# Locally Noetherian Schemes vs Noetherian Covers (Friday, October 08)

## Proof of Theorem

Recall that we were proving the following:

:::{.theorem title="?"}
$X\in \Sch$ is locally Noetherian iff for any affine open $U = \spec A \subseteq X$, $A$ is a Noetherian ring.
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:::{.remark}
Recall that we covered $X$ by $U_i$, had some affine open $U$ isomorphic to $\spec$ of a ring, and then covered each intersection $U \intersect U_i$ by distinguished opens which were $\spec R\localize{f_i} = D(f_i) = \ts{\mfp\not\ni f_i}$. 
Then $R\localize{f_i}$ is Noetherian iff $\Union_{i\in I} D(f_i) = \spec R$, which implies $\Intersect_{i\in I} V(f_i) = \emptyset$, and thus $\not\exists p\in \spec R$ prime with $p\ni f_i$ for all $i\in I$.
Then $\gens{f_i \st i\in I} = \gens{1}$.

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:::{.proposition title="?"}
$\spec R$ is quasicompact.
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:::{.proof title="of proposition"}
Let $\mcu \covers \spec R$, so $\spec R = \Union_{i\in I} U_i$, then we want to find a finite subcover.
Take $\ts{D(f_{ij})} \covers U_i$; it suffices to find a finite subcover of the refined cover by distinguished opens, so reduce to $U_i = D(f_i)$ for each $i$.
Using the argument from the above remark, $\gens{f_i \st i\in I} = \gens{1}$ since this is a cover.
But then there exists a finite sum $\sum_{j=1}^N a_j f_{ij} = 1$ for some $a_j\in R$, so $\ts{f_{ij} }_{j=1}^N = \gens{1}$ which implies that $\Union_{j=1}^N D(f_{ij}) = \spec R$.
:::

:::{.remark}
Applying the proposition above, we can find a finite set $\ts{f_i}$ such that $\gens{f_i \st i\in I} = \gens{1}$ with each $R\localize{f_i}$ Noetherian.
We'll use the following:
:::

:::{.lemma title="?"}
Let $J\normal R$ and $\phi_i: R\to R\localize{f_i}$ by the canonical localization morphism.
Setting $\gens{f_1,\cdots, f_s} = \gens{1}$,
\[
J = \intersect_{i=1}^s \phi_i \qty{ \phi_i\inv(J) R\localize{f_i} } 
.\]

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:::{.proof title="?"}
Note that $\phi_i\qty{ \phi_i\inv(J) R\localize{f} } \neq J$ generally, e.g. if $f\in J$.
So that $J$ is contained in the right-hand side is clear.
For the reverse containment, let $b \in \Intersect_i \phi_i\inv(\phi_i(J) R\localize{f_i} )$.
Then $\phi_i(b)\in \phi_i(J) R\localize{f_i}$ for all $i$, so $b \sim a_i / f_i^{n_i}$ in the localization for some $a_i \in J$.

Since $\ts{f_i}$ is finite, assume that that $n_i = n$ for some uniform $n$, e.g. $n = \max \ts{n_i}$.
Then $b \sim a_i/f_i^N$, so there exist $m_i$ such that $f_i^{m_i}(f_i^N b - a_i) = 0$ in the original ring $R$.
So now pick $M \da \max\ts{m_i}$ to obtain $f_i^M(f_i^N - a_i) = 0$.

Now a trick: use that $f_i^{M+N}b\in J$ for all $i$, and the claim is that $\gens{f_i^{M+N}}_{i\in I} = \gens{1}$.
More generally, raising all generators of a unit ideal to a fixed power still generates the unit ideal.
This follows from writing $1 = \sum_{i=1}^r c_i f_i \implies 1 = 1^M = \qty{\sum c_i f_i}^M$, so choose $M$ large enough so that some $f_i$ occurs with an exponent of at least $m+n$, e.g. choosing $M = r(m+n)$.

:::{.example title="?"}
If $1 = \gens{x, y}$, then $\gens{x^2, y^2} = 1$ by taking  $ax + by = 1$ and (say) expanding $(ax+by)^4=1$ and noting that any term is divisible by either $x^2$ or $y^2$.
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Now writing $\sum_{i=1}^r c_i f_i^{m+n} = 1$, we get $\sum_{i=1}^r c_i f_i^{m+n} \in J$, and thus $b\in J$.
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:::{.remark}
We now know that the $R\localize{f_i}$ are Noetherian.
Let $J_1 \subseteq J_2 \subseteq \cdots$ be an ascending chain of ideals in $R$, we'll show it stabilizes.
Since the $R\localize{f_i}$ are Noetherian, there is an ascending chain $J_1 R\localize{f_i} \subseteq J_2 R\localize{f_i} \subseteq \cdots$ in $R\localize{f_i}$, which is Noetherian and thus stabilizes.
So for some $N = N(i) \gg 0$, $J_k R\localize{f_i} = J_{k+1} R\localize = \cdots$ for all $k\geq N$.
But there are only finitely many $f_i$, so we can choose some uniformly large $\tilde N\gg 0$ not depending on $i$ where $J_k = J_{k+1} = \cdots$ for all $k\geq \tilde N$ by applying the lemma.
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:::{.remark}
On applying the lemma: use that
\[
\phi_i\inv(J_k R\localize{f_i}) &= \phi_i\inv(J_{k+1}R\localize{f_i}) \quad \forall k\geq N \\
&\implies
\intersect_i \phi_i\inv(J_k R\localize{f_i}) = \intersect_i \phi_i\inv(J_{k+1}R\localize{f_i}) \quad \forall k\geq N \\
&\implies
J_{k} = J{k+1}\quad \forall k\neq N
.\]
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## Other Properties

:::{.example title="A scheme that is not quasicompact"}
Let $X = \ZZ$ with the discrete topology (so every set is open) and set $\OO_X(U) = \Set(U, k)$ to be the local ring of arbitrary set functions.
Then for $p\in X$, the stalks are $\OO_{X, p} \OO_X(\ts{P}) = \Set(p, k) = k$, which is local.
This is a locally ringed space, since it's locally isomorphic to $\spec R$: we can take an open cover of such, or find a neighborhood where it holds, but $p \in \ts{p}$ which is open and letting $\mcf \da \ro{\OO_X}{\ts p}$, we have $(\ts{p}, \mcf) \cong \spec k$.
Then $X = \Union_{p\in X} \ts{p}$ is an open cover with no finite subcover.

So $\ZZ$ with the discrete topology is not $\spec R$ with the Zariski topology for any ring.
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:::{.exercise title="?"}
$X \da \spec \bar{\QQ}[t] = \ts{\gens 0} \union \ts{\gens{t-a_i} \st i\in I}$ where $I$ is a countable enumeration of $\bar{\QQ}$.
Is this quasicompact?
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:::{.exercise title="?"}
Consider $R \da \prod_{n\in \ZZ} \CC$, then there is a set map $\ts{I\normal R} \iso \mcp(\ZZ)$, given by sending any subset to the ideal $\CC \oplus \CC \oplus \cdots$ which are zeroed out at entries according to the complement of $S$.

What is $\spec R$, and what is the topology?
Is $\spec R$ quasicompact?

> Consider $I \da \ts{(a_i) \st a_i = 0\quad i\geq N \gg 0}$, which forms an ideal.
> Is $I$ prime? Are there prime ideals not containing $I$?

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