^*: A(X) = \frac{k[x, y]}{\left\langle{y^2 - x^3}\right\rangle} &\to A({\mathbb{A}}^1) = k[t] \\
x & \mapsto t^2 \\
y & \mapsto t^3
,\end{align*}
but even though \(f\) is a bijective morphism, it's not an isomorphism
of rings: this can be seen from the fact that
\(t\not \in \operatorname{im}f^*\).

\end{warnings}

\hypertarget{categorical-products}{%
\subsection{Categorical Products}\label{categorical-products}}

\begin{quote}
Review of introductory category theory.
\end{quote}

We'll define a category \(\mathrm{AffVar}_k\) whose objects are affine
varieties over \(k\) and morphisms in \(\hom(X, Y)\) will be morphisms
of ringed spaces. There is a contravariant functor \(A\) into
reduced\footnote{An algebra is \textbf{reduced} iff it has no nonzero
  nilpotent elements.} finitely generated \(k{\hbox{-}}\)algebras which
sends \(X\) to \(A(X)\) and sends morphisms \(f:X\to Y\) to their
pullbacks \(f^*:A(Y) \to A(X)\).

\begin{quote}
Review of the universal property of the product.
\end{quote}

\begin{remark}

If we have \(X,Y\) affine varieties, we take \(X\times Y\) to be the
categorical product instead of the underlying product of topological
spaces. We have
\begin{align*}
A(X\times Y) \cong A(X) \otimes_k A(Y) \cong \frac{ k[x_1, \cdots, x_n, y_1, \cdots, y_m]} { I(X) \otimes 1 + 1 \otimes I(Y) }
.\end{align*}
This recovers the product, since we have

\begin{center}
\begin{tikzcd}
Z \ar[dr, dotted, "{\exists! H = (f, g)}"]\ar[rrd, bend left, "f"]\ar[rdd, bend right, "g"] &                       & \\
                                                                                            & X\times Y\ar[r]\ar[d] & X  \\
                                                                                            & Y                     & \\
\end{tikzcd}
\end{center}

\end{remark}

\begin{remark}

Products of spaces are sent to the tensor product of
\(k{\hbox{-}}\)algebras, i.e.~pullbacks are sent to pushouts.

\end{remark}

\begin{remark}

Note that the groupoid associated to a group does not have products:
there can only be one element, but the outer triangles will not
necessarily simultaneously commute.

\end{remark}

\hypertarget{prevarieties-thursday-october-15}{%
\section{Prevarieties (Thursday, October
15)}\label{prevarieties-thursday-october-15}}

\begin{quote}
This corresponds to the end of Chapter 4.
\end{quote}

Recall that we had a proposition: morphisms between affine varieties are
in bijection with \(k{\hbox{-}}\)algebra morphisms between their
coordinate rings. As a result, we'll redefine an \emph{affine variety}
to be a ringed space isomorphic to an affine variety (using the previous
definition of affine variety). This provides a way of saying when affine
varieties embedded in different ways are the ``same''.

\begin{example}

\({\mathbb{A}}^2\) vs \(V(x) \subset {\mathbb{A}}^n\). In fact, the map
\begin{align*}  
f: {\mathbb{A}}^2 &\to {\mathbb{A}}^3
(y,z) &\mapsto (0, y, z)
.\end{align*}
This is continuous and the pullback of regular functions are again
regular.

\end{example}

\begin{remark}

With the new definition, there is a bijection between affine varieties
up to isomorphisms and finitely generated \(k{\hbox{-}}\)algebras up to
algebra isomorphism.

\end{remark}

\begin{proposition}[Distinguished opens are ringed spaces]

Let \(D(f) \subset X\) be a distinguished open, then \(D(f)\) is a
ringed space. This follows because \((X, {\mathcal{O}}_X)\) is a ringed
space, and we can restrict the structure sheaf to any open subset of
\(X\).

\end{proposition}

\begin{proof}

Set
\begin{align*}  
Y \coloneqq\left\{{(x, t) \in X\times{\mathbb{A}}^1 {~\mathrel{\Big|}~}tf(x) = 1}\right\} \subset X\times{\mathbb{A}}^1
.\end{align*}
This is an affine variety, since
\(Y = V(I + \left\langle{ft-1}\right\rangle)\). This is isomorphic to
\(D(f)\) by the map
\begin{align*}  
Y &\to D(f) \\
{\left[ {x, t} \right]} &\mapsto x
.\end{align*}
with inverse
\begin{align*}
D(f)  &\to Y \\
x     &\mapsto {\left[ {x, {1\over f(x)} } \right]}
\end{align*}

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-10-15-09-50-03.png}
\caption{Image}
\end{figure}

Note that \(\pi: X\times{\mathbb{A}}^1 \to X\) is regular, using prop
3.8 in \autocite{AndreasGathmann515}, if the coordinates of a map are
regular functions, then the entire map is a morphism of ringed spaces.
We can then note that \(1\over f(x)\) is regular on \(D(f)\), since
\(f\neq 0\) on this set.

\end{proof}

\begin{example}

\({\mathbb{A}}^2 \setminus\left\{{0}\right\}\) is not an affine variety.
Note that this is also not a distinguished open. We showed on a HW
problem that the regular functions on
\({\mathbb{A}}^2\setminus\left\{{0}\right\}\) are \(k[x, y]\), which are
also the regular functions on \({\mathbb{A}}^2\). So there is a map
inducing a pullback
\begin{align*}  
\iota: {\mathbb{A}}^2\setminus\left\{{0}\right\}&\to {\mathbb{A}}^2 \\ \\
\iota^*: k[x, y]  &\xrightarrow{\sim} k[x, y]
.\end{align*}
Note that \(\iota^*\) is an isomorphism on the space of regular
functions, but \(\iota\) itself is not an isomorphism of topological
spaces. Why? \(i^{-1}\) is not defined at zero.

\end{example}

\hypertarget{prevarieties-chapter-5}{%
\subsection{Prevarieties (Chapter 5)}\label{prevarieties-chapter-5}}

\begin{definition}[Prevariety]

A \textbf{prevariety} is a ringed spaced \(X\) with a finite open cover
by affine varieties. This is a topological space \(X\) with an open
cover \(\left\{{U_i}\right\}_{i=1}^n \rightrightarrows X\) such that
\((U_i, { \left.{{{\mathcal{O}}_X}} \right|_{{U_i}} } )\) is isomorphic
to an affine variety. We'll call \({\mathcal{O}}_X\) the sheaf of
\emph{regular functions} and \(U_i\subset X\) \emph{affine open sets}.

\end{definition}

One way to construct prevarieties from affine varieties is by
\emph{gluing}:

\begin{definition}[Glued Spaces]

Let \(X_1, X_2\) be prevarieties which are themselves actual varieties,
and let \(U_{12} \subset X_1, U_{21} \subset X_2\) be opens with
\(f: U_{12} \to U_{21}\) an isomorphism of ringed spaces.

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-10-15-10-08-59.png}
\caption{Image}
\end{figure}

As a set, take \(X = X_1 {\coprod}X_2/\sim\) where \(a\sim f(a)\) for
all \(a\in U_{12}\). As a topological space, \(U \subset X\) is open iff
\(U_i \coloneqq U\cap X_i\) are open in \(X_i\). As a ringed space, we
take
\begin{align*}
{\mathcal{O}}_X(U) \coloneqq\left\{{\phi: U\to k {~\mathrel{\Big|}~}{ \left.{{\phi}} \right|_{{U_i}} } \in {\mathcal{O}}_{X_i}}\right\}
\end{align*}

\end{definition}

\begin{example}

The prototypical example is \({\mathbb{P}}^1_{/k}\) constructed from two
copies of \({\mathbb{A}}^1_{/k}\). Set
\(X_1 = {\mathbb{A}}^1, X_2 = {\mathbb{A}}^2\), with
\(U_{12} \coloneqq D(x) \subset X_1\) and
\(U_{21} \coloneqq D(y) \subset X_2\). Then let
\begin{align*}  
f: U_{12} &\to U_{21} \\
x & \mapsto {1\over x}
.\end{align*}
This defines a regular function on \(U_{12}\) so defines a morphism
\(U_{12} \xrightarrow{\sim} {\mathbb{A}}^1\).

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-10-15-10-20-32.png}
\caption{Gluing two affine lines to obtain projective space.}
\end{figure}

Over \({\mathbb{C}}\), topologically this yields a sphere

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-10-15-10-23-24.png}
\caption{Complex projective space is a sphere.}
\end{figure}

Given a ringed space \(X = X_1\cup X_2\) with a structure sheaf
\({\mathcal{O}}_X\), what is \({\mathcal{O}}_X(X)\)? By definition, it's
\begin{align*}  
{\mathcal{O}}_X(X) \coloneqq\left\{{\phi: X\to k {~\mathrel{\Big|}~}{ \left.{{\phi}} \right|_{{X_1}} }, { \left.{{\phi}} \right|_{{X_2}} } \text{ are regular} }\right\}
.\end{align*}

Then if \({ \left.{{\phi}} \right|_{{X_1}} } = f(x)\) and
\({ \left.{{\phi}} \right|_{{X_2}} } = g(y)\), we have \(y=1/x\) on the
overlap and so
\begin{align*}
{ \left.{{f(x)}} \right|_{{D(x)}} } = { \left.{{g(1/x)}} \right|_{{D(x)}} }
\end{align*}
Since \(f, g\) are rational functions agreeing on an infinite set,
\(f(x) = g(1/x)\) both being polynomial forces \(f = g = c\) for some
constant \(c \in k\). Thus \({\mathcal{O}}_X(X) = k\).

What about \({\mathcal{O}}_X(X_1)\)? This is just \(k[x]\), and
similarly \({\mathcal{O}}_X(X_2) = k[y]\). We can also consider
\({\mathcal{O}}_X(X_1\cap X_2) = D(x) \subset X\), so this yields
\(k[x, 1/x]\). We thus have a diagram

\begin{center}
\begin{tikzcd}
                         & {\mathcal{O}}_X(X_1) = k[x] \ar[rd, "x\mapsto x"]   & \\
{\mathcal{O}}_X(X) \ar[ru] \ar[rd] &                                           & {\mathcal{O}}_X(X_1\cap X_2) = k[x, 1/x] \\
                         & {\mathcal{O}}_X(X_2) = k[y] \ar[ru, "y\mapsto 1/x"] &
\end{tikzcd}
\end{center}

\end{example}

\hypertarget{prevarieties-tuesday-october-20}{%
\section{Prevarieties (Tuesday, October
20)}\label{prevarieties-tuesday-october-20}}

\hypertarget{gluing-two-opens-in-a-prevariety}{%
\subsection{Gluing Two Opens in a
Prevariety}\label{gluing-two-opens-in-a-prevariety}}

Recall that a \emph{prevariety} is a ringed space that is locally
isomorphic to an affine variety, where we recall that
\((X, {\mathcal{O}}_X)\) is \emph{locally isomorphic} to an affine
variety iff there exists an open cover \(U_i \rightrightarrows X\) such
that \((U_i, {\mathcal{O}}_{U_i})\). We found one way of producing
these: the gluing construction. Given two ringed spaces
\((X_1, {\mathcal{O}}_{X_1})\) and \((X_2, {\mathcal{O}}_{X_2})\) and
open sets \(U_{12} \in X_1\) and \(U_{21} \in X_2\) and an isomorphism
\((U_{12}, {\mathcal{O}}_{U_{12}}) \xrightarrow{f} (U_{21}, {\mathcal{O}}_{U_{21}})\),
we defined

\begin{itemize}
\tightlist
\item
  The topological space as \(X_1 {\coprod}_f X_2\)
\item
  The sheaf of rings as
  \({\mathcal{O}}_X = \left\{{\phi:U\to k {~\mathrel{\Big|}~}{ \left.{{\phi}} \right|_{{U\cap X_i}} } \text{ is regular for } i=1,2 }\right\}\).
\end{itemize}

\begin{example}

\({\mathbb{P}}^1_{/k} = X_1 \cup X_2\) where
\(X_1 \cong X_2 \cong {\mathbb{A}}^1_{/k}\). Take
\(U_{12} \coloneqq D(x)\) and \(U_{21} \coloneqq D(y)\) with
\begin{align*}  
f: U_{12} &\to U_{21} \\
x &\mapsto y \coloneqq{1\over x}
.\end{align*}

\begin{figure}
\centering
\includegraphics{figures/image_2020-10-20-09-41-55.png}
\caption{Supposing \(\operatorname{ch}(k) \neq 2\). Note that for
\({\mathbb{C}}\) this recovers \(S^2\) in the classical topology.}
\end{figure}

\end{example}

\begin{example}

Let \(X_i = {\mathbb{A}}^1\) and \(U_{12} = D(x), U_{21} = D(y)\) with
\begin{align*}  
f: U_{12} &\to U_{21} \\
x &\mapsto y\coloneqq x
.\end{align*}

\begin{figure}
\centering
\includegraphics{figures/image_2020-10-20-09-44-41.png}
\caption{Line with the doubled origin.}
\end{figure}

Define \(X\coloneqq X_1 {\coprod}_f X_2\), then
\({\mathcal{O}}_X = \left\{{\phi: X\to k {~\mathrel{\Big|}~}{ \left.{{\phi}} \right|_{{X_i}} } \text{ is regular}}\right\} \cong k[x]\).

\end{example}

\hypertarget{more-general-gluing}{%
\subsection{More General Gluing}\label{more-general-gluing}}

Now we want to glue more than two open sets. Let \(I\) be an indexing
set for prevarieties \(X_i\). Suppose that for an ordered pair
\((i, j)\) we have open sets \(U_{ij} \subset X_i\) and isomorphisms
\(f_{ij}: U_{ij} \xrightarrow{\sim} U_{ji}\) such that

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(f_{ji} = f_{ij}^{-1}\)
\item
  \(f_{jk} \circ f_{ij} = f_{ik}\) (cocycle condition)
\end{enumerate}

\begin{figure}
\centering
\includegraphics{figures/image_2020-10-20-09-54-14.png}
\caption{Opens with isomorphisms.}
\end{figure}

Then the gluing construction is given by

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(X\coloneqq{\coprod}X_i/\sim\) where \(x\sim f_{ij}(x)\) for all
  \(i,j\) and all \(x\in U_{ij}\).
\item
  \({\mathcal{O}}_x(U) \coloneqq\left\{{\phi:U\to k {~\mathrel{\Big|}~}{ \left.{{\phi}} \right|_{{U\cap X_i}} } \in {\mathcal{O}}_{X_i} }\right\}\).
\end{enumerate}

Every prevariety arises from the gluing construction applied to affine
varieties \(X_i\), since a prevariety \((X, {\mathcal{O}}_X)\) by
definition has an open affine cover
\(\left\{{X_i}\right\} \rightrightarrows X\) and \(X\) is the result of
gluing the \(X_i\) by the identity.

\begin{example}

Let \(X_1 = X_2 = X_3 \coloneqq{\mathbb{A}}^2/k\). Glue by the following
instructions:

\begin{center}
\begin{tikzcd}
{\left[ {x_1, x_2} \right]} \in X_1 
\ar[rdd, "{{\left[ {y_1, y_2} \right]} = {\left[ { {1\over x_1}, {x_2\over x_1} } \right]}}"']
\ar[rr, "{{\left[ {z_1, z_2} \right]} = {\left[ { {1\over x_2}, {x_1\over x_2} } \right]} }"]
& & 
{\left[ {z_1, z_2} \right]} \in X_3 \ar[ldd, "f"] \\ \\
& 
{\left[ {y_1, y_2} \right]} \in X_2 \\ 
& 
\end{tikzcd}
\end{center}

where \(f\) is a map with whatever formula is required to make the
diagram commute. Here

\begin{itemize}
\tightlist
\item
  \(D(x_1 x_2) \subseteq X_1\)
\item
  \(U_{12} = D(x_1)\)
\item
  \(U_{21} = D(x_2)\).
\end{itemize}

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-10-20-10-13-56.png}
\caption{Yields \({\mathbb{P}}^2\)}
\end{figure}

Here \(X_1 = [1: y/x: z/x]\), \(X_2 = [x/y: 1: z/y]\).

\end{example}

\begin{example}

From Gathmann 5.10, open and closed subprevarieties. Let \(X\) be a
prevariety and suppose \(U\subset X\) is open. Then
\((U, {\mathcal{O}}_U)\) is a prevariety where
\({\mathcal{O}}_U = { \left.{{{\mathcal{O}}_X}} \right|_{{U}} }\). How
can we write \(U\) as (locally) an affine variety? Since the \(U_i\) are
covered by distinguished opens \(D_{ij}\) in \(X_i\) where
\(X = \cup X_i\) with \(X_i\) affine varieties, we can write
\(U = \bigcup_i U_i = \bigcup_{i, j} D_{ij}\).

\end{example}

\begin{example}

Let \(Y\subset X\) be a closed subset of a prevariety \(X\). We need to
define \({\mathcal{O}}_Y(U)\) for all \(U\subset Y\) open, so we set
\begin{align*}  
{\mathcal{O}}_Y(U) = \left\{{\phi: U\to k {~\mathrel{\Big|}~}\forall p\in U, \, \exists V_p \text{ with } p\in V_p 
  \mathrel{\stackunder[2pt]{\stackon[4pt]{$\subseteq$}{$\scriptscriptstyle$}}{
  $\scriptscriptstyle\text{open}$}}
 X \text{ and } \psi\in {\mathcal{O}}_X(V_p) \text{ s.t. } { \left.{{\psi}} \right|_{{U\cap V}} } = \phi  }\right\}
.\end{align*}

What's the picture?

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-10-20-10-29-25.png}
\caption{Sheaf for a closed subset.}
\end{figure}

\end{example}

\begin{exercise}[?]

Show that this is a prevariety.

\end{exercise}

\begin{remark}

If \(U\subset X\) is an open subprevariety or \(Y\subset X\) is a closed
subprevariety, then the inclusions are morphisms. We'd need to show that
a pullback of a function is regular, but this is set up by definition.

\end{remark}

\begin{remark}

Define \(\tilde {\mathcal{O}}_X(U)\) as the set of \emph{all} functions
\(U\to k\). Then the inclusion
\((X, {\mathcal{O}}_X) \hookrightarrow(X, \tilde {\mathcal{O}}_X)\)
given by the identity on \(X\) is a morphism, but the identity in the
reverse direction is not.

