# Noetherian Schemes (Wednesday, October 06)

## Proof of integral iff irreducible and reduced

:::{.remark}
Recall: $X\in \Sch$ is integral $\iff$ $X$ is irreducible and reduced, which are both defined on sections in terms of commutative algebra on the corresponding rings. 
:::

:::{.proof title="?"}
**Irreducible and reduced $\implies$ integral**:
By contrapositive, assume $\OO_X(U)$ is not a domain, so $fg = 0$ in $\OO_X(U)$.
A local ring need not be domain.
However, the germ $f_p g_p \da \Res(U, \mfp)(fg) = 0$ in the stalk $\OO_{X, \mfp}$.
If $(a/s)(b/t) = 0 \in \mfp$, then either $a/s$ or $b/t$ is in $\mfp$, so $f_p$ or $g_p$ is in $\mfp$.
Note that the following sets are closed:
\[
U_1 \da \ts{\mfm \in U \st f_p \in \mfm} \subseteq U, \qquad
U_2 \da \ts{\mfm\in U \st g_p \in \mfm} \subseteq U
.\]

We can write $U = U_1 \union U_2$, so if $X$ is irreducible then $U$ is irreducible, so some $U_i = U$, say $U_1$.
So take an open affine $V \subseteq U_1$ with $\ro{f}{V}\neq 0$, using the sheaf property.
Writing $V = \spec R$, we have $\ro{f}{V} \in \OO_X(V) = R$, and the stalk $f_p\in p$ for all $p\in R$.
Then $f\in p$ for all $p\in \spec R$, thus in their intersection, and so $f\in \nilrad{R}$.
Since $f\neq 0$, this contradicts that $X$ is not reduced.
$\contradiction$
:::

:::{.remark}
Recall that *Noetherian rings* are those that satisfy the ACC, or equivalently that all ideals are finitely generated (e.g. are finitely generated $k\dash$algebras).
Also recall that a *Noetherian space* is a space where every descending sequence of closed sets stabilizes.
:::

## Locally and globally Noetherian schemes

:::{.definition title="Noetherian rings and spaces"}
$X\in \Sch$ is **locally Noetherian** $\iff$ there exists an affine open cover with $U_i = \spec A_i$ for $A_i$ Noetherian.

$X\in \Sch$ is **(globally) Noetherian** $\iff$ $X$ is locally Noetherian and is additionally quasicompact.[^quasicompact_noeth]

[^quasicompact_noeth]: 
Every open cover has a finite subcover.

:::

:::{.example title="Non-Noetherian rings can produce Noetherian spaces"}
The hypothesis of being a Noetherian space isn't enough in general.
Consider the ring of **Puiseux series** studied by Newton, 
\[
R \da \Union_{n\geq 1} k\formalpowerseries{t^{1\over n}}
.\]
Then $\spec R$ has $2$ points:
\[
\spec R = \ts{ \mfp, \gens 0 }, \qquad \mfp \da \gens{\ts{t^r }_{r\in \QQ_{\geq 0}}}
.\]
Here $\gens{0}$ has closure containing $\mfp$, so $\mfp$ is a generic point.
This is a valuation ring, just not a DVR, and is a Noetherian topological space since there are only two closed sets.
However, $R$ is not Noetherian, since there is an infinite chain of ideals:
\[
\ts{I_j}_{j\geq 1}: \quad \gens{t} \subsetneq \gens{t^{1\over 2}} \subsetneq \gens{t^{1\over 3}} \subsetneq \gens{t^{1\over 3}} \subsetneq \cdots
.\]
However, $V(I_j) = V(I_{k})$ for all $j, k$, and all equal to $V(\gens p)$, so $\spec R$ is a Noetherian space!

:::

:::{.fact title="(fun!)"}
When $k=\kbar$,
\[
\ff\qty{\Union_{n\geq 1} k \fps{t^{1\over n} }} = \Union_{n\geq 1} k\laurentseries{t^{1\over n}} = \bar{k\laurentseries{t}}
.\]
:::

:::{.remark}
There are many theorems of the form "a scheme is locally *something*".
Here we required an open affine cover by $\spec R$ for $R$ Noetherian rings.
The following two conditions will thus be equivalent:
:::

:::{.slogan}
A scheme locally has property $P$ if either

- Property $P$ holds on every affine open $U \subseteq X$, or equivalently
- There exists some affine open cover $\mcu \covers X$ satisfying property $P$.

:::

## Locally Noetherian iff covered by Noetherian rings

:::{.theorem title="Locally Noetherian iff affine charts are Noetherian rings"}
$X\in \Sch$ is locally Noetherian $\iff$ for any affine open $U = \spec A \subseteq X$, $A$ is a Noetherian ring.
:::

:::{.proof title="of theorem"}
$\impliedby$:
Definitional, just apply the hypothesis to some affine open cover.

$\implies$:
The more nontrivial direction.


:::{.fact title="from ring theory"}
The localization of any Noetherian ring is again Noetherian
:::

Let $U \subseteq X$ be an affine open, so $U = \spec B$, and let $\mcu \covers X$ be an affine cover:

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Consider $U \intersect \mcu_i \subseteq \mcu_i$ open, which can be covered by distinguished open sets.
So write $U \intersect \mcu_i = \Union_j V_{ij}$ with $V_{ij} = D(f_{ij}) \subseteq \spec A_{i}$. 
Then $U$ is covered by $\spec (A_i)_{f_{ij}}$, i.e. spectra of local Noetherian rings.
Can we conclude that $B$ is Noetherian from this?
This will follow from the fact that we can further decompose $V_{ij} = \Union W_{ijk}$ where $W_{ijk} = D_B(f_{ijk})$.

So we want to show the following ring-theoretic statement: let $B\in \Ring$ and $\ts{g_i} \subseteq B$ be a collection such that $\spec B = \Union \spec B_{g_i}$ with each $B_{g_i}$ Noetherian, then $B$ is necessarily Noetherian.
Equivalently, we need $\gens{g_i} = \gens {1}$, which corresponds to $\intersect V(g_i) = \emptyset$.
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