# Monday, October 25

## Length

:::{.remark}
A correction from last time: we said $\CC = \QQbar(t_j \st j\in J)$ for some uncountable set of generators $J$.
Noting that $R \tensor_\ZZ \FF_p = R/pR$, which is zero if ${1\over p}\in R$, so $\CC \tensor_\ZZ \FF_p = 0$.
However, this doesn't happen for $\ZZbar(t_j \st j\in J)$, so passing to a ring given by adjoining coefficients of equations is still a reasonable thing to do.

Last time: for $X \mapsvia{f} Y$, the fiber over $p\in Y$ was $X\fiberprod{Y} \spec k(p)$ where $k(p) \da \OO_{Y, p}/\mfm_p$, sometime denoted $f\inv(p)$, the *scheme-theoretic fiber*.
:::

:::{.example title="?"}
Consider intersecting a parabola with a family of lines:

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

Then there is a map $\spec \CC[x, y]/\gens{y-x^2} \to \spec \CC[y]$ corresponding to a ring map $\CC[y] \to \CC[x,y]/\gens{y-x^2}$.
We showed the scheme theoretic fiber over $y=c_0$ is precisely $\spec \CC[x] / \gens{c_0 - x^2} \cong \CC\sumpower{2}$ if $c_0\neq 0$, and $\spec \CC[x]/\gens{x^2}$ if $c_0 = 0$.
The former has no nilpotents while the latter does, so the fibers are reduced away from $c_0 = 0$.
:::

:::{.definition title="Length"}
If $R\in \kalg^\fg$ with $\krulldim(R) = 0$, then $\length(R) \da \dim_k R < \infty$.
$\spec R$ is called a length $l$ scheme over $k$.
:::

:::{.remark}
Note that $\dim_\CC \CC[x, y] = \infty$, but has Krull dimension 2.
Most $k\dash$algebras will have infinite $k\dash$dimension in this setting.
:::

:::{.remark}
If $R$ is reduced and $k = \kbar$, one can prove that $R = k\sumpower{\ell}$ and $\spec R = \Disjoint_{i\leq \ell} \spec k$ where $\ell$ is the number of reduced points.
:::

:::{.example title="?"}
Take
\[
\spec \CC \oplus \CC[x]/\gens{x^3} \oplus \CC[x, y]/\gens{x^2, xy, y^3}
.\]
The terms have dimension $1,3,4$ respectively, yielding a 0-dimensional length 8 scheme.
Note that $\spec R = \spec R_\red$, and reducing this ring yields $\CC\sumpower{3}$.
:::

:::{.remark}
Let $Y\in \Sch^\irr\slice k$ and $X \mapsvia{f} Y$, then $Y$ has a generic point $\sqrt 0 \in \spec A_i$ for some cover $\ts{\spec A_i}\covers Y$.
This corresponds to the irreducible closed subset $Y$ itself, and yields a unique open generic point $Y_\generic$
One can then take the fiber $f\inv(Y_\generic)$ -- what fiber product is this?
Check that $Y_\generic = A_i\localize{\sqrt 0} / \sqrt 0 = \ff(A_i)$ is exactly the fraction field.
Note that if $\spec B_{ij} \subseteq \spec A_i$ a distinguished open, we have $\ff(B_{ij}) \subseteq \ff(A_i)$, so this doesn't depend on the choice of the affine open $\spec A_i$.
:::

:::{.example title="?"}
Consider $\spec \ZZ[x, y]/\gens{y^2 = x^3-1} \in \Sch\slice \ZZ$, which comes equipped with a map to $\spec \ZZ$.
The generic fiber is the base change to $\spec \QQ$:

\begin{tikzcd}
	{\spec \QQ[x, y] / \gens{y^2=x^3-1}} && X \\
	\\
	{\spec \QQ} && {\spec \ZZ}
	\arrow[from=3-1, to=3-3]
	\arrow[from=1-1, to=3-1]
	\arrow[from=1-3, to=3-3]
	\arrow[from=1-1, to=1-3]
	\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJYIl0sWzAsMCwiXFxzcGVjIFxcUVFbeCwgeV0gLyBcXGdlbnN7eV4yPXheMy0xfSJdLFswLDIsIlxcc3BlYyBcXFFRIl0sWzIsMiwiXFxzcGVjIFxcWloiXSxbMiwzXSxbMSwyXSxbMCwzXSxbMSwwXSxbMSwzLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=)