\end{remark}

\hypertarget{separatedness-thursday-october-22}{%
\section{Separatedness (Thursday, October
22)}\label{separatedness-thursday-october-22}}

\begin{example}

Consider \({\mathbb{A}}^1\), whose polynomial functions are \(k[x]\).
Consider now \(D(x) \subset {\mathbb{A}}^1\), which is isomorphic to the
affine variety \(V(xy-1)\). Then the regular functions on \(D(x)\) are
given by
\begin{align*}  
A(D(x)) = 
\frac{ k[x, y] }{ \left\langle{xy-1}\right\rangle } \cong k[x, x^{-1} ]
.\end{align*}

\end{example}

\hypertarget{products-of-prevarieties}{%
\subsection{Products of Prevarieties}\label{products-of-prevarieties}}

Recall that a \emph{prevariety} is a ringed space
\((X, {\mathcal{O}}_X)\) such that \(X\) has a finite open cover by
affine varieties
\((U_i, { \left.{{{\mathcal{O}}_X}} \right|_{{U_i}} })\), and a
\emph{morphism} of prevarieties is a morphism of ringed spaces. We saw
that one can construct prevarieties by gluing finite collections of
prevarieties or affine varieties along open sets, and all prevarieties
arise this way. Similar to varieties, the product \(P\) of prevarieties
\(X, Y\) will satisfy a universal property:

\begin{center}
\begin{tikzcd}
P \arrow[rrd, "\pi_X", bend left] \arrow[rdd, "\pi_Y"', bend right] &                                                     &   \\
                                                                    & Z \arrow[d] \arrow[r] \arrow[lu, "\exists !", dashed] & X \\
                                                                    & Y                                                   &  
\end{tikzcd}
\end{center}

\href{https://tikzcd.yichuanshen.de/\#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAAUQBfU9TXfIRQBGUsKq1GLNgC1uvEBmx4CRAExiJ9Zq0QgAGvL7LBRUWq1TdIAJrcJMKAHN4RUADMAThAC2SUSA4EEgAzDwe3n6IAUFIauEgXr7+1LGIZCAMWGDWUHRwABaOINTa0noAOhUwAB5YcDgICUlRGWkaIABGMGBQoRll1lVoWAD6hs2RSG3BiCHU3b1IALQhA1Zsw2N21Ax03Qzs-CpCIJ5YTgU49lxAA}{Link
to Diagram}

\begin{proposition}[?]

The product is unique up to unique isomorphism, i.e.~there is a unique
isomorphism between any two products.

\end{proposition}

\begin{proof}

Standard exercise in category theory.

\end{proof}

\hypertarget{constructing-the-product-of-prevarieties}{%
\subsubsection{Constructing the product of
prevarieties}\label{constructing-the-product-of-prevarieties}}

\begin{example}

Consider \({\mathbb{A}}^1 \times {\mathbb{A}}^1\), then the product is
(and should be) \({\mathbb{A}}^2\), but \({\mathbb{A}}^2\) does not have
the product topology. For example, one problem is that the Zariski open
set \(D(x-y)\) is not covered by products of open sets in
\({\mathbb{A}}^1\).

\end{example}

This happens because the Zariski topology is too weak. Strategy to fix:
use gluing. Let \(X, Y\) be prevarieties and
\(\left\{{U_i}\right\}, \left\{{V_i}\right\}\) be open affine covers of
\(X\) and \(Y\) respectively. We can construct the product
\(U_i \times V_j \subset {\mathbb{A}}^{n+m}\), which is an affine
variety and satisfies the universal property for products. We then glue
two such products \(U_{i_1} \times V_{j_1}\) and
\(U_{i_2} \times V_{j_2}\) along their common open subset in
\(\qty{U_{i_1}\cap U_{i_2} }\cap\qty{V_{j_1} \cap V_{j_2}} \subseteq X\times Y\).

Let
\(\tilde U \coloneqq U_{i_1} \cap U_{i_2} \times V_{j_1} \cap V_{j_2}\),
we then need that

\begin{align*}  
\qty{ \tilde U, { \left.{{ {\mathcal{O}}_{U_{i_1} \times V_{j_1}} }} \right|_{{\tilde U}} } } \cong
\qty{ \tilde U, { \left.{{ {\mathcal{O}}_{U_{i_2} \times V_{j_2}} }} \right|_{{\tilde U}} } }
.\end{align*}
This follows from the universal property of products, since the open set
\((U\times V, { \left.{{ {\mathcal{O}}_{X\times Y} }} \right|_{{U\times V}} } )\)
is a categorical product of ringed spaces, and the identity provides a
unique isomorphism. By the gluing construction, this produces a ringed
space \((X\times Y, {\mathcal{O}}_{X\times Y})\), we just need to check
that this satisfies the universal property. We have projections
\(\pi_X, \pi_Y\) set-theoretically, which restrict to morphisms on every
\(U_i \times V_j\). For any prevariety \(Z\), we get a unique set map
\(h:Z\to X\times Y\) which commutes, so it suffices to check that \(h\)
is a morphism of ringed spaces.

So consider \(h^{-1} (U_i \times V_j) \subset Z\), which is an open
subset of \(Z\) given by \(f^{-1}(U) \times f^{-1}(V)\). Take an open
cover and let \(W\) be an element in it. We can then restrict \(f\) and
\(g\) to get \({ \left.{{f}} \right|_{{W}} }:W\to U_i\) and
\({ \left.{{g}} \right|_{{W}} }:W\to V_j\) and their product is a
morphism of ringed spaces. So \(Z\) is covered by open sets for which
\(h\) is a morphism of ringed spaces, making \(h\) itself a morphism.

What was the point of constructing the product? We want some notion
analogous to being Hausdorff to distinguish spaces like
\({\mathbb{P}}^1/k\) from the line with the doubled origin. The issue is
that these spaces with the Zariski topology are never Hausdorff. So we
make the following definition:

\begin{definition}[Separated]

A prevariety is \textbf{separated} iff the diagonal morphism
\begin{align*}  
\Delta_X: X &\to X\times X \\
x &\mapsto (\operatorname{id}_X \times\operatorname{id}_X)(x) \coloneqq(x, x)
\end{align*}
is a closed embedding.

\end{definition}

\begin{definition}[Variety]

A \textbf{variety} is a separated prevariety.

\end{definition}

\hypertarget{general-varieties-and-completeness-tuesday-october-27}{%
\section{General Varieties and Completeness (Tuesday, October
27)}\label{general-varieties-and-completeness-tuesday-october-27}}

Recall the following:

\begin{itemize}
\item
  An \emph{affine variety} is given by
  \(X = V(I) \subset {\mathbb{A}}^n/k\), and we have sheaves of rings of
  regular functions \({\mathcal{O}}_X\) on \(X\).
\item
  A \emph{prevariety} is a ringed space that is covered by finitely many
  affine spaces.
\item
  A \emph{morphism} of prevarieties \(f:X\to Y\) is a continuous map
  such that the pullbacks of regular functions are regular, i.e.~for all
  \(\phi \in {\mathcal{O}}_X(U)\) we have
  \(f^* \phi \in {\mathcal{O}}_X(f^{-1} (U) )\).
\item
  We can form a category \(\operatorname{PreVar}_k\) of prevarieties
  over \(k\), where we have several important constructions

  \begin{enumerate}
  \def\labelenumi{\arabic{enumi}.}
  \item
    Gluing
  \item
    Products: Given \(X, Y\), there is a unique prevariety \(X\times Y\)
    such that

    \begin{center}
    \begin{tikzcd}
      Z\ar[drr, bend left , "f_x"] \ar[rdd, bend right, "f_y"] \ar[rd, "\exists ! h", dotted] &                                         & \\
                                                                                           & X\times Y\ar[r, "\pi_X"]\ar[d, "\pi_Y"] & X \\
                                                                                           & Y                                       &
      \end{tikzcd}
    \end{center}
  \end{enumerate}
\item
  We had an analogue of being Hausdorff: the diagonal \(\Delta_X\) is
  closed.
\end{itemize}

\begin{example}[The line with double origins is not separated]

Glue \(D(x) \subset {\mathbb{A}}^1\) to \(D(y) \subset {\mathbb{A}}^1\)
by the isomorphism
\begin{align*}  
D(x) & \xrightarrow{\sim} D(y) \\
x &\mapsto y
.\end{align*}
This yields an affine line with two origins:

\begin{figure}
\centering
\includegraphics[width=2.60417in,height=\textheight]{figures/image_2020-10-27-09-47-27.png}
\caption{Line with two origins.}
\end{figure}

Consider the product:

\begin{figure}
\centering
\includegraphics[width=2.60417in,height=\textheight]{figures/image_2020-10-27-09-51-10.png}
\caption{Product of lines with two origins}
\end{figure}

Since the diagonal is given by
\(\Delta_X = \left\{{(x, x) {~\mathrel{\Big|}~}x\in X}\right\}\), we
have the following situation in blue:

\begin{figure}
\centering
\includegraphics[width=2.60417in,height=\textheight]{figures/image_2020-10-27-09-52-26.png}
\caption{The diagonal has four copies of the origin.}
\end{figure}

We claim that \(\Delta_X\) is not closed, and for example
\((0, 0') \in \mkern 1.5mu\overline{\mkern-1.5mu\Delta\mkern-1.5mu}\mkern 1.5mu_X\).
Consider \(U\times U' \subset X\times X\) where \(U, U'\) are the two
copies of \({\mathbb{A}}^1\) in \(X\). This is an affine open set, since
it's isomorphic to \({\mathbb{A}}^1\times{\mathbb{A}}^1\). If
\(\Delta_X\) were closed, then
\(S \coloneqq\Delta_X \cap(U\times U') = \left\{{(x, x) {~\mathrel{\Big|}~}x\neq 0}\right\}\)
would be closed in \(U\times U'\):

\begin{figure}
\centering
\includegraphics[width=2.60417in,height=\textheight]{figures/image_2020-10-27-10-01-15.png}
\caption{Open diagonal in a product.}
\end{figure}

This is because any polynomial vanishing on \(S\) must vanish at
\((0, 0)\), so \(S\) is an affine variety. But then
\(V(I(S)) = \Delta_{{\mathbb{A}}^1}\).

\end{example}

\begin{lemma}[Affine varieties and affine subvarieties are varieties]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  Any affine variety is a variety.
\item
  Open and closed subprevarieties of a variety \(X\) are themselves
  varieties.
\end{enumerate}

Thus it makes since to consider \emph{open} and \emph{closed
subvarieties}.

\end{lemma}

\begin{proof}[of a]

We need to check that \(\Delta_X \subset X^2\) is closed for any affine
\(X\subset {\mathbb{A}}^n\). Note that we can write.
\begin{align*}
\Delta_X = X^2 \cap V\qty{ \left\{{x_j - y_j {~\mathrel{\Big|}~}1\leq j \leq n}\right\}} \subset {\mathbb{A}}^n \times{\mathbb{A}}^2
\end{align*}

\end{proof}

\begin{proof}[of b]

Let \(\iota:Y\to X\) be the inclusion of either an open or closed
subset. Then we have a morphism \((\iota, \iota): Y^2 \to X^2\) by the
universal property. Then \(\Delta_Y = (\iota, \iota)^{-1} (\Delta_X)\),
so is closed by the continuity of \((\iota, \iota)\) and the fact that
\(\Delta_X\). Thus \(Y\) is a variety.

\end{proof}

\hypertarget{properties-of-varieties}{%
\subsection{Properties of Varieties}\label{properties-of-varieties}}

\begin{proposition}[Properties of Varieties]

Let \(f, g: X\to Y\) be morphisms of prevarieties and assume \(Y\) is a
variety.

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  The graph of \(f\), given by
  \(\Gamma_f \coloneqq\left\{{(x, f(x)) {~\mathrel{\Big|}~}x\in X}\right\}\),
  is closed in \(X\times Y\).
\item
  The set \(\left\{{x\in X{~\mathrel{\Big|}~}f(x) = g(x)}\right\}\) is
  closed in \(X\).
\end{enumerate}

\end{proposition}

\begin{proof}[of a]

Consider the product morphism
\((f, \operatorname{id}): X\times Y \to Y^2\). Since \(\Delta_Y\) is
closed, \((f, \operatorname{id})^{-1} (\Delta_Y)\) is closed, and is the
locus where \(f(x) = y\), so this is \(\Gamma_f\).

\end{proof}

\begin{proof}[of b]

Consider \((f, g): X\to Y^2\). Since \(\Delta_Y \subset Y^2\) is closed,
\begin{align*}
(f, g)^{-1}(\Delta_Y) = \left\{{x\in X {~\mathrel{\Big|}~}f(x) = g(x)}\right\} \subset X
\end{align*}

is closed.

\end{proof}

\hypertarget{chapter-6-projective-varieties}{%
\subsection{Chapter 6: Projective
Varieties}\label{chapter-6-projective-varieties}}

Note that affine varieties of positive dimension over \({\mathbb{C}}\)
are not compact in the classical topology, but \emph{are} compact in the
Zariski topology. Similarly, they are Hausdorff classically, but not in
the Zariski topology. We want to find notions equivalent to being
Hausdorff and compact that coincide with these notions in the classical
topologies but generalize to varieties. The fix for being Hausdorff case
was ``separatedness'', and the fix for compactness will be the
following:

\begin{definition}[Complete]

A variety \(X\) is \textbf{complete} iff for any variety \(Y\) the
projection map
\begin{align*}
\pi_Y:X\times Y\to Y
\end{align*}
is a closed\footnote{Recall that this means that \(\pi_Y(U)\) is closed
  whenever \(U\) is closed.} map.

\end{definition}

\begin{example}[$\AA^1$ is not complete]

Let \(X = Y={\mathbb{A}}^1\) and set
\(Z \coloneqq V(xy-1)\subset X\times Y\). Then
\(\pi_Y(Z) = D(y) \subset Y \subset {\mathbb{A}}^1\) is not closed.

\end{example}

\hypertarget{projective-space-homogeneous-polynomials-thursday-october-29}{%
\section{Projective Space \& Homogeneous Polynomials (Thursday, October
29)}\label{projective-space-homogeneous-polynomials-thursday-october-29}}

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

\begin{definition}[Projective Space]

Let \(n\in {\mathbb{N}}\), and define \textbf{projective
\(n{\hbox{-}}\)space} over \(k\) by\\
\begin{align*}  
{\mathbb{P}}^n_{/k} = \left\{{\text{lines through the origin in } k^{n+1}}\right\}
.\end{align*}

\end{definition}

\begin{remark}

For notation, given \(L\in {\mathbb{P}}^n/k\), it is spanned by any
nonzero points \({\left[ {x_0, \cdots, x_n} \right]} \in L\), and \(L\)
is uniquely determined by this point up to scaling by elements in
\(k^{\times}\). In this case, we write
\(L = {\left[ {x_0: \cdots : x_n} \right]} = {\left[ {\lambda x_0: \cdots : \lambda x_n} \right]}\).
We can then alternatively define
\({\mathbb{P}}^n{_/k} \coloneqq\qty{ k^{n+1}\setminus\left\{{0}\right\}} / \sim\)
where we mod out by scalar multiplication \(x\sim \lambda x\) for
\(\lambda\in k^{\times}\). We call \([x_1 : \cdots : x_n]\) the
\emph{homogeneous coordinates} on \({\mathbb{P}}^n/k\).

\end{remark}

\begin{remark}

Consider the map
\begin{align*}  
{\mathbb{A}}^n \to {\mathbb{P}}^n \\
{\left[ {x_1, \cdots, x_n} \right]} &\mapsto {\left[ {1: x_1 : \cdots : x_n} \right]}
.\end{align*}
This is injective. Conversely, consider
\begin{align*}  
"D(x_0)" \subset {\mathbb{P}}^n \coloneqq\left\{{{\left[ {x_0 : \cdots : x_n} \right]} {~\mathrel{\Big|}~}x_0\neq 0}\right\}
.\end{align*}
This is a well-defined subset of \({\mathbb{P}}^n\), since it only
depends on the equivalence class of a point. In this case, there is a
unique \(\lambda(x_0, \cdots, x_n)\), namely \(\lambda = 1/x_0\), such
that each point in this set is of the form
\({\left[ {1: {x_1\over x_0} : \cdots : {x_n \over x_0}} \right]}\),
yielding a copy of \({\mathbb{A}}^n\subset {\mathbb{P}}^n\) given by
points \({\left[ {{x_1\over x_0}, \cdots, {x_n\over x_0}} \right]}\).
What is its complement? It's given by
\(\left\{{{\left[ {0: x_1: \cdots : x_n} \right]}}\right\} \subset {\mathbb{P}}^n\),
which is equal (as a set) to a copy of \({\mathbb{P}}^{n-1}\) defined by
the set of lines in \(k^n\) defined by \(x_0 = 0\).

\end{remark}

\begin{example}[?]

Note that \({\mathbb{P}}^1\) contains a copy of \({\mathbb{A}}^1\) where
\(x_0 \neq 0\) and a second copy where \(x_1 \neq 0\), yielding maps
\begin{align*}  
f_1: {\mathbb{P}}^1 &\to {\mathbb{A}}^1 \\
{\left[ {x_0: x_1} \right]} &\mapsto {\left[ {{x_0 \over x_1}} \right]} \\ \\
f_2: {\mathbb{P}}^1 &\to {\mathbb{A}}^1 \\
{\left[ {x_0: x_1} \right]} &\mapsto {\left[ {{x_1 \over x_0}} \right]}
,\end{align*}
since every point in \({\mathbb{P}}^1\) corresponds to some line in
\({\mathbb{A}}^2\), and thus has either \(x_0\neq 0\) or \(x_1 \neq 0\).
These two copies cover \({\mathbb{P}}^1\), and the ``transition map'' is
inversion.

