# Monday, October 18

## Dimension

:::{.question}
If $X = \spec A$ is affine and $U \subset \realize{X}$ is open, is the inclusion $U \injects X$, represented say by $\spec A' \injects \spec A$, represented by a ring map $A\to A'$?
:::

:::{.definition title="Dimension"}
For $X\in \Sch$, write $\dim X \da \dim_\Top \realize{X}$ as the topological dimension of the underlying space, which is the length of the longest chain of irreducible closed subsets
\[
\emptyset \subsetneq Z_0 \subseteq Z_1 \subsetneq \cdots \subsetneq Z_n \subseteq \realize{X}
,\]
where equality at the end is possible if $\realize{X}$ is irreducible.
:::

:::{.example title="?"}
\envlist

- $\dim \spec k = 0$
- $\dim \spec \ZZpadic$: consider $\emptyset \subsetneq \pt \subseteq \spec \ZZpadic$, where $\pt$ is a generic point, so $\dim \spec \ZZpadic = 1$.
- $\dim \PP^n\slice k = \dim \AA^n\slice k = n$.

:::

:::{.example title="?"}
If $X = \spec A$ is affine for $A$ then $\dim X = \krulldim A$ is the Krull dimension of the ring $A$.
This follows because irreducible closed subsets of $\spec A$ biject with prime ideals of $A$.
Why is this true?

$\impliedby$:

Suppose $p\containedin A$ is prime, then note that $V(p) = \ts{q \in \spec A \st q \contains p}$.
If $V(p) = V(I) \union V(J) = V(IJ)$, then $p \contains IJ$ so $p$ contains one of $I, J$.
But then $V(p) = V(I)$ wlog, so $V(p)$ is an irreducible closed subset.

$\implies$:
We can reverse almost all of these implications:

- $V(p) = V(IJ)$
- $\iff p\contains IJ$
- $\iff p \subseteq I$ or $p \subseteq J$
- $\iff V(p) = V(I)$ or $V(J)$.

Note that bijections preserve strict containments, so we have correspondences on chains:
\[
\emptyset \subsetneq Z_0 \subsetneq \cdots \subsetneq Z_n \subset X = \spec A \\
\iff \\
\gens{1} \supsetneq p_0 \supsetneq \cdots \supsetneq p_n
.\]

:::

:::{.remark}
So we can use that $\krulldim \kxn = n$ to show $\dim \AA^n\slice k = n$.
For $\PP^n\slice k$, use that any maximal chain contains a point $\ts{z_0}$, so choosing such a point and intersecting $z_i$ with the embedded copy of $\AA^n\slice k\injects \PP^n\slice k$.
Then use that there is a chain $\gens{0} \subsetneq \gens{x_1} \subsetneq \gens{x_1, x_2} \cdots \subsetneq \gens{x_1,\cdots, x_n}$, so $\dim X \geq n$.
For the reverse inequality: this is hard! 
See Atiyah-MacDonald's discussion of regular systems of parameters.
:::

:::{.definition title="Codimension"}
The **codimension** $\codim(Z, X)$ for $Z \subseteq X$ a closed irreducible subset is the length of the longest chain starting at $Z$:
\[
Z = Z_0 \subsetneq Z_1 \subsetneq \cdots \subsetneq Z_n \subset X
.\]
:::

:::{.fact}
For $X = \spec A$ and $A\in \kalg^\fg$, there is a formula
\[
\dim(Z) + \codim(X, Z) = \dim (X)
.\]
:::

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

:::{.example title="?"}
$\dim \spec \ZZ = 1$, instead of having dimension zero!
This is because there's always a chain $0 \to \gens{p} \to \ZZ$ for any prime.
An analogy here is a curve $\spec k[x, y] / \gens{f(x, y)}$:


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

One can similarly do this for $\OO_K$ the ring of integers in a number field $K$ and get $\dim \spec \OO_K = 1$.
This leads to a good theory of divisors (free modules on codimension 1 subvarieties) and the Picard group, so a useful geometrization of number theory.
:::

## Fiber Products

:::{.remark}
Perhaps the most important construction in schemes!
Picks up intersection multiplicities.
:::

:::{.definition title="Fiber products"}
Let $X, Y \in \Sch\slice{S}$ then $X\fiberprod{S} Y \in \Sch\slice{S}$ is an $S\dash$scheme equipped with morphisms of $S\dash$schemes onto $X, Y$ satisfying a universal property.
For any $Z$ with maps to $X$ and $Y$, there is a unique $\theta$ making the following diagram commute:

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

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMSwwLCJaIl0sWzAsMywiWSJdLFsyLDMsIlgiXSxbMSw0LCJTIl0sWzEsMV0sWzEsMiwiWFxcZnB7U30gWSJdLFsxLDNdLFsyLDNdLFswLDFdLFswLDJdLFswLDUsIlxcZXhpc3RzICFcXHRoZXRhIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzUsMV0sWzUsMl1d)

:::

:::{.remark}
Note that on the ring side, this yields a tensor product over $S$.
:::