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

\begin{tikzcd}
	{\spec \CC[x, y]/\gens{y-x^2}} && {\spec \CC[y]} \\
	\\
	{\spec \CC(y)[x]/\gens{y-x^2}} && {\spec \CC(y) = \ts{\pt}}
	\arrow[from=3-3, to=1-3]
	\arrow[from=3-1, to=3-3]
	\arrow[from=1-1, to=1-3]
	\arrow[from=3-1, to=1-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgXFxDQ1t4LCB5XS9cXGdlbnN7eS14XjJ9Il0sWzAsMiwiXFxzcGVjIFxcQ0MoeSlbeF0vXFxnZW5ze3kteF4yfSJdLFsyLDIsIlxcc3BlYyBcXENDKHkpID0gXFx0c3tcXHB0fSJdLFsyLDAsIlxcc3BlYyBcXENDW3ldIl0sWzIsM10sWzEsMl0sWzAsM10sWzEsMF1d)

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

:::{.example title="?"}
Consider two lines projecting onto the $y$ axis, say $(x-1)(x+1) = 0$, then this splits/factors over the generic point.
:::

## Separated/Proper Morphisms


:::{.definition title="Diagonal"}
Let $X\in \Sch\slice S$ with structure map $X \mapsvia{f} S$, then the **diagonal** $\Delta: X\to X\fiberpower{S}{2}$ is the following induced map:

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

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMSwxLCJYXFxmaWJlcnBvd2Vye1N9ezJ9Il0sWzMsMSwiWCJdLFsxLDMsIlgiXSxbMywzLCJTIl0sWzAsMCwiWCJdLFsyLDMsImYiXSxbMSwzLCJmIiwyXSxbMCwxXSxbMCwyXSxbMCwzLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbNCwwLCJcXERlbHRhIiwxXSxbNCwxXSxbNCwyXV0=)

:::


:::{.definition title="Separated"}
A structure map $X \mapsvia{f} S$ is **separated** if the diagonal $\Delta: X\to X\fiberpower{S}{2}$ is a closed embedding.
$X$ itself is separated if $\Delta: X\to X\fiberpower{\spec \ZZ}{2}$ is separated.
:::


:::{.warnings}
The usual "Hausdorff iff diagonal is closed" depends on a separation axiom!
Which will often not hold in AG: for example, $\spec R$ is separated for any ring but never Hausdorff.
:::


:::{.proposition title="?"}
Any morphism in $\Aff\Sch$ is separated.
:::


:::{.proposition title="?"}
Consider:

\begin{tikzcd}
	{\spec B} \\
	\\
	&& {\spec (B\tensorpower{A}{2})} && {\spec B} \\
	\\
	&& {\spec B} && {\spec A}
	\arrow[from=3-5, to=5-5]
	\arrow[from=5-3, to=5-5]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-3, to=5-3]
	\arrow[from=1-1, to=3-5]
	\arrow[from=1-1, to=5-3]
	\arrow["\Delta"{description}, from=1-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiwyLCJcXHNwZWMgKEJcXHRlbnNvcnBvd2Vye0F9ezJ9KSJdLFs0LDIsIlxcc3BlYyBCIl0sWzIsNCwiXFxzcGVjIEIiXSxbNCw0LCJcXHNwZWMgQSJdLFswLDAsIlxcc3BlYyBCIl0sWzEsM10sWzIsM10sWzAsMV0sWzAsMl0sWzQsMV0sWzQsMl0sWzQsMCwiXFxEZWx0YSIsMV1d)

Then there is a ring morphism
\[
\Delta^* B\tensorpower{A}{2} &\to B && \in \CRing\\
b_1\tensor b_2 &\mapsto b_1 b_2
.\]
Since this is surjective, $\Delta$ is a closed immersion.
:::

:::{.example title="A classic non-example"}
Let $X$ be $\AA^1$ with the doubled origin, so $X = \AA^1\disjoint_f \AA^1$ where for $U\da \AA^1\smz$, we glue by $\id_U$.
Taking the product $X\fiberprod{\spec k} X$ yields the following:

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

Note that $(0, 0') \not\in \Delta(X)$ is not closed, but $(0, 0')\in \bar{\Delta(X)}$ is in its closure.
:::