\end{example}

\begin{remark}

More generally, there are \(n+1\) projection
\({\mathbb{P}}^n \twoheadrightarrow{\mathbb{A}}^n\) given by dividing by
the \(j\)th coordinate, and the union of their images is the entire
space. The gluing construction gives \({\mathbb{P}}^n\) the structure of
a prevariety: we can consider \(D(x_j) \subset {\mathbb{P}}^n\) where
each has the structure of a ringed space
\(({\mathbb{A}}^n, {\mathcal{O}}_{{\mathbb{A}}^n})\). We have
\(D(x_i) \cap D(x_j) \subset D(x_i)\), which has coordinate
\(\left\{{ x_k/x_i,\, k\neq i }\right\}\), and similarly
\(D(x_i) \cap D(x_j) \subseteq D(x_j)\) with coordinates
\(\left\{{ x_k/x_j, \, k\neq j}\right\}\). Their intersection is
\(D(x_i / x_j) \cong {\mathbb{A}}^{n-1}\).

\end{remark}

\begin{example}[?]

Consider \({\mathbb{P}}^1\), then \(D(x_0) \cong {\mathbb{A}}^1\) with
which contains a copy of \({\mathbb{A}}^1\) with coordinate ring
\(k[x_1 / x_0]\) and a subset \(D(x_1 / x_0)\) with coordinate ring
\(k[y, y^{-1}]\), and similarly, \(D(x_1) \cong {\mathbb{A}}^1\) has
coordinate ring \(k[x_0\over x_1]\) and contains \(D(x_0/ x_1)\) with
coordinate ring \(k[z, z^{-1}]\). Consider their overlap
\(D(x_0) \cap D(x_1)\). \todo[inline]{Might be mistakes here.}

When do \(y, z\) denote the same point in \({\mathbb{P}}^1\)? When
\(y = z^{-1}\). We can conclude that the \(n+1\) copies
\(D(x_i) \subset {\mathbb{P}}^n\) are affine varieties isomorphic as
ringed spaces on the overlaps, so the gluing construction makes
\({\mathbb{P}}^n\) a prevariety.

\end{example}

\hypertarget{homogeneous-polynomials}{%
\subsection{Homogeneous Polynomials}\label{homogeneous-polynomials}}

\begin{definition}[Homogeneous Degrees of Polynomials]

A polynomial \(f\) is \textbf{homogeneous of degree \(d\)} if every
monomial in \(f\) has total degree \(d\).

\end{definition}

\begin{example}[?]

The polynomial
\begin{align*}  
f(x_0, x_1, x_2) = x_0^3 + x_1 x_2^2 + x_0 x_1 x_2
.\end{align*}
has homogeneous degree 3.

\end{example}

\begin{remark}

If \(f\) is homogeneous of degree \(d\), then for all
\(\lambda \in k^{\times}\),
\begin{align*}  
f(\lambda x_0, \cdots, \lambda x_n) = \lambda^d f(x_0, \cdots, x_n)
.\end{align*}

If \(f\) is homogeneous, \(V(f) \subset {\mathbb{P}}^n\) is a
well-defined subset, since
\begin{align*}
f(x_0, \cdots, x_n) = 0 \iff \lambda^d f(x_0, \cdots, x_n) = 0 \iff f(\lambda x_0, \cdots, \lambda x_n) = 0
\end{align*}

\end{remark}

\begin{definition}[Graded Rings]

A \textbf{graded ring} \(R\) is a ring \(R\) with abelian subgroups
\(R_d \subset R\) with

\begin{itemize}
\tightlist
\item
  \(R = \bigoplus_{d\geq 0} R_d\), and
\item
  For all \(f\in R_d\) and \(g\in R_{d'}\), we have \(fg \in R_{d+d'}\)
  and \(R_d + R_{d} \subset R_d\).
\end{itemize}

\end{definition}

\hypertarget{projective-nullstellensatz-tuesday-november-02}{%
\section{Projective Nullstellensatz (Tuesday, November
02)}\label{projective-nullstellensatz-tuesday-november-02}}

\hypertarget{quotients-of-graded-rings}{%
\subsection{Quotients of Graded Rings}\label{quotients-of-graded-rings}}

\begin{proposition}[Quotients of graded rings by homogeneous ideals are again graded]

If \(R\) is a graded ring and \(I{~\trianglelefteq~}R\) is a homogeneous
ideal, then \(R/I\) is a graded ring.

\end{proposition}

\hypertarget{cones-and-projectivization}{%
\subsection{Cones and
Projectivization}\label{cones-and-projectivization}}

\begin{definition}[Cones]

An affine variety \(X \subseteq {\mathbb{A}}^{n+1}\) is a \textbf{cone}
iff

\begin{itemize}
\tightlist
\item
  \(\mathbf{0} \in X\)
\item
  \(kX \subseteq X\)
\end{itemize}

\end{definition}

\begin{remark}

This says that \(X\) is the origin and a union of lines through the
origin. For the following definitions, we define a map
\begin{align*}  
\pi: {\mathbb{A}}^{n+1}\setminus\left\{{0}\right\}&\to {\mathbb{P}}^n \\
{\left[ {x_0, \cdots, x_n} \right]} &\mapsto {\left[ {x_0 : \cdots :x_n} \right]}
.\end{align*}

\end{remark}

\begin{definition}[Projectivization of a Cone]

For a cone \(X \subseteq {\mathbb{A}}^{n+1}\), the
\textbf{projectivization} of \(X\) is defined as
\begin{align*}  
{\mathbb{P}}(X) \coloneqq\pi(X\setminus\left\{{0}\right\}) = \left\{{ {\left[ {x_0: \cdots : x_n } \right]} \in {\mathbb{P}}^n {~\mathrel{\Big|}~}{\left[ {x_0, \cdots, x_n} \right]} \in X }\right\} \subseteq {\mathbb{P}}^n
.\end{align*}

\end{definition}

\begin{definition}[Cone Over a Projective Variety]

For a projective variety \(X \subseteq {\mathbb{P}}^n\), the
\textbf{cone over \(X\)} is the cone defined by
\begin{align*}  
C(X) \coloneqq\left\{{0}\right\} \cup\pi^{-1}(X) = \left\{{0}\right\} \cup\left\{{ {\left[ {x_0, \cdots, x_n} \right]} {~\mathrel{\Big|}~}{\left[ {x_0: \cdots : x_n} \right]} \in X }\right\} \subseteq {\mathbb{A}}^{n+1}
.\end{align*}

\end{definition}

\begin{remark}

We have
\begin{align*}  
{\mathbb{P}}V_a(S) = V_p(S) &\text{and}& C(V_p(S)) = V_a(S)
.\end{align*}

\end{remark}

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

\begin{proposition}[Projective Nullstellensatz Construction]

Define
\begin{align*}  
V_p(J) &\coloneqq\left\{{\mathbf{x} \in {\mathbb{P}}^n {~\mathrel{\Big|}~}f(\mathbf{x}) = 0 \text{ for all homogeneous } f\in J}\right\} \subseteq {\mathbb{P}}^n \\
I_p(X) &\coloneqq\left\langle{ f \in k[x_0, \cdots, x_n] \text{ homogeneous } {~\mathrel{\Big|}~}f(\mathbf{x}) = 0\,\, \forall x\in X}\right\rangle {~\trianglelefteq~}k[x_0, \cdots, x_n]
.\end{align*}

\end{proposition}

\todo[inline]{Missing some info, fill in.}

\hypertarget{projective-spaces-thursday-november-05}{%
\section{Projective Spaces (Thursday, November
05)}\label{projective-spaces-thursday-november-05}}

We defined
\({\mathbb{P}}^n_{/k} \coloneqq\qty{ k^{n+1}\setminus\left\{{0}\right\}} /\sim\)
where \(x\sim \lambda x\) for all \(x\in k^{\times}\), which we
identified with lines through the origin in \(k^{n+1}\). We have
homogeneous coordinates \(p = [x_0: \cdots : x_n]\). We say an ideal is
\emph{homogeneous} iff for all \(f\in I\), the homogeneous part
\(f_d\in I\) for all \(d\). In this case
\(V_p(I) \subset {\mathbb{P}}^{n}_{/k}\) defined as the vanishing locus
of all homogeneous elements of \(I\) is well-defined, and we think of
this as the ``projective version'' of a vanishing locus. Similarly we
defined \(I_p(S)\) defined as the ideal generated by all homogeneous
\(f\in k[x_1, \cdots, x_{n}]\) such that \(f(x) = 0\) for all
\(x\in S\).

\begin{remark}

Observe that \(V_a(I)\) defined as the cone over \(V_p(I)\) is the set
of points in
\({\mathbb{A}}^{n+1}\setminus\left\{{0}\right\}\cup\left\{{0}\right\}\)
which map to \(V_p(I)\).

\end{remark}

We have an alternative definition of a cone in \({\mathbb{A}}^{n+1}\),
characterized as a closed subset \(C\) which is closed under scaling, so
\(kC\subseteq C\). The following proposition(s) show that these notions
are related.

\begin{proposition}

\envlist

\begin{itemize}
\item
  If \(S\subset k[x_1, \cdots, x_{n}]\) is a set of homogeneous
  polynomials, then \(V_a(S)\) is a cone since it is closed and closed
  under scaling. This follows from the fact that
  \(f(x) = 0 \iff f(\lambda x) = 0\) for \(\lambda \in k^{\times}\) when
  \(f\) is homogeneous.
\item
  If \(C\) is a cone, then its affine ideal \(I_a(C)\) is homogeneous.
\end{itemize}

\end{proposition}

\begin{proof}[?]

Let \(f\in I_a(C)\), then \(f(x) = 0\) for all \(x\in C\). Since \(C\)
is closed under scaling, \(f(\lambda x) = 0\) for all \(x\in C\) and
\(\lambda \in k^{\times}\). Decompose \(f = \sum_d f_d\) into
homogeneous pieces, then
\begin{align*}  
x\in C \implies 0 = f(\lambda x) = \sum \lambda^d f_d(x) 
.\end{align*}

Fixing \(x\in C\), the quantities \(f_d(x)\) are constants, so the
resulting polynomial in \(\lambda\) vanishes for all \(\lambda\). But
since \(k\) is infinite, this forces \(f_d(x) = 0\) for all \(d\), which
shows that \(f_d \in I_a(C)\).

\end{proof}

\begin{lemma}[?]

There is a bijective correspondence
\begin{align*}  
\left\{{\substack{\text{Cones}}}\right\} 
&\iff
\left\{{\substack{\text{Projective Varieties}}}\right\} \\
{\mathbb{A}}^{n+1} \supset X &\mapsto {\mathbb{P}}X\subset {\mathbb{P}}^n \\
{\mathbb{A}}^{n+1} \supset CX &\mapsfrom X\subset {\mathbb{P}}^n \\
.\end{align*}

\end{lemma}

\begin{proof}[?]

\({\mathbb{P}}V_a(S) = V_p(S)\) for any set \(S\) of homogeneous
polynomials, and \(C(V_p(S)) = V_a(S)\), where \(V_p(S)\) is a cone by
part (a) of the previous proposition. Conversely, every cone is the
variety associated to some homogeneous ideal.

\end{proof}

\hypertarget{projective-nullstellensatz-1}{%
\subsection{Projective
Nullstellensatz}\label{projective-nullstellensatz-1}}

\begin{definition}[Irrelevant Ideal]

The homogeneous ideal
\(I_0 \coloneqq(x_0, \cdots, x_n) \subset k[x_1, \cdots, x_{n}]\) is
denoted the \textbf{irrelevant ideal}. This corresponds to the origin in
\({\mathbb{A}}^{n+1}\), which does not correspond to any point in
\({\mathbb{P}}^n\).

\end{definition}

\begin{proposition}[Projective Nullstellensatz]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  For all \(X\subseteq {\mathbb{P}}^n\),
  \begin{align*}
  V_p(I_p(X)) = X
  \end{align*}
\item
  For all homogeneous ideal \(J\subset k[x_1, \cdots, x_{n}]\) such that
  (importantly) \(\sqrt{J} \neq I_0\),
  \begin{align*}
  I_p(V_p(J)) = \sqrt J
  \end{align*}
\end{enumerate}

\end{proposition}

\begin{proof}[of a]

\(\supset\): If we let \(I\) denote the ideal of all homogeneous
polynomials vanishing on \(X\), then this certainly contains \(X\).

\(\subset\): This follows from part (b), since \(X = V_p(J)\) implies
that
\begin{align*}
(V_p I_p V_p)(J) = V_p(\sqrt J) = V_p(J) = X
\end{align*}
since taking roots of homogeneous polynomials doesn't change the
vanishing locus.

\end{proof}

\begin{proof}[of b]

That \(I_p(V_p(J)) \supset \sqrt J\) is obvious, since \(f\in \sqrt{J}\)
vanishes on \(V_p(J)\). \todo[inline]{Check}

It remains to show \(\sqrt{J} \subset I_p(V_p(J))\) , but we can write
\(I_p(V_p(J))\) as
\(\left\langle{f \in k[x_1, \cdots, x_{n}]}\right\rangle\) the set of
homogeneous polynomials vanishing on \(V_p(S)\), which is equal to those
vanishing on \(V_a(J) \setminus\left\{{0}\right\}\). But since
\(I_p(\cdots)\) is closed, this is equal to the \(f\) that vanish on
\(\mkern 1.5mu\overline{\mkern-1.5muV_a(J)\setminus\left\{{0}\right\}\mkern-1.5mu}\mkern 1.5mu\),
which is only equal to \(V_a(J)\) iff
\(V_a(J) \neq \left\{{0}\right\}\).

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-11-05-10-10-38.png}
\caption{Projective Varieties as Cones in \({\mathbb{A}}^n\)}
\end{figure}

By the affine Nullstellensatz,
\begin{align*}  
V_a(J) = \left\{{0}\right\} \iff \sqrt{J} = I_0
.\end{align*}

Thus
\begin{align*}
I_p(V_p(J)) = \left\langle{f {~\mathrel{\Big|}~}\text{homogeneous vanishing on }V_a(J)}\right\rangle
\end{align*}
Using the fact that \(V_a(J)\) is a cone, its ideal is homogeneous and
thus generated by homogeneous polynomials by part (b) of the previous
proposition. Thus
\begin{align*}  
I_p(V_p(J)) = I_a(V_a(J)) = \sqrt J
,\end{align*}
where the last equality follows from the affine Nullstellensatz.

\end{proof}

\begin{corollary}[?]

There is an order-reversing bijection
\begin{align*}  
\left\{{\substack{\text{Projective varieties} \\ X\subset {\mathbb{P}}^n}}\right\}
&\iff
\left\{{\substack{\text{Homog non-irrelevant radical ideals} \\ J \in k[x_1, \cdots, x_{n}]}}\right\} \\
X &\mapsto I_p(X) \\
V_p(J) &\mapsfrom J
.\end{align*}

\end{corollary}

\begin{remark}

A better definition of a cone over \(X\subset {\mathbb{P}}^n_{/k}\) is
\begin{align*}  
C(X) &\coloneqq\mkern 1.5mu\overline{\mkern-1.5mu\pi^{-1}(X)\mkern-1.5mu}\mkern 1.5mu \subset {\mathbb{A}}^{n+1}_{/k} \\
\text{where} \\
\pi: {\mathbb{A}}^{n+1}\setminus\left\{{0}\right\}&\to {\mathbb{P}}^n \\
{\left[ {x_0, \cdots, x_n} \right]} &\mapsto {\left[ {x_0: \cdots: x_n} \right]}
.\end{align*}

\end{remark}

\hypertarget{projective-coordinate-ring}{%
\subsection{Projective Coordinate
Ring}\label{projective-coordinate-ring}}

\begin{definition}[Homogeneous / Projective Coordinate Ring]

Given \(X\subset {\mathbb{P}}^n\) a projective variety, the
\textbf{projective coordinate ring} of \(X\) is given by
\begin{align*}  
S(X) \coloneqq k[x_1, \cdots, x_{n}] / I_p(X)
.\end{align*}

\end{definition}

\begin{remark}

This is a graded ring since \(I_p(X)\) is homogeneous. This follows
since the quotient of a graded ring by a homogeneous ideal yields a
grading on the quotient.

\end{remark}

\begin{remark}

We have relative versions of everything. Projective subvarieties of
projective varieties are given by \(Y\subset X\subset {\mathbb{P}}^n\)
where \(X\) is a projective variety. We have a topology on \(X\) where
the closed subsets are projective subvarieties.

\end{remark}

\begin{remark}

Given \(J\subset S(X)\), where \(S(X)\) is the projective coordinate
ring of \(X\) and has a grading, we can take \(V_p(J) \subset X\).
Conversely, given a set \(Y\subset S(X)\), we can take
\(I_p(Y) \subset S(X)\) those homogeneous elements vanishing on \(Y\).
Thus there is an order-reversing bijection
\begin{align*}  
\left\{{\substack{\text{Projective subvarieties } \\ Y\subset X}}\right\}
\iff
\left\{{\substack{\text{Homogeneous radical ideals} \\ I \neq I_0 {~\trianglelefteq~}S(X)}}\right\}
\end{align*}
and
\(S(X) = k[x_1, \cdots, x_{n}]/J \subset \mkern 1.5mu\overline{\mkern-1.5muI_0\mkern-1.5mu}\mkern 1.5mu\).

\end{remark}

\begin{remark}

Every nontrivial homogeneous ideal \(J\) contains the irrelevant ideal
\(I_0\). Why? Suppose \(f\in J\setminus I_0\) and \(f_0\neq 0\). Then
\(f_0\in J\) but \(f_0\in k\subset k[x_1, \cdots, x_{n}]\), implying
that \(1\in J\) and thus \(J = \left\langle{1}\right\rangle\).

\end{remark}

\todo[inline]{Check?}

\begin{remark}

It is sometimes useful to know that a projective variety is cut out by
homogeneous polynomials all of \emph{equal} degree, so
\(X = V (f_1, \cdots, f_m)\) with each \(f_i\) homogeneous of degree
\(d_i\). Then there is some maximum degree \(d\). We can write
\begin{align*}  
V(f_1) &= V(x_0^k f_1, \cdots, x_n^k f_1 ) \qquad \forall k\geq 0 \\
X &= \bigcap V(f_1) \cup V(x_i)
.\end{align*}
This follows because \(V\) of a product is a union of the vanishing
loci, but \(\bigcap V(x_i) = \emptyset\). The equality follows because
for all points
\({\left[ {x_0, \cdots, x_n} \right]} \in {\mathbb{P}}^n\), some \(x_i\)
is nonzero.

