# Dimension (Tuesday, September 08)

Review: we discussed irreducible components.
Recall that the *Zariski topology* on an affine variety $X$ has affine subvarieties as closed sets, and a *Noetherian space* has no infinitely decreasing chains of closed subspaces.
We showed that any Noetherian space has a decomposition into irreducible components $X = \union X_i$ 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 $\kx{n}$ is a Noetherian ring, this chain terminates, so affine varieties are Noetherian.

## Dimension

:::{.definition title="Dimensions"}
Let $X$ be a topological space.

1. The **dimension** $\dim X \in \NN\union\ts{\infty}$ is either $\infty$ or the length $n$ of the longest chain of *irreducible* closed subsets $\emptyset \neq Y_0 \subsetneq \cdots \subsetneq Y_n \subset X$ where $Y_n$ need not be equal to $X$.[^explaining_chain]



2. The **codimension** of $Y$ in $X$, $\codim_X(Y)$, for an irreducible subset $Y\subseteq X$ is the length of the longest chain $Y\subset Y_0 \subsetneq Y_1 \cdots \subset X$.


[^explaining_chain]: 
Note that we count the number of nontrivial strict subset containments in this chain.

:::


:::{.example}
Consider $\AA^1/k$, 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 $\ts{a}, \AA^1/k$, since an affine variety is irreducible iff its coordinate ring is a domain and $A(\AA^1/k) = k[x]$.
We can check

\begin{tikzcd}
	{\emptyset} & {\ts{a}} & {\AA^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}

which is of length $1$, since there is one nontrivial containment $Y_1 \subsetneq Y_2$, and so $\dim(\AA^1/k) = 1$.
:::


:::{.example}
Consider $V(x_1 x_2) \subset \AA^2/k$, 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 $k[x_1, x_2] / \gens{x_1} \cong k[x_2]$ is a domain, so $V(x_i)$ are irreducible.
So we can have a chain
\[  
\emptyset \subsetneq \ts{a} \subsetneq V(x_1) \subset X
,\]
where $a$ is any point on the $x_2\dash$axis, so $\dim(X) = 1$.

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

:::{.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 **false**!
:::


:::{.example}
Take $X = \NN$ and define a topology by setting closed subsets be the sets $\ts{0, \cdots, n}$ as $n$ ranges over $\NN$, along with $\NN$ itself.
Is $X$ Noetherian? 
Check descending chains of closed sets:

\[  
\NN \supsetneq \ts{0, \cdots, N} \supsetneq \ts{0, \cdots, N-1} \cdots
,\]

which has length at most $N$, so it terminates and $X$ is Noetherian.
But note that all of these closed subsets $X_N \da \ts{0, \cdots, N}$ are irreducible.
Why?
If $X_n = X_i \union X_j$ 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
$\dim(\NN) = \infty$.
:::

:::{.remark}
Let $X$ be an affine variety. 
There is a correspondence
\[  
\correspond{\text{Chains of irreducible closed subsets} \\ Y_0 \subsetneq \cdots \subsetneq Y_n \text{ in } X}
\correspond{\text{Chains of prime ideals} \\ P_0\supsetneq \cdots \supsetneq P_n \text{ in } A(X)}
.\]
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 $A(Y) \da A(X) / I$ is a domain, which precisely correspond to prime ideal in $A(X)$.
:::

We thus make the following definition:

:::{.definition title="Krull Dimension"}
The *Krull dimension* of a ring $R$ is the length $n$ of the longest chain of prime ideals
\[  
P_0 \supsetneq P_1 \supsetneq \cdots \supsetneq P_n
.\]

:::

:::{.remark}
This uses the key fact from commutative algebra: a finitely generated $k\dash$algebra $M$ satisfies

1. $M$ has finite $k\dash$dimension
2. If $M$ is a domain, every maximal chain has the same length.
:::

:::{.remark}
From scheme theory: for any ring $R$, there is an associated topological space $\spec R$ given by the set of prime ideals in $R$, where the closed sets are given by 
\[  
V(I) = \ts{\text{Prime ideals } \mfp \normal R \st I\subseteq \mfp }
.\]

If $R$ is a Noetherian ring, then $\spec(R)$ is a Noetherian space.
:::


:::{.example}
Using the fact above, let's compute $\dim \AA^n/k$.
We can take the following chain of prime ideals in $\kx{n}$:
\[  
0 \subsetneq \gens{x_1} \subsetneq \gens{x_1, x_2} \cdots \subsetneq \gens{x_1, \cdots, x_n}
.\]

By applying $V(\wait)$ we obtain
\[  
\AA^n/k \supsetneq \AA^{n-1}/k \cdots \supsetneq \AA^0/k = \ts{0} \supsetneq \emptyset
,\]
where we know each is irreducible and closed, and it's easy to check that these are maximal:

If there were an ideal $\gens{x_1, x_2} \subset P \subset \gens{x_1, x_2, x_3}$, then take $P\intersect k[x_1, x_2, x_3] / \gens{x_1, x_2}$ which would yield a polynomial ring in $k[x_1]$.
But we know the only irreducible sets in $\AA^1/k$ are a point and the entire space.

So this is a chain of maximal length, implying $\dim \AA^n/k = n$.
:::