\end{remark}

\hypertarget{homogenization-and-dehomogenization-tuesday-november-10}{%
\section{Homogenization and Dehomogenization (Tuesday, November
10)}\label{homogenization-and-dehomogenization-tuesday-november-10}}

Last time: projective varieties \(V(f_i) \subset {\mathbb{P}}^n_{/k}\)
with \(f_i\) homogeneous. We proved the projective nullstellensatz: for
any projective variety \(X\), we have \(V_p(I_p(X))\) and for any
homogeneous ideal \(I\) with \(\sqrt{I} \neq I_0\) the irrelevant ideal,
\(I_p(V_p(I)) = \sqrt{I}\). Recall that
\(I_0 = \left\langle{x_0, \cdots, x_n}\right\rangle\). We had a notion
of a projective coordinate ring,
\(S(X) \coloneqq k[x_1, \cdots, x_{n}] / I_p(X)\), which is a graded
ring since \(I_p(X)\) is a homogeneous ideal.

\begin{remark}

Note that \(S(X)\) is not a ring of functions on \(X\): e.g.~for
\(X= {\mathbb{P}}^n\), \(S(X) = k[x_1, \cdots, x_{n}]\) but \(x_0\) is
not a function on \({\mathbb{P}}^n\). This is because
\(f\qty{{\left[ {x_0: \cdots : x_n} \right]}} = f\qty{{\left[ {\lambda x_0: \cdots : \lambda x_n} \right]}}\)
but \(x_0\neq \lambda x_0\). It still makes sense to ask when \(f\) is
zero though, so \(V_p(f)\) is a well-defined object.

\end{remark}

\hypertarget{dehomogenization}{%
\subsection{Dehomogenization}\label{dehomogenization}}

\begin{definition}[Dehomogenization of functions and ideals]

Let \(f\in k[x_1, \cdots, x_{n}]\) be a homogeneous polynomial, then we
define its \textbf{dehomogenization} as
\begin{align*}  
f^i \coloneqq f(1, x_1, \cdots, x_n) \in k[x_1,\cdots, x_n]
.\end{align*}

For a homogeneous ideal, we define
\begin{align*}  
J^i \coloneqq\left\{{f^i {~\mathrel{\Big|}~}f\in J}\right\}
.\end{align*}

\end{definition}

\begin{example}

The dehomogenization is usually not homogeneous. Take
\begin{align*}  
f &\coloneqq x_0^3 + x_0 x_1^2 + x_0 x_1 x_2 + x_0^2 + x_1 \\
\implies f^i &= 1  +x_1^2 + x_1 x_2 + x_1
,\end{align*}
which has terms of mixed degrees.

\end{example}

\begin{remark}

\begin{align*}  
(fg)^i &= f^i g^i \\
(f+g)^i &= f^i + g^i
.\end{align*}

In other words, evaluating at \(x_0 = 1\) is a ring morphism.

\end{remark}

\hypertarget{homogenization}{%
\subsection{Homogenization}\label{homogenization}}

\begin{definition}[Homogenization of a function]

Let \(f\in k[x_1, \cdots, x_{n}]\), then the \textbf{homogenization} of
\(f\) is defined by
\begin{align*}  
f^h \coloneqq x_0^d f\qty{ {x_1 \over x_0}, \cdots, {x_n \over x_0} }
\end{align*}
where \(d\coloneqq\deg(f)\).

\end{definition}

\begin{example}[?]

Set
\begin{align*}  
f(x_1, x_2) &\coloneqq 1 + x_1^2 + x_1 x_2 + x_2^3 \\
\implies f^h(x_0, x_1, x_2) &= x_0^3 + x_0 x_1^2 + x_0 x_1 x_2 + x_2^3
,\end{align*}
which is a homogeneous polynomial of degree \(3\). Note that
\((f^h)^i = f\).

\end{example}

\begin{example}[?]

It need not be the case that \((f^i)^h = f\). Take
\(f = x_0^3 + x_0 x_1 x_2\), then \(f^i = 1 + x_1 x_2\) and
\((f^i)^h = x_0^2 + x_1 x_2\). Note that the total degree dropped, since
everything was divisible by \(x_0\).

\end{example}

\begin{remark}

\begin{align*}  
(f^i)^h = f \iff x_0 \nmid f
.\end{align*}

\end{remark}

\begin{definition}[Homogenization of an ideal]

Given \(J\subset k[x_1, \cdots, x_{n}]\), define its
\textbf{homogenization} as
\begin{align*}  
J^h \coloneqq\left\{{f^h {~\mathrel{\Big|}~}f\in J}\right\}
.\end{align*}

\end{definition}

\begin{example}

This is not a ring morphism, since \((f+g)^h \neq f^h + g^h\) in
general. Taking \(f = x_0^2 + x_1\) and \(g= -x_0^2 + x_2\), we have
\begin{align*}  
f^h + g^h &= x_0 x_1 + x_0 x_2 \\ 
(f+g)^h &= x_1 + x_2
.\end{align*}

\end{example}

What is the geometric significance?

\begin{proposition}[Geometric significance of homogenization]

Set
\begin{align*}
U_0 \coloneqq\left\{{{\left[ {x_0: \cdots :x_n} \right]} \in {\mathbb{P}}^n_{/k} {~\mathrel{\Big|}~}x_0 \neq 0 }\right\} \cong {\mathbb{A}}^n_{/k}
\end{align*}
with coordinates
\({\left[ {{x_1\over x_0} : \cdots : {x_n \over x_0}} \right]}\). Then
\(U_0\) with the subspace topology is equal to \({\mathbb{A}}^n\) with
the Zariski topology.

\end{proposition}

\begin{proof}[?]

If we define the Zariski topology on \({\mathbb{P}}^n\) as having closed
sets \(V_p(I)\), we would want to check that
\({\mathbb{A}}^n\cong U_0 \subset {\mathbb{P}}^n\) is closed in the
subspace topology. This amounts to showing that \(V_p(I) \cap U_0\) is
closed in \({\mathbb{A}}^n \cong U_0\). We can check that
\begin{align*}  
V_p\qty{f {~\mathrel{\Big|}~}f\in I} = \left\{{\mathbf{x} \coloneqq{\left[ {x_0:\cdots:x_n} \right]} {~\mathrel{\Big|}~}f(\mathbf{x}) = 0 \,\, \forall f\in I}\right\}
.\end{align*}
Intersecting with \(U_0\) yields
\begin{align*}  
V_p\qty{f{~\mathrel{\Big|}~}f\in I} \bigcap U_0 = \left\{{{\left[ {x_1:\cdots:x_n} \right]} {~\mathrel{\Big|}~}f(\mathbf{x}) = 0,\, x_0\neq 0}\right\}
.\end{align*}
Equivalently, we can rewrite this set \(S\) as
\begin{align*}
S = \left\{{{\left[ {x_1:\cdots:x_n} \right]} {~\mathrel{\Big|}~}f\qty{{\left[ {1, {x_1 \over x_0}, \cdots,{x_n \over x_0} } \right]}} = 0,\, f \text{ homogeneous}}\right\}
\end{align*}
Since these are coordinates on \({\mathbb{A}}^1\), we have
\(V_p(I) \cap U_0 = V_a(I^i)\) which is closed. Conversely, given a
closed set \(V(I)\), we can write this as \(V(I) = U_0 \cap V_p(I^h)\).

\end{proof}

\begin{corollary}[Projective space is irreducible]

\({\mathbb{P}}^n_{/k}\) is irreducible of dimension \(n\).

\end{corollary}

\begin{proof}[?]

This follows from the fact that \({\mathbb{P}}^n\) is covered by
irreducible topological spaces of dimension \(n\) with nonempty
intersection, along with a fact from the exercises.

\end{proof}

\begin{example}[?]

Consider \(f(x_1, x_2) = x_1^2 - x_2^2 - 1\) and consider
\(V(f) \subset {\mathbb{A}}^2_{/{\mathbb{C}}}\):

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-11-10-10-10-42.png}
\caption{The variety \(V(x_1^2 - x_2^2 - 1)\)}
\end{figure}

Note that for real projective space, we can view this as a sphere with
antipodal points identified. We can thus visualize this in the following
way:

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-11-10-10-12-20.png}
\caption{Projective 2-space as sphere with anitpodal points identified}
\end{figure}

We can normalize the \(x_0\) coordinate to one, hence the plane. We can
also project \(V(f)\) from the plane onto the sphere, mirroring to
antipodal points:

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-11-10-10-14-09.png}
\caption{Projecting a variety onto a sphere}
\end{figure}

This misses some points on the equator, since we aren't including points
where \(x_0 = 0\). Consider the homogenization
\(V(f^h) \subset {\mathbb{P}}^2_{/{\mathbb{C}}}\):
\begin{align*}  
V(f^h) = V(x_1^2 - x_2^2 - x_0^2)
.\end{align*}
Then
\begin{align*}  
V(f^h) \cap V(x_0) = \left\{{{\left[ {0:x_1:x_2} \right]} {~\mathrel{\Big|}~}f^h(0, x_1, x_2) = 0 }\right\} = \left\{{{\left[ {0:1:1} \right]}, {\left[ {0:1:-1} \right]}}\right\}
,\end{align*}
which can be seen in the picture as the points at infinity:

\begin{figure}
\centering
\includegraphics[width=5.72917in,height=\textheight]{figures/image_2020-11-10-10-19-19.png}
\caption{Homogenization and points at infinity}
\end{figure}

Note that the equator is
\(V(x_0) = {\mathbb{P}}^2_{/{\mathbb{C}}}\setminus U_0 \cong {\mathbb{P}}^2\setminus{\mathbb{A}}^2\).
So we get a circle of points at infinity,
i.e.~\(V(x_0) = {\mathbb{P}}^1 = \left\{{{\left[ {0:v_1:v_2} \right]}}\right\}\).

\end{example}

\begin{example}[?]

Consider \(V(f)\) where \(f\) is a line in
\({\mathbb{A}}^2_{/{\mathbb{C}}}\), say
\(f(x_1, x_2) = ax_1 + bx_2 + c\). This yields
\(f^h = ax_1 + bx_2 + cx_0\) and we can consider
\(V(f^h) \cong {\mathbb{P}}^2_{{\mathbb{C}}}\). We know
\({\mathbb{P}}^1_{{\mathbb{C}}}\) is topologically a sphere and
\({\mathbb{A}}^1_{/{\mathbb{C}}}\) is a point:

\begin{figure}
\centering
\includegraphics[width=2.60417in,height=\textheight]{figures/image_2020-11-10-10-26-40.png}
\caption{\({\mathbb{P}}^1_{{\mathbb{C}}}\)}
\end{figure}

The points at infinity correspond to
\begin{align*}  
V(f^h) = V(f^h) \cap V(x_0) = \left\{{{\left[ {0:-b:a} \right]}}\right\}
,\end{align*}
which is a single point not depending on \(c\).

\end{example}

\begin{remark}

\({\mathbb{P}}^2_{/k}\) for any field \(k\) is a \textbf{projective
plane}, which satisfies certain axioms:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  There exists a unique line through any two distinct points,
\item
  Any two distinct lines intersect at a single point.
\end{enumerate}

A famous example is the \emph{Fano plane}:

\begin{figure}
\centering
\includegraphics{figures/image_2020-11-10-10-29-34.png}
\caption{Fano Plane}
\end{figure}

Why is this true? \({\mathbb{P}}^2_{/k}\) is the set of lines in
\(k^3\), and the lines in \({\mathbb{P}}^2_{/k}\) are the vanishing loci
of homogeneous polynomials and also planes in \(k^3\), since any two
lines determine a unique plane and any two planes intersect at the
origin.

\end{remark}

\begin{proposition}[?]

Let \(J\subset k[x_1, \cdots, x_{n}]\) be an ideal. Let
\(X \coloneqq V_a(J) \subset {\mathbb{A}}^n\) where we identify
\({\mathbb{A}}^n = U_0 \subset {\mathbb{P}}^n\). Then the closure
\(\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu \subset {\mathbb{P}}^n\)
is given by
\(\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu = V_p(J^h)\).
In particular,
\begin{align*}
V_a(J) = V_p(J^h)
\end{align*}

\end{proposition}

\begin{proof}[?]

\(\supseteq\): It's clear that \(V_p(J^h)\) is closed and contains
\(V_a(J)\).

\(\subseteq\) Let \(Y\supseteq X\) be closed; we want to show that
\(Y\supseteq V_p(J^h)\). Since \(Y\) is closed, \(Y = V_p(J')\) where
\(J'\) is some homogeneous ideal. Any element \(f'\in J'\) satisfies
\(f' = x^d f\) for some maximal \(d\) where \(x_0^d f\) vanishes on
\(X\). We also have \(f=0\) on \(X\) since \(X\subset U_0\). We can
compute
\begin{align*}  
f\in I_a(X) = I_a(V_a(J)) = \sqrt J
,\end{align*}
so \(f^m\in J\). Then \((f^h)^m \in J^h\) for some \(m\), and this
\(f^h \in \sqrt{J^h}\). So \(J'\subset \sqrt J\), which shows that
\(V_p(J') \supseteq V_p(J^h)\) as desired.

\end{proof}

\begin{definition}[Projective Closure]

The \textbf{projective closure} of \(X = V_a(J)\) is the smallest closed
subset containing \(X\) and is given by
\begin{align*}  
\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu = V_p(J^h)
.\end{align*}

\end{definition}

\hypertarget{projective-closures-thursday-november-12}{%
\section{Projective Closures (Thursday, November
12)}\label{projective-closures-thursday-november-12}}

Recall that if \(f\in k[x_1, \cdots, x_{n}]\) is a homogeneous degree
\(d\) polynomial, then
\begin{align*}
f^i \coloneqq f(1, x_1, \cdots, x_n) \in k[x_1,\cdots, x_n]
\end{align*}
is the dehomogenization of \(f\). Conversely,
\begin{align*}
f^h \coloneqq x_0^d f\qty{ {x_1 \over x_0}, \cdots, {x_n \over x_0} }
\end{align*}
is the homogenization. This is related to looking at the open subset
\(U_0 \coloneqq\left\{{ x\in {\mathbb{P}}^n_{/k} {~\mathrel{\Big|}~}x_0\neq 0}\right\} \subseteq {\mathbb{P}}^n_{/k}\),
where we found that \(U_0 \cong {\mathbb{A}}^n_{/k}\).

\begin{proposition}[Projective Closure]

Let \(V(I) \subset U_0\) be an affine variety, then
\(V(I) \subset {\mathbb{P}}^n_{/k}\) is given by
\begin{align*}
V(I^h) \coloneqq\left\{{f^h {~\mathrel{\Big|}~}f\in I}\right\}
,\end{align*}
the \textbf{projective closure}.

\end{proposition}

\begin{remark}

Projective varieties are better! They're closed in the classical
topology, and subsets of projective space and thus compact.

\end{remark}

\begin{warnings}[It doesn't suffice to just homogenize the individual generators of an ideal $I$]

Take \(J \coloneqq\left\langle{x_1, x_2 - x_1^2}\right\rangle\). We have
\(V(J) \subset {\mathbb{A}}^2\) given by \(\left\{{(0, 0)}\right\}\),
and by the proposition, \(V(J^h) = \left\{{[1:0:0]}\right\}\) since the
single point at the origin is closed in \({\mathbb{P}}^2\).

On the other hand,
\begin{align*}  
V_p(x_1, x_0 x_2 - x_1^2) = \left\{{[1:0:0], [0:0:1]}\right\} \subset {\mathbb{P}}^2
.\end{align*}

Note that \(x_2 \in J\), so this needs to be homogenized too.\footnote{It
  is possible to get around this issue computationally by using Gröbner
  bases, a special type of generating set for ideals.}

\end{warnings}

\begin{remark}

An aside: how do you implement algebraic geometry? For example, when is
\(\left\langle{f_i}\right\rangle = \left\langle{1}\right\rangle\)? This
is generally a somewhat difficult problem, since checking that their
corresponding varieties are equal isn't so tractable.

\end{remark}

\hypertarget{chapter}{%
\subsection{Chapter ?}\label{chapter}}

Goal: understand and define the sheaf of regular functions on projective
varieties. Given an open subset \(U\subset V_p(J)\), what are the
regular functions on it?

\begin{definition}[Regular Functions on Projective Varieties]

Let \(U\subset X\) be an open subset of a projective variety, and define
\begin{align*}  
{\mathcal{O}}_X(U) \coloneqq\left\{{\phi:U\to k {~\mathrel{\Big|}~}\phi \text{ is locally of the form }   {g_p \over f_p} \in S(X)_d }\right\}
.\end{align*}
i.e.~the functions in the homogeneous coordinate ring of the same degree
\(d\).

\end{definition}

\begin{remark}

Note that \(g_p/f_p\) is well-defined on \(V(f_p)^c\) since
\begin{align*}  
{ g_p(\lambda \mathbf{x}) \over  f_p(\lambda \mathbf{x})} 
= { \lambda^d g_p(\mathbf{x}) \over \lambda^d f_p(\mathbf{x}) }
= {g_p(\mathbf{x}) \over f_p (\mathbf{x})}
\end{align*}

\end{remark}

Recall that ``locally of the form \(\cdots\)'' means that for all
\(p\in U\), there exists an open neighborhood \(U_p\) on which
\({ \left.{{\phi}} \right|_{{U_p}} } = g_p / f_p\).

\begin{definition}[Projective Variety as a Ringed Space]

Note that if \(X\subset {\mathbb{P}}^n\) is closed, then \(X\cap U_i\)
is closed and thus an affine variety.
\begin{align*}  
\tilde {\mathcal{O}}_X(U) \coloneqq\left\{{\phi: U\to k {~\mathrel{\Big|}~}{ \left.{{\phi}} \right|_{{U\cap U_i}} } \text{ is a regular function} }\right\}
.\end{align*}

\end{definition}

\begin{proposition}[?]

These two definitions are equivalent.

\end{proposition}

\begin{proof}[?]

It suffices to check that
\({\mathcal{O}}_{X\cap U_i} = \tilde {\mathcal{O}}_{X\cap U_i}\) as
sheaves on \(X\cap U_i\), i.e.~checking on an open cover, since then
they'd both arise from the gluing construction. We have
\begin{align*}  
X\cap U_i = \left\{{[x_0 : \cdots: x_n] {~\mathrel{\Big|}~}x_i \neq 0 }\right\}
.\end{align*}

Let \(V\subset X\cap U_0\) be an open subset, we then want to show that
\({\mathcal{O}}_X(V)\) are the regular functions on \(V\) when \(V\) as
a subset of an affine variety. So let \(\phi\in {\mathcal{O}}_X(V)\), so
that locally \(\phi = g_p/f_p \in S(X)_d\) as a ratio of two homogeneous
polynomials. We want to know if \(\phi\) can be written as the ratio of
two polynomials in one additional variable, so we just dehomogenize to
obtain \(\phi = g^i_p / f^i_p\) locally where both are in
\(A(X\cap U_0)\). So \(\phi\) is a regular function on the open subset
\(V\) of the affine variety \(X\cap U_0\).\\

Conversely, suppose that \(\phi = g_p/f_p \in A(X\cap X_0)\) locally
around \(p\). It's not necessarily the case that
\(\phi = g^h_p / f^h_p\), but it is true that
\begin{align*}  
\phi = {x_0^d g_p^h \over f_p^h} = {g_p^h \over x_0^{-d} f_p^h}
,\end{align*}
where \(d = \deg f^h - \deg g^h\). This is locally a ratio of two
homogeneous polynomials of equal degree, so \({\mathcal{O}}_X\) and
\(\tilde {\mathcal{O}}_X\) define the same sheaf of functions on \(X\).

\end{proof}

\hypertarget{morphisms-of-projective-varieties}{%
\subsection{Morphisms of Projective
Varieties}\label{morphisms-of-projective-varieties}}

\begin{lemma}[?]

Let \(X\) be a projective variety and \(f_0, \cdots, f_m \in S(X)_d\).
Then on the open subset \(X\setminus V(f_i)\), there is a morphism
\begin{align*}  
f: U &\to {\mathbb{P}}^m \\
p &\mapsto {\left[ {f_0(p) : \cdots : f_m(p) } \right]}
.\end{align*}

\end{lemma}

\begin{proof}[?]

This map is well-defined, since letting \(p = [x_0: \cdots : x_n]\) we
have
\begin{align*}  
[\lambda x_0 : \cdots : \lambda x_n] 
&\mapsto 
[ \lambda^d f_0(p) : \cdots : \lambda^d f_m(p)] 
= f(p)
.\end{align*}

We need to check that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(f\) is continuous, and
\item
  The pullback of a regular function on any open set is again regular.
\end{enumerate}

\begin{claim}

\(f\) is continuous.

\end{claim}

Consider \(f^{-1}(V(h))\) with \(h\in k[y_0, \cdots, y_m]\) homogeneous.
We can check that
\begin{align*}  
f^{-1}(V_p(h)) = V_p(h(f_0, \cdots, f_m))
,\end{align*}
which is closed, so \(f\) is continuous.

\begin{claim}

\(f\) pulls back regular functions.

\end{claim}

Let \(h_1, h_2 \in S({\mathbb{P}}^m)\) be homogeneous polynomials of
equal degree in \(k[y_0, \cdots, y_m]\). Then on \(V(h_2)^c\), we have
\begin{align*}  
f^*\qty{h_1 \over h_2 } = {h_1(f_0, \cdots, f_m) \over h_2(f_0, \cdots, f_m)}
.\end{align*}
This is a ratio of homogeneous polynomials of equal degree in the
\(x_i\), the pullback is again locally homogeneous ratios of functions
of equal degree.

\end{proof}

\hypertarget{projections-and-embeddings-tuesday-november-17}{%
\section{Projections and Embeddings (Tuesday, November
17)}\label{projections-and-embeddings-tuesday-november-17}}

\hypertarget{projecting-from-a-point}{%
\subsection{Projecting From a Point}\label{projecting-from-a-point}}

We have
\({\mathbb{P}}^n \coloneqq{\mathbb{A}}^{n+1}\setminus\left\{{0}\right\}/ \sim\)
where \(x\sim \lambda x\), and projective varieties
\(V(I) \subset {\mathbb{P}}^n\) where
\(I {~\trianglelefteq~}k[x_0, \cdots, x_n]\) is a homogeneous ideal. We
defined a sheaf of rings \({\mathcal{O}}_X\) on \(X = V(I)\) by
\begin{align*}  
{\mathcal{O}}_X(U) \coloneqq\left\{{\phi: U\to k {~\mathrel{\Big|}~}\phi \text{ is locally a ratio of two homogeneous polynomials of equal degree}}\right\}
.\end{align*}
We showed that this was the same as the sheaf \(\tilde {\mathcal{O}}_X\)
defined by gluing ringed spaced
\((X \cap U_i, {\mathcal{O}}_{X\cap U_i})\) where \(U_i = D(x_i)\). We
also showed that \(S(X) \coloneqq k[x_0, \cdots, x_n] / I(X)\) is
homogeneous, i.e.~the quotient by a homogeneous ideal is again
homogeneous. Moreover, if
\(\left\{{f_i}\right\}_{i=0}^m \subseteq S(X)_d\) and
\(V(\left\{{f_i}\right\}) = \emptyset\). then the map
\begin{align*}  
(f_0, \cdots, f_m): X &\to {\mathbb{P}}^m \\
x &\mapsto [f_0(x), \cdots, f_m(x)]
.\end{align*}

Recall that a variety is separated iff \(\Delta \hookrightarrow X\) is
closed. Let \(A\in \operatorname{GL}_{n+1}(k)\) and define a map
\begin{align*}  
A: {\mathbb{P}}^n &\to {\mathbb{P}}^n \\
\begin{bmatrix}
x_0  \\
\vdots \\
x_n 
\end{bmatrix}
&\mapsto
A
\begin{bmatrix}
x_0  \\
\vdots \\
x_n 
\end{bmatrix}
.\end{align*}

This is a morphism because
\begin{align*}  
\begin{bmatrix}
- & \vec A_0 & - \\
- & \vdots & - \\
- & \vec A_n & - \\
\end{bmatrix}
\begin{bmatrix}
x_0  \\
\vdots \\
x_n 
\end{bmatrix}
=
\begin{bmatrix}
A_0 \cdot \mathbf{x}  \\
\vdots \\
A_n \cdots \mathbf{x} 
\end{bmatrix}
,\end{align*}
which are linear homogeneous polynomials.

Then \(V_p(A_i \cdot \mathbf{x}) = \emptyset\), and thus
\(V_a(A_i \cdot \mathbf{A}) = \left\{{0}\right\}\). So we should view
\(A\in \operatorname{PGL}_{n+1}(k)\). Note that this is a group, since
\(A^{-1}\) again forms a morphism. Thus
\(\operatorname{PGL}_{n+1}(k) \subset {\operatorname{Aut}}({\mathbb{P}}^n)\),
and it turns out that these are in fact equal.

\begin{definition}[Projection from a point]

Let \(a = [1: 0 : \cdots : 0] \in {\mathbb{P}}^n\), then there is a
morphism
\begin{align*}  
{\mathbb{P}}^n \setminus\left\{{a}\right\} &\to {\mathbb{P}}^{n-1} \\
[x_0: \cdots : x_n] &\mapsto [x_1: \cdots : x_n]
.\end{align*}
Note that this morphism does not extend to \({\mathbb{P}}^n\). More
generally, given any point \(p\in {\mathbb{P}}^n\), we can project from
it by making a linear change of coordinates to
\(p = [1: 0 : \cdots : 0]\).

\end{definition}

Let \(x\in {\mathbb{P}}^n\setminus\left\{{a}\right\}\), then there is a
unique line through \(a\) and \(x\). It can be described parametrically
as follows: writing \(x = [x_0: \cdots : x_n]\), we take the plane they
span and projectivize to obtain
\(s[x_0 : \cdots : x_n] + t [1: 0 : \cdots : 0]\) where we range over
\([s: t] \in {\mathbb{P}}^1\). In fact, this defines a morphism
\({\mathbb{P}}^1 \to {\mathbb{P}}^n\).

Consider now \({\mathbb{P}}^{n-1} = V(x_0)\), this copy of
\({\mathbb{P}}^{n-1}\) intersects any such line at a unique point:

\begin{figure}
\centering
\includegraphics[width=5.72917in,height=\textheight]{figures/image_2020-11-17-10-10-32.png}
\caption{Copy of \({\mathbb{P}}^{n-1}\) intersecting a line.}
\end{figure}

\begin{example}[?]

Consider \(X = V(x_0 x_2 - x_1^2) \subset {\mathbb{P}}^2\), which
defines a conic, and the projection
\({\mathbb{P}}^2 \setminus\left\{{[1:0:0]}\right\} \to {\mathbb{P}}^1\):

\begin{figure}
\centering
\includegraphics{figures/image_2020-11-17-10-13-40.png}
\caption{Projection from \(V(x_0 x_2 - x_1^2)\) onto \(V(x_0)\).}
\end{figure}

This morphism can be restricted to
\(\phi: X\setminus\left\{{[1:0:0]}\right\} \to {\mathbb{P}}^2\), and the
claim is that this morphism extends to all of \(X\). The secant lines
approach a tangent line at \([1:0:0]\), which \(V(x_0)\) at a unique
point. So we define
\begin{align*}  
\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu(x) \coloneqq
\begin{cases}
[x_1: x_2] & x \neq [1:0:0] \\
[x_0: x_1] & x \neq [0:0:1]
\end{cases}
.\end{align*}

This locally writes \(\phi\) as a morphism, so we only need to check
that they agree on the overlap. Note that on \(X\), we have
\([x_1: x_2] = [x_0 : x_1]\) wherever both are well-defined. In fact,
\(\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu\) is
an isomorphism, since an inverse can be explicitly written. Thus
\(X\cong {\mathbb{P}}^1\), and in fact all nondegenerate\footnote{Here
  nondegenerate means that if \(Q\) is a quadratic polynomials in
  \(x_0, x_1, x_2\), then \(Q\) does not factor as a product of linear
  factors.} conics are isomorphic to \({\mathbb{P}}^1\) as well. Note
that such a \(Q\) is a quadratic form, so \(Q(x) = B(x, x)\) for some
bilinear form, and \(Q\) is nondegenerate iff \(\det B \neq 0\) where
\(B_{ij} = B(e_i, e_j)\).

\end{example}

\hypertarget{the-segre-embedding}{%
\subsection{The Segre Embedding}\label{the-segre-embedding}}

\begin{definition}[Segre Embedding]

Letting \(N = (n+1)(m+1) - 1\), the \textbf{Segre embedding} is the
morphism
\begin{align*}  
f: {\mathbb{P}}^n \times{\mathbb{P}}^m &\to {\mathbb{P}}^N \\
([x_0: \cdots : x_n], [y_0: \cdots : y_m]) &\mapsto
[x_0 y_0 : \cdots : z_{ij} \coloneqq x_i y_j : x_n y_m]
.\end{align*}

\end{definition}

Note that \({\mathbb{P}}^n, {\mathbb{P}}^m\) are prevarieties and we
thus know how to construct their product as a prevariety. Check that
this is well-defined!

\begin{proposition}[Properties of the Segre embedding]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  The image \(X\) is a projective variety.
\item
  \(f: {\mathbb{P}}^n \times{\mathbb{P}}^m \to X\) is a morphism.
\end{enumerate}

\end{proposition}

\begin{proof}[of (a)]

It suffices to write polynomials in the coordinate \(z_{ij}\) that cut
out \(f({\mathbb{P}}^n \times{\mathbb{P}}^m)\). Given
\(z_{ij} = x_i y_j\), we have \(z_{ij} z_{kl} = z_{il} z_{kj}\) and
\((x_i y_j)(x_k y_l) = (x_i y_l)(x_k y_j)\). The former quadric
equations in \(z_{ij}\) variables vanish on
\(f({\mathbb{P}}^n \times{\mathbb{P}}^m)\).

\begin{claim}

\(V(z_{ij} z_{kl} - z_{il} z_{kj})\) works.

\end{claim}

Without loss of generality, we can assume \(z_{00} = 1\), in which case
\(z_{ij} z_{00} = z_{ij} = z_{i0} z_{0j}\) on \(X\). Setting
\(x_i = z_{i0}\) and \(y_j = z_{0j}\), we've now constructed a point in
the preimage, so \(f\) surjects onto \(X\).

\end{proof}

\begin{proof}[of (b)]

That \(f\) is a morphism to \({\mathbb{P}}^n\) is easy, and since
\(\operatorname{im}f \subset X\),
\(f: {\mathbb{P}}^n \times{\mathbb{P}}^m \to X\) is a morphism. On
\(D(z_{00}) \subset X\), the inverse described above is a morphism.
Since this works for any \(z_{ij}\), \(f^{-1}\) is well-defined and a
morphism, making \(f\) an isomorphism.

\end{proof}

\begin{example}[of the Segre embedding being isomorphic to a variety]

Take

\begin{align*}  
f: {\mathbb{P}}^1 \times{\mathbb{P}}^1 &\to {\mathbb{P}}^3 \\
\qty{ {\left[ {x_0: x_1} \right]}, {\left[ {y_0: y_1} \right]}} &\mapsto {\left[ {z_{00} : z_{01} : z_{10} : z_{11} } \right]} \coloneqq{\left[ {x_0 y_0: x_0 y_1 : x_1 y_0 : x_1 y_1} \right]}
.\end{align*}
Restricting to \(\operatorname{im}(f)\) yields an isomorphism to
\(X \subseteq {\mathbb{P}}^3\) given by the quadric\footnote{A
  \textbf{quadric} is the vanishing locus of a degree 4 polynomial.}
\(X = V(z_{00} z_{11} - z_{10} z_{01} )\).

\end{example}

\hypertarget{projective-varieties-thursday-november-19}{%
\section{Projective Varieties (Thursday, November
19)}\label{projective-varieties-thursday-november-19}}

\hypertarget{why-use-projective-varieties}{%
\subsection{Why use projective
varieties?}\label{why-use-projective-varieties}}

For e.g.~a manifold, there is a well-defined intersection pairing, and
the same way that \([\mu] \in H^1(T, {\mathbb{Z}}) = 1\) in the torus,
we have \([L]^2 = 1\) in \({\mathbb{P}}^2_{/{\mathbb{C}}}\), so every
two lines intersect in a unique point. Also, Bezout's theorem: any two
curves of degrees \(d, e\) in projective space intersect in \(d\cdot e\)
points. Also note that we have a notion of compactness that works in the
projective setting but not for affine varieties.

\hypertarget{projective-varieties-are-varieties}{%
\subsection{Projective Varieties are
Varieties}\label{projective-varieties-are-varieties}}

Last time: we saw the Segre embedding
\((\mathbf{x}, \mathbf{y})\mapsto [x_i y_j]\), which was an isomorphism
onto its image \(X = V(z_{ij}z_{kl} - z_{ik} z_{kj} )\), which exhibits
\({\mathbb{P}}^n \times{\mathbb{P}}^m\) as a projective variety.

\begin{example}[?]

For \({\mathbb{P}}^1 \times{\mathbb{P}}^1 \to {\mathbb{P}}^3\), its
image is \(X = V_p(xy - zw)\), which is a quadric (vanishing locus of a
degree 4 polynomial).

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-11-19-09-45-47.png}
\caption{\({\mathbb{P}}^1 \times{\mathbb{P}}^1\) is ruled.}
\end{figure}

The projection map has fibers, which induce a \textbf{ruling}\footnote{A
  family of copies of \({\mathbb{P}}^1\).} which we can see from the
real points:

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-11-19-09-46-38.png}
\caption{Its image, a quadric surface, is also ruled.}
\end{figure}

\end{example}

\begin{corollary}[?]

Every projective variety is a separated prevariety, and thus a variety.

\end{corollary}

\begin{proof}[?]

It suffices to show that \(\Delta_X \subset X\times X\) is closed. We
can write
\begin{align*}  
\Delta_{{\mathbb{P}}^n} = 
\left\{{
[x_0: \cdots: x_n], [y_0: \cdots : y_n] {~\mathrel{\Big|}~}
x_i y_j - x_j y_i = 0 \, \forall i, j
}\right\}
.\end{align*}
This says that \(\mathbf{x}, \mathbf{y}\) differ by scaling. We know
that
\(\Delta_{{\mathbb{P}}^n} \hookrightarrow{\mathbb{P}}^n \times{\mathbb{P}}^n\),
which is isomorphic to the Segre variety \(S_V\) in
\({\mathbb{P}}^{(n+1)^2 -1}\), and we can write \(z_{ij} = x_i y_j\) and
thus
\begin{align*}  
\Delta_{{\mathbb{P}}^n} = S_V \cap V(z_{ij} - z_{ji})
.\end{align*}
Note that the Segre variety is closed. The conclusion is that
\({\mathbb{P}}^n\) is a variety, and any closed subprevariety of a
variety is also a variety by taking
\(\Delta_{{\mathbb{P}}^n} \cap(X\times X) = \Delta_X\), which is closed
as the intersection of two closed subsets.

\end{proof}

\begin{definition}[Closed Maps]

Recall that a map \(f:X\to Y\) is topological spaces is \textbf{closed}
if whenever \(U \subset X\) is closed, then \(f(U)\) is closed in \(Y\).

\end{definition}

\begin{definition}[Complete Varieties]

A variety \(X\) is \textbf{complete} if the projection
\(\pi_Y: X\times Y \twoheadrightarrow Y\) is a closed map for any \(Y\).

\end{definition}

\begin{slogan}

Completeness is the analog of compactness for varieties.

\end{slogan}

\begin{proposition}[Projection maps from products of projective spaces are closed.]

The projection
\({\mathbb{P}}^n \times{\mathbb{P}}^m \to {\mathbb{P}}^m\) is closed.

\end{proposition}

\begin{proof}[?]

Let \(Z \subset {\mathbb{P}}^n \times{\mathbb{P}}^m\), and write
\(Z = V(f_i)\) with \(f_i \in S(S_V)\). Note that if the \(f_i\) are
homogeneous of degree \(d\) in \(z_{ij}\), the pulling back only the
isomorphism \({\mathbb{P}}^n\times{\mathbb{P}}^m \to S_V\) yields
\(z_{ij} = x_i y_j\) and polynomials \(h_i\) which are homogeneous
polynomials in \(x_i, y_j\) which have degree \(d\) in both the \(x\)
and \(y\) variables individually. Consider \(a\in {\mathbb{P}}^m\), we
want to determine if \(a\in \pi(Z)\) and show that this is a closed
condition. Note that \(a\not\in \pi(Z)\)

\begin{itemize}
\item
  \(\iff\) there does not exists an \(x\in {\mathbb{P}}^n\) such that
  \((x, a) \in Z\)
\item
  \(\iff\) \(V_p(f_i(x, a))_{i=1}^r = \emptyset\)
\item
  \(\iff\)
  \(\sqrt{\left\langle{f_i(x, a)}\right\rangle_{i=1}^r } = \left\langle{1}\right\rangle\)
  or the irrelevant ideal \(I_0\)
\item
  \(\iff\) there exist \(k_i \in {\mathbb{N}}\) such that
  \(x_i^{k_i} \in \left\langle{f_i(x, a)}\right\rangle_{i=1}^r\)
\item
  \(\iff\)
  \(k[x_1, \cdots, x_{n}]_k \subset \left\langle{f_i(x, a)}\right\rangle_{i=1}^r\)
  (where this is the degree \(k\) part)
\item
  \(\iff\) the map
  \begin{align*}  
  \Phi_a: k[x_1, \cdots, x_{n}]_{d - \deg f_2} \oplus \cdots \oplus k[x_1, \cdots, x_{n}]_{d - \deg f_r} &\to k[x_1, \cdots, x_{n}]_d \\
  (g_1, \cdots, g_r) &\mapsto \sum f_i(x, a) g_i (x, a)
  \end{align*}
  is surjective.
\end{itemize}

Recap: we have a closed subset of
\({\mathbb{P}}^n \times{\mathbb{P}}^m\), want to know its projection is
closed. We looked at points not in the closed set, this happens iff the
degree \(d\) part of the polynomial is not contained in the part where
we evaluate by \(a\). This reduces to a linear algebra condition: taking
arbitrary linear combinations yields a surjective map. Thus
\(a\in \pi(Z)\) iff \(\Phi_a\) is \emph{not} surjective.\\

Expanding in a basis, we can write \(\Phi_a\) as a matrix whose entries
are homogeneous polynomials in the coordinates of \(a\). Moreover,
\(\Phi_a\) is not surjective iff all \(d\times d\) determinants of
\(\Phi_a\) are nonzero (since this may not be square). This is a
polynomial condition, so \(a\in \pi(Z)\) iff a bunch of homogeneous
polynomials vanish, making \(\pi(Z)\) is closed.

\end{proof}

\begin{corollary}[$\PP^n$ is complete.]

The projection \(\pi: {\mathbb{P}}^n\times Y\to Y\) is closed for any
variety \(Y\) and thus \({\mathbb{P}}^n\) is complete.

\end{corollary}

\begin{proof}[?]

How to prove anything for varieties: use the fact that they're glued
from affine varieties, so prove in that special case. So first suppose
\(Y\) is affine. Let \(Z \subset {\mathbb{P}}^n \times Y\) be closed,
and consider
\(\mkern 1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu ss {\mathbb{P}}^m\)
and
\begin{align*}
\mkern 1.5mu\overline{\mkern-1.5muZ\mkern-1.5mu}\mkern 1.5mu \subset{\mathbb{P}}^n \times\mkern 1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu \subset{\mathbb{P}}^n \times{\mathbb{P}}^m
\end{align*}
as a closed subset. Then we know that the projection
\(\pi: {\mathbb{P}}^n \times{\mathbb{P}}^m \to {\mathbb{P}}^m\) is
closed, so
\(\pi(\mkern 1.5mu\overline{\mkern-1.5muZ\mkern-1.5mu}\mkern 1.5mu) \subset{\mathbb{P}}^m\)
is closed. But we can write
\begin{align*}
\pi(Z) = \pi(\mkern 1.5mu\overline{\mkern-1.5muZ\mkern-1.5mu}\mkern 1.5mu \cap{\mathbb{P}}^n \times Y) = \pi(\mkern 1.5mu\overline{\mkern-1.5muZ\mkern-1.5mu}\mkern 1.5mu) \cap Y
\end{align*}
which is closed. So \(\pi(Z)\) is closed in \(Y\), which proves this for
affine varieties.\\

Supposing now that \(Y\) is instead glued from affines, it suffices to
check that the set is closed in an open cover. So \(Z \subset X\) is
closed if when we let \(X = \cup U_i\), we can show \(Z \cap U_i\) is
closed. But this essentially follows from above.

\end{proof}

\begin{corollary}[Projective varieties are complete.]

Any projective variety is complete.

\end{corollary}

\begin{proof}[?]

If \(X \subset {\mathbb{P}}^n\) is closed and if
\({\mathbb{P}}^n \times Y\to Y\) is a closed map for all \(Y\), then
restricting to \(X\times Y\to Y\) again yields a closed map.

\end{proof}

\begin{corollary}[Images of varieties under morphisms are closed.]

Let \(f:X\to Y\) be a morphism of (importantly) \emph{varieties} and
suppose \(X\) is complete. Then \(f(X)\) is closed in \(Y\).

\end{corollary}

\begin{proof}[?]

Consider the graph of \(f\),
\begin{align*}
\Gamma_f = \left\{{(x, f(x))}\right\} \subset X\times Y
\end{align*}
From a previous proof, we know \(\Gamma_f\) is closed when \(Y\) is a
variety (by pulling back a diagonal). So \(\Gamma_f\) is closed in
\(X\times Y\), and thus \(\pi_Y(\Gamma_f) = f(X)\) is closed because
\(X\) is complete.

\end{proof}

The next result is an analog of the maximum modulus principle: if \(X\)
is a compact complex manifold, then any function that is holomorphic on
all of \(X\) is constant.

\begin{corollary}[Maximum modulus principle for varieties]

Let \(X\) be complete, then \({\mathcal{O}}_X(X) = k\), i.e.~every
global regular function is constant.

\end{corollary}

\begin{proof}[?]

Suppose \(\phi X\to {\mathbb{A}}^1\) is a regular function. Since
\({\mathbb{A}}^1 \subset {\mathbb{P}}^1\), extend \(\phi\) to a morphism
\(\widehat{\phi}: X\to PP^1\). By a previous corollary, \(\phi(X)\) is
closed, but \(\infty \not\in \phi(X)\) implies
\(\phi(X) \neq {\mathbb{P}}^2\), so \(\phi(X)\) is finite. Since \(X\)
is connected, \(\phi(X)\) is a point, making \(\phi\) a constant map.

\end{proof}

\hypertarget{embeddings-and-smoothness-tuesday-november-24}{%
\section{Embeddings and Smoothness (Tuesday, November
24)}\label{embeddings-and-smoothness-tuesday-november-24}}

\hypertarget{the-veronese-embedding}{%
\subsection{The Veronese Embedding}\label{the-veronese-embedding}}

\begin{definition}[Veronese Embedding]

Let \(n, d > 0\) and let \(f_0, \cdots, f_n\) be the monomials of degree
\(d\) in \(k[x_1, \cdots, x_{n}]\). There is a morphism
\begin{align*}  
{\mathbb{P}}^n \setminus V(f_0,\cdots , f_n) &\to {\mathbb{P}}^N \\
\mathbf{x} &\mapsto [f_0(\mathbf{x}), \cdots, f_N(\mathbf{x})]
,\end{align*}
where \(N+1\) is the number of monomials, and is equal to
\({n+d \choose d}\).

\end{definition}

\begin{remark}

It is true that \(V(f_0, \cdots, f_N) \neq \emptyset\), since
\(V(x_0^d, x_1^d, \cdots, x_n^d) = V(x_0, \cdots, x_n)\). This will be
the Veronese embedding, although we need to prove it is an embedding. On
an open set \(D(x_0) \subset {\mathbb{P}}^2\) one can define an inverse.
Suppose we have a coordinate \(z_j = x_i^{d-1} x_j\) and \(z_i = x_i^d\)
on \({\mathbb{P}}^N\). Then we can take the point
\begin{align*}  
{\left[ { {z_1 \over z_i}, \cdots, {z_i \over z_i}, \cdots, {z_n \over z_i} } \right]}
.\end{align*}
This defines an inverse on \(D(z_i)\). Since the open sets \(D(x_i)\)
cover \({\mathbb{P}}^N\), we have an inverse on the entire image.

\end{remark}

\hypertarget{exchanging-hypersurface-sections-for-hyperplane-sections}{%
\subsubsection{Exchanging Hypersurface Sections for Hyperplane
Sections}\label{exchanging-hypersurface-sections-for-hyperplane-sections}}

\begin{remark}

This embedding converts hypersurfaces of degree \(d\) into hyperplanes.
The Veronese is an isomorphism onto its image. Consider some arbitrary
degree \(d\) element of \(S({\mathbb{P}}^n)\). Consider
\(X \coloneqq V(\sum_{j=1}^N a_j f_j) \subset{\mathbb{P}}^n\), where
\(a_j\in k\), which is equal to \(\phi^{-1}(V(\sum_{j=1}^N a_j w_j ))\).

\todo[inline]{Probably not right.}

We have a picture: embedding
\({\mathbb{P}}^n\hookrightarrow{\mathbb{P}}^N\) in some curved way sends
a hypersurface to the intersection of a hyperplane with the embedded
image:

\begin{figure}
\centering
\includegraphics[width=5.72917in,height=\textheight]{figures/image_2020-11-24-09-48-27.png}
\caption{Embedding a hypersurface into a hyperplane.}
\end{figure}

\end{remark}

\begin{definition}[Hyperplane Sections]

Let \(X \subset{\mathbb{P}}^n\) be a projective variety. A
\textbf{hyperplane section} is the intersection of \(X\) with some
hyperplane \(H \coloneqq V(f)\) for \(f\) some linear homogeneous
polynomial.

\end{definition}

\begin{example}[of the Veronese embedding]

Let \(n=1\), then we get the embedding
\begin{align*}  
{\mathbb{P}}^1 &\hookrightarrow{\mathbb{P}}^d \\
[x_0: x_1] &\mapsto [x_0^d: x_0^{d-1}x_1 : \cdots : x_0 x_1^{d-1} : x_1^d]
.\end{align*}
Note that there are \(d+1\) such monomials, and not all can
simultaneously vanish. The image of this \({\mathbb{P}}^1\) is called
the \emph{twisted normal curve}.

\end{example}

\begin{example}[?]

Take
\begin{align*}  
{\mathbb{P}}^1 &\hookrightarrow{\mathbb{P}}^2 \\
[x_0 : x_1] &\mapsto [x_0^2 : x_0 x_1: x_1^2]
.\end{align*}
What homogeneous polynomials cut out \(\phi({\mathbb{P}}^1)\)? I.e.,
what is \(I(\phi({\mathbb{P}}^1)) \subset S({\mathbb{P}}^2)\)? Note that
\(w_0 w_2 - w_1^2 { \left.{{}} \right|_{{\phi({\mathbb{P}}^1)}} }\), so
this is an element. Is it a generator? I.e., given any
\(p\in V(w_0 w_2 - w_1^2)\) is of the form
\(p = [x_0^2 : x_0 x_1: x_1^2]\) for some \(x_), x_1 \in k\)? The answer
is yes, by choosing signs of \(\sqrt{w_0}, \sqrt{w_2}\).

\end{example}

\begin{example}[?]

Take
\begin{align*}  
\phi: {\mathbb{P}}^1 &\hookrightarrow{\mathbb{P}}^3 \\
[x_0: x_1] &\mapsto [x_0^3: x_0^2 x_1 : x_0 x_1^2: x_1^3]
.\end{align*}
What are some elements of this ideal?

\begin{itemize}
\tightlist
\item
  \(w_0 w_3 - w_1 w_2\)
\item
  \(w_0 w_2 - w_1^2\)
\item
  \(w_1 w_3 - w_2^2\)
\end{itemize}

Note that the first is not a \(k{\hbox{-}}\)linear combination of the
other two. There is also a pattern:
\(w_0/w_1 = w_1 / w_2 = w_2/w_3 = \cdots\). However, there will be
issues when the denominators are zero.

In this case, \(\phi({\mathbb{P}}^1)\) is the \emph{twisted cubic}. What
is
\(V(w_0 w_2 - w_1^2, w_1 w_3 - w_2^2) \setminus\phi({\mathbb{P}}^1)\)?
Note that being in \(\phi({\mathbb{P}}^1)\) means
\(w_1, w_2, w_3 \neq 0\), and similarly if \(w_0, w_1, w_2 \neq 0\). We
can conclude that
\(V(w_1, w_2) \subset V(w_0 w_2 - w_1^2, w_1 w_3 - w_2^2)\):

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-11-24-10-16-35.png}
\caption{Image}
\end{figure}

This variety has two components: the twisted cubic, and a line. This
variety has degree 4, since any generic hyperplane intersects it at 4
points. Why? Pulling back a hyperplane yields a cubic, which generally
vanishes at three points in affine space.

\end{example}

\begin{remark}

\(\phi({\mathbb{P}}^1)\) is a nice example of a curve in
\({\mathbb{P}}^3\) that can not be cut out by two homogeneous
polynomials.

\end{remark}

\begin{remark}

This is usually used to embed intersections like \(X\cap V(f)\) to
\(X\cap H\), exchanging a hypersurface section for a hyperplane section.
This is useful for induction:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Prove for \({\mathbb{P}}^n\).
\item
  Induction: If it's true for \(X \subset{\mathbb{P}}^n\), then it's
  true for \(X \cap H\) for some hyperplane \(H \subset{\mathbb{P}}^N\).
\end{enumerate}

This will prove it for any projective variety by taking
\(X = V(f_1, \cdots, f_n)\) and embedding.

\end{remark}

\hypertarget{the-tangent-space-chapter-10}{%
\subsection{The Tangent Space (Chapter
10)}\label{the-tangent-space-chapter-10}}

Motivation: we want to distinguish between things like \(V(xy)\) and
\(V(xy-1)\). Over \({\mathbb{C}}\), we can distinguish these: one is a
complex manifold, and the other is not.

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-11-24-10-31-47.png}
\caption{\(V(xy-1)\) is a manifold in \({\mathbb{C}}^2\)}
\end{figure}

This means we want each point to have a neighborhood biholomorphic to a
disc.

\begin{definition}[Tangent Space]

Let \(a\in X\) be a point on a variety \(X\). Choose an affine open set
containing \(a\) and a chart such that \(a\) is the origin, then define
\begin{align*}  
T_a X \coloneqq V(f_1 {~\mathrel{\Big|}~}f\in I(X))
,\end{align*}
where \(f_1\) denotes the linear part of \(f\).

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-11-24-10-38-10.png}
\caption{Tangent space at a point.}
\end{figure}

\end{definition}

\begin{remark}

Since \(0=a\), any \(f\in I(X)\) has no constant term -- otherwise \(f\)
would not vanish at the origin.

\end{remark}

\begin{example}[?]

Consider \(T_{(1, 1)} V(xy-1)\). First translate \((1, 1)\) to the
origin, so
\begin{align*}
T_{(1, 1)} V(xy-1) = T_{(0, 0)} = V((x-1)(y-1) - 1) = T_{(0, 0)} V(xy-x-y) = V(-x-y)
\end{align*}
On the other hand, \(T_{(0, 0)} V(xy) = V(0) = {\mathbb{C}}^2\).

\begin{figure}
\centering
\includegraphics[width=5.72917in,height=\textheight]{figures/image_2020-11-24-10-41-13.png}
\caption{Image}
\end{figure}

\end{example}

\hypertarget{the-tangent-space-and-smoothness-tuesday-december-01}{%
\section{The Tangent Space and Smoothness (Tuesday, December
01)}\label{the-tangent-space-and-smoothness-tuesday-december-01}}

\begin{definition}[Tangent Space]

The \textbf{tangent space} \(T_p X\) of a variety \(X\) at a point
\(p\in X\) is defined as
\begin{align*}
V\qty{\left\{{f_1 {~\mathrel{\Big|}~}f\in I(U_i),\, U_i \ni p = 0 \text{ affine } }\right\}}
\end{align*}
where \(f_1\) denotes the degree 1 part.

\begin{figure}
\centering
\includegraphics[width=5.72917in,height=\textheight]{figures/image_2020-12-01-09-40-28.png}
\caption{Image}
\end{figure}

\end{definition}

\begin{remark}

We've really only defined it for affine varieties and \(p=0\), but this
is a local definition. Note that this is also not a canonical
definition, since it depends on the affine chart \(U_i\).

\end{remark}

\hypertarget{computing-tangent-spaces}{%
\subsection{Computing Tangent Spaces}\label{computing-tangent-spaces}}

\begin{example}[?]

Consider
\begin{align*}
T_0 V(xy) = V(f_1 {~\mathrel{\Big|}~}f\in \left\langle{xy}\right\rangle) = V(0) = {\mathbb{A}}^2
,\end{align*}
since every polynomial in this ideal has degree at least 2. Letting
\(X = V(xy)\), note that we could embed
\(X\hookrightarrow{\mathbb{A}}^3\) as \(X\cong V(xy, z)\). In this case
we have
\begin{align*}
T_0 X = V(f_1 {~\mathrel{\Big|}~}f\in \left\langle{xy, z}\right\rangle) = V(z) \cong {\mathbb{A}}^2
\end{align*}
So we get a vector space of a different dimension from this different
affine embedding, but \(\dim T_0 X\) is the same.

\end{example}

\begin{example}[?]

Let \(X = V_p(xy-z^2) \subset {\mathbb{P}}^2\), which is a projective
curve. What is \(T_p X\) for \(p = [0:1:0]\)? Take an affine chart
\(\left\{{y\neq 0}\right\} \cap X\), noting that
\(\left\{{y\neq 0}\right\} \cong {\mathbb{A}}^2\). We could dehomogenize
the ideal
\({ \left.{{\left\langle{xy-z^2}\right\rangle}} \right|_{{y=1}} } = \left\langle{x-z^2}\right\rangle\).
Thus \(X \cap D(y) = V(x-z^2) \subset {\mathbb{A}}^2\) and the point
\([0:1:0] \in X\) gives \((0, 0)\) in this affine chart. Then
\begin{align*}
T_p X = V(f_1 {~\mathrel{\Big|}~}f\in \left\langle{x-z^2}\right\rangle) = V(x)
\end{align*}
Then \(f = (x-z^2)g\) implies that \(f_1 = (xg)_1 = g_0 x\), the
constant term of \(g\) multiplied by \(x\), since \(z^2\) kills any
degree 1 part of \(g\). So \(T_p X\) is a line.

\end{example}

\begin{example}[?]

Take \(X\) to be the union of the coordinate axes in \({\mathbb{A}}^3\):

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-12-01-09-54-30.png}
\caption{The coordinate axes in \({\mathbb{A}}^3\).}
\end{figure}

Then \(I(X) = \left\langle{xy, yz, xz}\right\rangle\) and
\begin{align*}
T_0 X = V(f_1 {~\mathrel{\Big|}~}f\in I(X)) = V(0) = {\mathbb{A}}^3
\end{align*}
since the minimal degree of any such polynomial is 2. Note that
\(\dim X = 1\) but \(\dim T_0 X = 3\)

\end{example}

\begin{example}[?]

Take \(Y = V(xy(x-y)) \subset {\mathbb{A}}^2\):

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-12-01-09-59-06.png}
\caption{\(V(xy(x-y))\) in \({\mathbb{A}}^2\).}
\end{figure}

Then \(T_0 X = V(0) = {\mathbb{A}}^2\).

\end{example}

\begin{remark}

Note that \(X\) and \(Y\) both consists of 3 copies of
\({\mathbb{A}}^1\) intersecting at a single point.

\begin{figure}
\centering
\includegraphics[width=5.72917in,height=\textheight]{figures/image_2020-12-01-10-01-58.png}
\caption{Comparing \(X\) and \(Y\).}
\end{figure}

Note that \(\dim T_0 X = 3\) but \(\dim T_0 Y = 2\), and interestingly
\(X\not\cong Y\) as affine varieties. There is a bijective morphism that
is not invertible.

\end{remark}

\begin{remark}

We will prove that \(\dim T_p X\) is invariant under choice of affine
embedding.

\end{remark}

\begin{example}[How to compute $T_{(1,0,0)} V(xy, yz, xz)$]

First move \((1,0,0)\) to the origin, yielding
\(T_{(0,0,0)} V((x+1)y, yz, (x+1) z)\). This is a different choice of
affine embedding into \({\mathbb{A}}^3\) which sends
\((1,0,0) \mapsto (0,0,0)\). Taking the vanishing locus of linear parts,
it suffices to take the linear parts of the generators, which yields the
\(x{\hbox{-}}\)axis \(V(y, z)\), making the dimension of the tangent
space 1.

\end{example}

\hypertarget{identifying-the-cotangent-space-as-mathfrakmmathfrakm2}{%
\subsection{\texorpdfstring{Identifying the Cotangent Space as
\(\mathfrak{m}/\mathfrak{m}^2\)}{Identifying the Cotangent Space as \textbackslash mathfrak\{m\}/\textbackslash mathfrak\{m\}\^{}2}}\label{identifying-the-cotangent-space-as-mathfrakmmathfrakm2}}

\begin{lemma}[The tangent space is given by $I/I^2$]

Let \(X \subset{\mathbb{A}}^n\) be an affine variety and let
\(0 = p\in X\). Then
\begin{align*}
T_0(X)^\vee\coloneqq\hom_k(T_0X, k) \cong I_X(p) / I_X(p)^2
\end{align*}

\end{lemma}

\begin{remark}

Note that the hom involves an affine embedding, but the quotient of
ideals does not. We know that the category of affine varieties is
equivalent to the category of reduced \(k{\hbox{-}}\)algebras, since the
points of \(X\) biject with the maximal ideals of the coordinate ring
\(A(X)\). \(I_X(p)\) is the maximal ideal in \(A(X)\) of regular
functions vanishing at \(p\).

\end{remark}

\begin{proof}[?]

Consider the map
\begin{align*}  
\phi: I_X(p) &\to T_0(X)^\vee\\
\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu &\mapsto { \left.{{f_1}} \right|_{{T_0(X)}} }
.\end{align*}
E.g. given
\(\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_1 - \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_2^2 \in A(X)\),
we first lift to \(x_1 - x_2^2 \in A({\mathbb{A}}^n)\), restrict to the
linear part \(x_1\), then restrict to \(T_0(X)\). Note that
\(I_X(p) = \left\langle{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_1, \cdots, \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_n}\right\rangle \in k[x_1, \cdots, x_n]/I(X)\),
and we need to check that this well-defined since there is ambiguity in
choosing the above lift.

\begin{claim}

\(\phi\) is well-defined.

\end{claim}

Consider two lifts \(f, f'\) of
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu\in A(X) = k[x_1, \cdots, x_n]/I(X)\).
Then \(f - f'\in I(X)\), so \((f - f')_1 = f_1 - f_1'\) is the linear
part of some element in \(I(X)\). The definition of \(T_0(X)\) was the
vanishing locus of linear parts of elements in \(I(X)\), which contains
\(f_1 - f_1'\), and thus
\({ \left.{{\qty{f_1 - f_1'} }} \right|_{{T_0(X)}} } = 0\). So
\(f_1 = f_1'\) on \(T_0(X)\).

\begin{claim}

\(I_X(p)^2 \to 0\).

\end{claim}

We know
\(I_X(p) = \left\langle{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_1, \cdots, \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_n}\right\rangle\),
and so
\(I_X(p)^2 = \left\langle{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_i \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_j}\right\rangle\).
Giving any
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu\in I_X(p)^2\),
we can lift this to some \(f\in \left\langle{x_i x_j}\right\rangle\), in
which case \(f_1 = 0\).

So \(\phi\) descends to
\begin{align*}
\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu: I_X(p) / I_X(p^2) &\to T_0(X)^\vee\\
\end{align*}

\begin{claim}

\(\phi\) is injective and surjective.

\end{claim}

That \(\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu\)
is surjective follows from the fact that if
\(\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_1, \cdots, \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_n \in I_X(p)\),
then the restrictions
\({ \left.{{x_1}} \right|_{{T_0X}} }, \cdots, { \left.{{x_n}} \right|_{{T_0X}} }\)
are in
\(\operatorname{im}\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu\)
These elements generate \(T_0(X)^\vee\), since
\(T_0(X) \subset{\mathbb{A}}^n\). For injectivity, suppose
\(\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu) = 0\),
then \({ \left.{{f_1}} \right|_{{T_0(X)}} } = 0\), so \(f_1\) is the
linear part of some \(f' \in I(X)\). Then \(f' \in I(X)\) and \(f, f'\)
have the same linear part \(f_1\), and \(f-f'\) has no linear part. Thus
\(f-f'\in \left\langle{x_i x_j}\right\rangle\), which implies that
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu' \in I_P(X)^2\)
and
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu \equiv \mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu' \in I_p(X) / I_p(X)^2\).
But \(f' \equiv 0\) since \(f'\in I(X)\).

\end{proof}

\begin{remark}

So for \(X\) an affine variety, the cotangent space has a more intrinsic
description, and we can recover the tangent space by dualizing:
\begin{align*}  
T_p(X) \coloneqq\qty{\mathfrak{m}_p/\mathfrak{m}_p^2 }^\vee
\end{align*}
where \(\mathfrak{m}_p = I_X(p)\) is the maximal ideal of regular
functions vanishing at \(p\). So how can we get rid of the word affine?
Given \(X\) any variety, we can define
\(T_p(X) \coloneqq\qty{\mathfrak{m}/\mathfrak{m}^2}^\vee\) where
\(\mathfrak{m}\) is the maximal ideal of the local ring
\({\mathcal{O}}_{X, p}\). This allows us to work on affine patches and
localize. Moreover, this will be left invariant under the localization.

\end{remark}

\hypertarget{smoothness-thursday-december-03}{%
\section{Smoothness (Thursday, December
03)}\label{smoothness-thursday-december-03}}

We showed last time that if \(X\) is an affine variety, then
\(T_p X = V\qty{f_1 {~\mathrel{\Big|}~}f\in I(X)}\) for
\(p = \mathbf{0} \in {\mathbb{A}}^n\), and we showed this is naturally
isomorphic to \(\qty{\mathfrak{m}_p /\mathfrak{m}_p^2}\). Then there was
a claim that generalizing this definition to an arbitrary variety \(X\)
involved taking \(\mathfrak{n}_p \leq {\mathcal{O}}_{X, p}\), a maximal
ideal in this local ring of germs of regular functions, given by
\(\left\{{(U, \phi) {~\mathrel{\Big|}~}p\in U, \, \phi\in {\mathcal{O}}_{X}(U)}\right\}\).
In this case, \(T_p = \qty{\mathfrak{n}_p/\mathfrak{n}_p^2}\). To prove
this, it suffices to show that
\(\mathfrak{m}_p/\mathfrak{m}_p^2 \cong \mathfrak{n}_p/\mathfrak{n}_p^2\).
Note that for any affine open \(U_i \ni p\), we have
\({\mathcal{O}}_{X, p} = {\mathcal{O}}_{U_i, p}\).

When \(X\) is affine, we have
\({\mathcal{O}}_{X, p} = A(X)_{\mathfrak{m}_p} \coloneqq\left\{{f/g {~\mathrel{\Big|}~}f\in A(X), g\not\in \mathfrak{m}_p}\right\}/\sim\).
Note that this localization makes sense, since the complement of a
maximal ideal is multiplicatively closed since it is prime. The
equivalence relation was \(f/g = f'/g'\) if there exists an
\(s\not\in \mathfrak{m}_p\) such that \(s(fg' - f'g) = 0\). We want to
show that
\(\mathfrak{m}_p / \mathfrak{m}_p^2 = \mathfrak{m}_p A(X)_{\mathfrak{m_p}} / \mathfrak{m}_p A(X)_{\mathfrak{m_p}}^2\),
i.e.~this doesn't change when we localize. In other words, we want to
show that
\({\mathfrak{m}}_p / {\mathfrak{m}}_p^2 \cong S^{-1} {\mathfrak{m}}_[ / (S^{-1} {\mathfrak{m}}_p)^2\).

Let \(f\in S\) so \(f(p) \neq 0\). Then
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu\in A(X) / \mathfrak{m}_p \cong K\)
is a nonzero element in a field and thus invertible. Thus
\(c\coloneqq 1/\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu\)
is an element of \(K^{\times}\), and for all \(g\in \mathfrak{m}_p\) we
have \(g/f \cong cg\) in \(\mathfrak{m}_p / \mathfrak{m}_p^2\). So
multiplying by elements of \(S\) is invertible in
\(\mathfrak{m}_p / \mathfrak{m}_p^2\). Thus
\(S^{-1} \qty{\mathfrak{m}_p / \mathfrak{m}_p^2} \cong \mathfrak{m}_p / \mathfrak{m}_p^2\),
where the LHS is isomorphic to
\(S^{-1} \mathfrak{m}_p / \qty{S^{-1}\mathfrak{m}_p^2}\).

\hypertarget{defining-smoothness}{%
\subsection{Defining Smoothness}\label{defining-smoothness}}

\begin{definition}[Smooth/Regular Varieties]

A connected variety \(X\) is \textbf{smooth} (or \textbf{regular}) if
\(\dim T_p X = \dim X\) for all \(p\in X\). More generally, an arbitrary
(potentially disconnected) variety is smooth if every connected
component is smooth.

\end{definition}

\begin{example}[?]

\({\mathbb{A}}^n\) is smooth since \(T_p {\mathbb{A}}^n = k^n\) for all
points \(p\), which has dimension \(n\).

\end{example}

\begin{example}[?]

\({\mathbb{A}}^n {\coprod}{\mathbb{A}}^{n-1}\) is also smooth since each
connected component is smooth.

\end{example}

\begin{definition}[Singular Varieties]

A variety that is not smooth is \textbf{singular} at \(p\) if
\(\dim T_p X \neq \dim X\).

\end{definition}

\begin{fact}

\(\dim T_p X\geq \dim X\) for \(X\) equidimensional, i.e.~every
component has the same dimension. This rules out counterexamples like
the following in \({\mathbb{A}}^3\):

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-12-03-10-05-17.png}
\caption{Union of Plane and Axis}
\end{figure}

\end{fact}

\begin{example}[?]

Consider \(X\coloneqq V(y^2 - x^3) \subset {\mathbb{A}}^2\):

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-12-03-10-07-39.png}
\caption{\(V(y^2 - x^2)\)}
\end{figure}

Note that \(\dim T_0 X = 2\) is easy to see since it's equal to
\(V\qty{f_1 {~\mathrel{\Big|}~}f\in \left\langle{y^2 - x^3}\right\rangle} = V(0) = k^2\).
Thus \(p\neq 0\) are smooth points and \(p=0\) is the unique singular
point. So \(X\) is not smooth, but \(X\setminus\left\{{0}\right\}\) is.

\end{example}

\begin{definition}[Regular Ring]

A local ring \(R\) over a field \(k\) is \textbf{regular} iff
\(\dim_k \mathfrak{m}/\mathfrak{m}^2 = \dim R\), the length of the
longest chain of prime ideals. Note that we'll add the additional
assumption that \(R/\mathfrak{m} \cong k\).

\end{definition}

\begin{remark}

A variety \(X\) is thus smooth iff
\(\dim_k \mathfrak{m}_p / \mathfrak{m}_p^2 = \dim_p X = \dim {\mathcal{O}}_{X, p}\).

\end{remark}

\begin{theorem}[A hard theorem from commutative algebra (Auslander–Buchsbaum, 1940s)]

A regular local ring is a UFD.

\end{theorem}

\begin{corollary}[?]

Each connected component of a smooth variety is irreducible.

\end{corollary}

\begin{proof}[?]

If a connected component\(X\) is not irreducible, then there exists a
point \(p\in X\) such that \({\mathcal{O}}_{X, p}\) is not a domain, and
thus a nonzero pair \(f, g \in {\mathcal{O}}_{X, p}\) such that
\(fg=0\). These exist by simply taking an indicator function on each
component. So \(0\) doesn't have a unique factorization. So
\({\mathcal{O}}_{X, p}\) is not regular, and thus
\(\dim T_p X > \dim_p X\), which is a contradiction.

\end{proof}

\begin{remark}

How can we check if a variety \(X\) is smooth then? Just checking
dimensions from the definitions is difficult in general.

\end{remark}

\hypertarget{checking-smoothness}{%
\subsection{Checking Smoothness}\label{checking-smoothness}}

\begin{proposition}[Jacobi Criterion]

Let \(p\in X\) an affine variety embedded in \({\mathbb{A}}^n\), and
suppose \(I(X) = \left\langle{f_1, \cdots, f_r}\right\rangle\). Then
\(X\) is smooth at \(p\) \(\iff\) the matrix
\(\qty{{\frac{\partial f}{\partial x_j}\,}}\Big|_{p}\) has rank
\(n - \dim X\).

\end{proposition}

\begin{example}[?]

Is \(V(x^2 - y^2 - 1) \subset {\mathbb{A}}^2\) smooth? We have
\(I(X) = \left\langle{f_1}\right\rangle \coloneqq\left\langle{x^2 - y^2 - 1}\right\rangle\),
so let \(p\in X\). Then consider the matrix
\begin{align*}  
\begin{bmatrix}
J \coloneqq
{\frac{\partial f}{\partial x}\,} & {\frac{\partial f}{\partial y}\,} \\
\end{bmatrix} = 
\begin{bmatrix}
2x & -2y \\
\end{bmatrix}
.\end{align*}
We want to show that at any \(p\in X\), we have
\(\operatorname{rank}(J) = 1\). This is true for \(p\neq (0, 0)\), but
this is not a point in \(X\).

\end{example}

\begin{example}[?]

Consider \(X \coloneqq V(y^2 - x^3 + x^2) \subset {\mathbb{A}}^2\):

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-12-03-10-32-44.png}
\caption{Image}
\end{figure}

Then
\(I(X) = \left\langle{y^2 - x^3 + x^2}\right\rangle = \left\langle{f}\right\rangle\),
and
\begin{align*}  
J = 
\begin{bmatrix}
2y & -3x^2 + 2x \\
\end{bmatrix}
.\end{align*}
Then \(\operatorname{rank}(J) = 0\) at \(p = (0, 0)\), which is a point
in \(X\), so \(X\) is not smooth.

\end{example}

\begin{example}[?]

Consider \(X\coloneqq V(x^2 + y^2, 1+z^3) \subset{\mathbb{A}}^3\), then
\(I(X) = \left\langle{x^2 + y^2, 1 + z^3}\right\rangle \left\langle{f, g}\right\rangle\)
which is clearly a radical ideal.

We then have
\begin{align*}  
J = 
\begin{bmatrix}
f_x & f_y & f_z \\
g_x & g_y & g_z \\
\end{bmatrix}
=
\begin{bmatrix}
2x & 2y & 0 \\
0 & 0 & 3z^2
\end{bmatrix}
,\end{align*}
and thus
\begin{align*}  
\operatorname{rank}(J) = 
\begin{cases}
0 & x=y=z=0 \\
1 & x=y=0 \text{ xor } z=0 \\
2 & \text{else}.
\end{cases}
\end{align*}

We can check that \(\dim X = 1\) and
\(\operatorname{codim}_{{\mathbb{A}}^3} X = 3-1 = 2\), so a point
\((x,y,z) \in X\) is smooth iff \(\operatorname{rank}(J) = 2\). The
singular locus is where \(x=y=0\) and \(z= \zeta_6\) is any generator of
the 6th roots of unity, i.e.~\(\zeta_6, \zeta_6^3, \zeta_6^5\), along
with the point \(0\). Note that \(z=0\) is not a point on \(X\), since
\(1+z^3\neq 0\) in this case.

Thus the singular locus is
\(V(x^2 + y^2) = V((x+iy)(x-iy)) \cap V(1+z^3)\), which results in 3
singular points after intersecting:

\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-12-03-10-42-21.png}
\caption{Image}
\end{figure}

Note that it doesn't matter that \(V(1+z^3)\) was intersected here, as
long as it's anything that intersects the \(z{\hbox{-}}\)axis
nontrivially we will still get something singular.

\end{example}

\hypertarget{appendix-commutative-algebra}{%
\section*{Appendix: Commutative
Algebra}\label{appendix-commutative-algebra}}
\addcontentsline{toc}{section}{Appendix: Commutative Algebra}

\hypertarget{useful-algebra-facts}{%
\subsection{Useful Algebra Facts}\label{useful-algebra-facts}}

\begin{fact}

\envlist

\begin{itemize}
\tightlist
\item
  \(\mathfrak{p}{~\trianglelefteq~}R\) is prime \(\iff R/\mathfrak{p}\)
  is a domain.
\item
  \(\mathfrak{p}{~\trianglelefteq~}R\) is maximal
  \(\iff R/\mathfrak{p}\) is a field.
\item
  Maximal ideals are prime.
\item
  Prime ideals are radical.
\item
  If \(R\) is a PID and
  \(\left\langle{f}\right\rangle {~\trianglelefteq~}R\) is generated by
  an irreducible element \(f\), then \(\left\langle{f}\right\rangle\) is
  maximal
\end{itemize}

\end{fact}

\begin{proposition}[Finitely generated polynomial rings are Noetherian]

A polynomial ring \(k[x_1, \cdots, x_{n}]\) on finitely many generators
is Noetherian. In particular, every ideal
\(I{~\trianglelefteq~}k[x_1, \cdots, x_{n}]\) has a finite set of
generators and can be written as
\(I = \left\langle{f_1, \cdots, f_m}\right\rangle\).

\end{proposition}

\begin{proof}[?]

A field \(k\) is both Artinian and Noetherian, since it has only two
ideals and thus any chain of ideals necessarily terminates. By Hilbert's
basis theorem (\cref{thm:hilbert_basis}), \(k[x_1, \cdots, x_{n}]\) is
thus Noetherian.

\end{proof}

\begin{proposition}[Properties and Definitions of Ideal Operations]

\begin{align*}  
I+J   &\coloneqq\left\{{f+g {~\mathrel{\Big|}~}f\in I,\, g\in J}\right\} \\
IJ    &\coloneqq\left\{{\sum_{i=1}^N f_i g_i {~\mathrel{\Big|}~}f_i\in I,\, g_i\in J, N\in {\mathbb{N}}}\right\} \\
I+J   = \left\langle{1}\right\rangle 
      &\implies I\cap J = IJ && \text{(coprime or comaximal)}
\left\langle{a}\right\rangle + \left\langle{b}\right\rangle = \left\langle{a, b}\right\rangle
.\end{align*}

\end{proposition}

\begin{theorem}[Noether Normalization]\label{thm:noether_normalization}

Any finitely-generated field extension \(k_1 \hookrightarrow k_2\) is a
finite extension of a purely transcendental extension, i.e.~there exist
\(t_1, \cdots, t_\ell\) such that \(k_2\) is finite over
\(k_1(t_1, \cdots, t_\ell)\).

\end{theorem}

\begin{theorem}[Hilbert's Basis Theorem]\label{thm:hilbert_basis}

If \(R\) is a Noetherian ring, then \(R[x]\) is again Noetherian.

\end{theorem}

\hypertarget{appendix-course-exercises}{%
\section{Appendix: Course Exercises}\label{appendix-course-exercises}}

\hypertarget{problem-set-1}{%
\subsection{Problem Set 1}\label{problem-set-1}}

\begin{exercise}[Gathmann 1.19]

Prove that every affine variety \(X\subset {\mathbb{A}}^n/k\) consisting
of only finitely many points can be written as the zero locus of \(n\)
polynomials.

\begin{quote}
Hint: Use interpolation. It is useful to assume at first that all points
in \(X\) have different \(x_1{\hbox{-}}\)coordinates.
\end{quote}

\end{exercise}

\begin{exercise}[Gathmann 1.21]

Determine \(\sqrt{I}\) for
\begin{align*}  
I\coloneqq\left\langle{x_1^3 - x_2^6,\, x_1 x_2 - x_2^3}\right\rangle {~\trianglelefteq~}{\mathbb{C}}[x_1, x_2]
.\end{align*}

\end{exercise}

\begin{exercise}[Gathmann 1.22]

Let \(X\subset {\mathbb{A}}^3/k\) be the union of the three coordinate
axes. Compute generators for the ideal \(I(X)\) and show that it can not
be generated by fewer than 3 elements.

\end{exercise}

\begin{exercise}[Gathmann 1.23: Relative Nullstellensatz]

Let \(Y\subset {\mathbb{A}}^n/k\) be an affine variety and define
\(A(Y)\) by the quotient
\begin{align*}  
\pi: k[x_1,\cdots, x_n] \to A(Y) \coloneqq k[x_1, \cdots, x_n]/I(Y)
.\end{align*}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  Show that \(V_Y(J) = V(\pi^{-1}(J))\) for every
  \(J{~\trianglelefteq~}A(Y)\).
\item
  Show that \(\pi^{-1} (I_Y(X)) = I(X)\) for every affine subvariety
  \(X\subseteq Y\).
\item
  Using the fact that \(I(V(J)) \subset \sqrt{J}\) for every
  \(J{~\trianglelefteq~}k[x_1, \cdots, x_n]\), deduce that
  \(I_Y(V_Y(J)) \subset \sqrt{J}\) for every
  \(J{~\trianglelefteq~}A(Y)\).
\end{enumerate}

Conclude that there is an inclusion-reversing bijection
\begin{align*}  
  \left\{{\substack{\text{Affine subvarieties}\\ \text{of } Y}}\right\} \iff \left\{{\substack{\text{Radical ideals} \\ \text{in } A(Y)}}\right\}
  .\end{align*}

\end{exercise}

\begin{exercise}[Extra]

Let \(J {~\trianglelefteq~}k[x_1, \cdots, x_n]\) be an ideal, and find a
counterexample to \(I(V(J)) =\sqrt{J}\) when \(k\) is not algebraically
closed.

\end{exercise}

\hypertarget{problem-set-2}{%
\subsection{Problem Set 2}\label{problem-set-2}}

\begin{exercise}[Gathmann 2.17]

Find the irreducible components of
\begin{align*}  
X = V(x - yz, xz - y^2) \subset {\mathbb{A}}^3/{\mathbb{C}}
.\end{align*}

\end{exercise}

\begin{exercise}[Gathmann 2.18]

Let \(X\subset {\mathbb{A}}^n\) be an arbitrary subset and show that
\begin{align*}  
V(I(X)) = \mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu
.\end{align*}

\end{exercise}

\begin{exercise}[Gathmann 2.21]

Let \(\left\{{U_i}\right\}_{i\in I} \rightrightarrows X\) be an open
cover of a topological space with \(U_i \cap U_j \neq \emptyset\) for
every \(i, j\).

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  Show that if \(U_i\) is connected for every \(i\) then \(X\) is
  connected.
\item
  Show that if \(U_i\) is irreducible for every \(i\) then \(X\) is
  irreducible.
\end{enumerate}

\end{exercise}

\begin{exercise}[Gathmann 2.22]

Let \(f:X\to Y\) be a continuous map of topological spaces.

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  Show that if \(X\) is connected then \(f(X)\) is connected.
\item
  Show that if \(X\) is irreducible then \(f(X)\) is irreducible.
\end{enumerate}

\end{exercise}

\begin{exercise}[Gathmann 2.23]

Let \(X\) be an affine variety.

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  Show that if \(Y_1, Y_2 \subset X\) are subvarieties then
  \begin{align*}  
  I(\mkern 1.5mu\overline{\mkern-1.5muY_1\setminus Y_2\mkern-1.5mu}\mkern 1.5mu) = I(Y_1): I(Y_2)
  .\end{align*}
\item
  If \(J_1, J_2 {~\trianglelefteq~}A(X)\) are radical, then
  \begin{align*}  
  \mkern 1.5mu\overline{\mkern-1.5muV(J_1) \setminus V(J_2)\mkern-1.5mu}\mkern 1.5mu = V(J_1: J_2)
  .\end{align*}
\end{enumerate}

\end{exercise}

\begin{exercise}[Gathmann 2.24]

Let \(X \subset {\mathbb{A}}^n,\, Y\subset {\mathbb{A}}^m\) be
irreducible affine varieties, and show that
\(X\times Y\subset {\mathbb{A}}^{n+m}\) is irreducible.

\end{exercise}

\hypertarget{problem-set-3}{%
\subsection{Problem Set 3}\label{problem-set-3}}

\begin{exercise}[Gathmann 2.33]

Define
\begin{align*}  
X \coloneqq\left\{{M \in \operatorname{Mat}(2\times 3, k) {~\mathrel{\Big|}~}{\operatorname{rank}}M \leq 1}\right\} \subseteq {\mathbb{A}}^6/k
.\end{align*}

Show that \(X\) is an irreducible variety, and find its dimension.

\end{exercise}

\begin{exercise}[Gathmann 2.34]

Let \(X\) be a topological space, and show

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  If \(\left\{{U_i}\right\}_{i\in I} \rightrightarrows X\), then
  \(\dim X = \sup_{i\in I} \dim U_i\).
\item
  If \(X\) is an irreducible affine variety and \(U\subset X\) is a
  nonempty subset, then \(\dim X = \dim U\). Does this hold for any
  irreducible topological space?
\end{enumerate}

\end{exercise}

\begin{exercise}[Gathmann 2.36]

Prove the following:

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  Every noetherian topological space is compact. In particular, every
  open subset of an affine variety is compact in the Zariski topology.
\item
  A complex affine variety of dimension at least 1 is never compact in
  the classical topology.
\end{enumerate}

\end{exercise}

\begin{exercise}[Gathmann 2.40]

Let
\begin{align*}  
R = k[x_1, x_2, x_3, x_4] / \left\langle{x_1 x_4 - x_2 x_3}\right\rangle 
\end{align*}
and show the following:

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(R\) is an integral domain of dimension 3.
\item
  \(x_1, \cdots, x_4\) are irreducible but not prime in \(R\), and thus
  \(R\) is not a UFD.
\item
  \(x_1 x_4\) and \(x_2 x_3\) are two decompositions of the same element
  in \(R\) which are nonassociate.
\item
  \(\left\langle{x_1, x_2}\right\rangle\) is a prime ideal of
  codimension 1 in \(R\) that is not principal.
\end{enumerate}

\end{exercise}

\begin{exercise}[Problem 5]

Consider a set \(U\) in the complement of \((0, 0) \in {\mathbb{A}}^2\).
Prove that any regular function on \(U\) extends to a regular function
on all of \({\mathbb{A}}^2\).

\end{exercise}

\hypertarget{problem-set-4}{%
\subsection{Problem Set 4}\label{problem-set-4}}

\begin{exercise}[Gathmann 3.20]

Let \(X\subset {\mathbb{A}}^n\)be an affine variety and \(a\in X\). Show
that
\begin{align*}  
{\mathcal{O}}_{X, a} = {\mathcal{O}}_{{\mathbb{A}}^n, a} / I(X) {\mathcal{O}}_{A^n,a}
,\end{align*}
where \(I(X) {\mathcal{O}}_{{\mathbb{A}}^n, a}\) denotes the ideal in
\({\mathcal{O}}_{{\mathbb{A}}^n, a}\) generated by all quotients \(f/1\)
for \(f\in I(X)\).

\end{exercise}

\begin{exercise}[Gathmann 3.21]

Let \(a\in {\mathbb{R}}\), and consider sheaves \(\mathcal{F}\) on
\({\mathbb{R}}\) with the standard topology:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\mathcal{F} \coloneqq\) the sheaf of continuous functions
\item
  \(\mathcal{F} \coloneqq\) the sheaf of locally polynomial functions.
\end{enumerate}

For which is the stalk \(\mathcal{F}_a\) a local ring?

\begin{quote}
Recall that a local ring has precisely one maximal ideal.
\end{quote}

\end{exercise}

\begin{exercise}[Gathmann 3.22]

Let \(\phi, \psi \in \mathcal{F}(U)\) be two sections of some sheaf
\(\mathcal{F}\) on an open \(U\subseteq X\) and show that

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  If \(\phi, \psi\) agree on all stalks, so
  \(\mkern 1.5mu\overline{\mkern-1.5mu(U, \phi)\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5mu(U, \psi)\mkern-1.5mu}\mkern 1.5mu \in \mathcal{F}_a\)
  for all \(a\in U\), then \(\phi\) and \(\psi\) are equal.
\item
  If \(\mathcal{F} \coloneqq{\mathcal{O}}_X\) is the sheaf of regular
  functions on some irreducible affine variety \(X\), then if
  \(\psi = \phi\) on one stalk \(\mathcal{F}_a\), then \(\phi = \psi\)
  everywhere.
\item
  For a general sheaf \(\mathcal{F}\) on \(X\), (b) is false.
\end{enumerate}

\end{exercise}

\begin{exercise}[Gathmann 3.23: Geometry of a Certain Localization]

Let \(Y\subset X\) be a nonempty and irreducible subvariety of an affine
variety \(X\), and show that the stalk \({\mathcal{O}}_{X, Y}\) of
\({\mathcal{O}}_X\) at \(Y\) is a \(k{\hbox{-}}\)algebra which is
isomorphic to the localization \(A(X)_{I(Y)}\).

\end{exercise}

\begin{exercise}[Gathmann 3.24]

Let \(\mathcal{F}\) be a sheaf on \(X\) a topological space and
\(a\in X\). Show that the stalk \(\mathcal{F}_a\) is a \emph{local
object}, i.e.~if \(U\subset X\) is an open neighborhood of \(a\), then
\(\mathcal{F}_a\) is isomorphic to the stalk of
\({ \left.{{ \mathcal{F} }} \right|_{{U}} }\) at \(a\) on \(U\) viewed
as a topological space.

\end{exercise}

\hypertarget{problem-set-5}{%
\subsection{Problem Set 5}\label{problem-set-5}}

\begin{exercise}[Gathmann 4.13]

Let \(f:X\to Y\) be a morphism of affine varieties and
\(f^*: A(Y) \to A(X)\) the induced map on coordinate rings. Determine if
the following statements are true or false:

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(f\) is surjective \(\iff f^*\) is injective.
\item
  \(f\) is injective \(\iff f^*\) is surjective.
\item
  If \(f:{\mathbb{A}}^1\to{\mathbb{A}}^1\) is an isomorphism, then \(f\)
  is \emph{affine linear}, i.e.~\(f(x) = ax+b\) for some \(a, b\in k\).
\item
  If \(f:{\mathbb{A}}^2\to{\mathbb{A}}^2\) is an isomorphism, then \(f\)
  is \emph{affine linear}, i.e.~\(f(x) = Ax+b\) for some
  \(a \in \operatorname{Mat}(2\times 2, k)\) and \(b\in k^2\).
\end{enumerate}

\end{exercise}

\begin{exercise}[Gathmann 4.19]

Which of the following are isomorphic as ringed spaces over
\({\mathbb{C}}\)?

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  \(\mathbb{A}^{1} \backslash\{1\}\)
\item
  \(V\left(x_{1}^{2}+x_{2}^{2}\right) \subset \mathbb{A}^{2}\)
\item
  \(V\left(x_{2}-x_{1}^{2}, x_{3}-x_{1}^{3}\right) \backslash\{0\} \subset \mathbb{A}^{3}\)
\item
  \(V\left(x_{1} x_{2}\right) \subset \mathbb{A}^{2}\)
\item
  \(V\left(x_{2}^{2}-x_{1}^{3}-x_{1}^{2}\right) \subset \mathbb{A}^{2}\)
\item
  \(V\left(x_{1}^{2}-x_{2}^{2}-1\right) \subset \mathbb{A}^{2}\)
\end{enumerate}

\end{exercise}

\begin{exercise}[Gathmann 5.7]

Show that

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  Every morphism
  \(f:{\mathbb{A}}^1\setminus\left\{{0}\right\}\to {\mathbb{P}}^1\) can
  be extended to a morphism
  \(\widehat{f}: {\mathbb{A}}^1 \to {\mathbb{P}}^1\).
\item
  Not every morphism
  \(f:{\mathbb{A}}^2\setminus\left\{{0}\right\}\to {\mathbb{P}}^1\) can
  be extended to a morphism
  \(\widehat{f}: {\mathbb{A}}^2 \to {\mathbb{P}}^1\).
\item
  Every morphism \({\mathbb{P}}^1\to {\mathbb{A}}^1\) is constant.
\end{enumerate}

\end{exercise}

\begin{exercise}[Gathmann 5.8]

Show that

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  Every isomorphism \(f:{\mathbb{P}}^1\to {\mathbb{P}}^1\) is of the
  form
  \begin{align*}  
  f(x) = {ax+b \over cx+d} && a,b,c,d\in k
  .\end{align*}
  where \(x\) is an affine coordinate on
  \({\mathbb{A}}^1\subset {\mathbb{P}}^1\).
\item
  Given three distinct points \(a_i \in {\mathbb{P}}^1\) and three
  distinct points \(b_i \in {\mathbb{P}}^1\), there is a unique
  isomorphism \(f:{\mathbb{P}}^1 \to {\mathbb{P}}^1\) such that
  \(f(a_i) = b_i\) for all \(i\).
\end{enumerate}

\end{exercise}

\begin{proposition}[?]

There is a bijection
\begin{align*}
\left\{{ \text { morphisms } X \rightarrow Y }\right\} 
&\stackrel{1: 1}{\leftrightarrow}
\left\{{  K{\hbox{-}}\text{algebra morphisms } \mathscr{O}_{Y}(Y) \rightarrow \mathscr{O}_{X}(X) }\right\} \\
f &\mapsto f^{*}
\end{align*}

\end{proposition}

\begin{exercise}[Gathmann 5.9]

Does the above bijection hold if

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(X\) is an arbitrary prevariety but \(Y\) is still affine?
\item
  \(Y\) is an arbitrary prevariety but \(X\) is still affine?
\end{enumerate}

\end{exercise}

\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


\printbibliography[title=Bibliography]


\end{document}
