# Thursday, February 24

## Going up, going down

:::{.theorem title="Going Up I"}
Let $A \subseteq B\in \CRing$ with $B$ integral over $A$.
Then for every $p\in \spec A$ there is a $q\in \spec B$ with $p = A \intersect q$.
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:::{.remark}
Idea: $\spec B\to \spec A$ has finite fibers.
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:::{.lemma title="?"}
If $A \injects B$ is an integral extension, then $A\in \Field\iff B\in \Field$.
:::

:::{.proof title="of lemma"}
$\implies$: 
Last time.

$\impliedby$:
Given $x\in A$ we have $x\inv \in A$.
Then if $f(x) = x^{-n} + \sum_{0\leq k \leq n-1} a_k x^{-k}$, with $f(x) = 0$, then $x^{n-1}f(x) = 0 \implies x\inv = -\sum a_k x^k \in A$.
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:::{.proof title="of Going Up"}
Note $A\localize{p} \injects B\localize{p}$ is still integral, and let $q'\in \mspec B\localize{p} \neq \emptyset$.
Write $p$ for the maximal ideal of $A$, then (claim) $A\localize{p} \intersect q' = p$.
STS $A\localize{p} \intersect q'$ is maximal, since these are local rings, and so it's ETS $A\localize{p}/(A\localize{p} \intersect q') \in \Field$.
Since this is integral iff $B\localize{p}/q'$ is integral, which it is, by the lemma this is a field too.

:::{.exercise title="?"}
Show tat taking the preimage of $q'$ in $B$ works.
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:::

:::{.theorem title="Going Up II: partial lifts of chains of primes extend"}
Given $A \injects B$ integral and 
\[
p_1 &\subseteq p_2 \subseteq \cdots \subseteq p_n \in \spec A \\
q_1 &\subseteq q_2 \subseteq \cdots \subseteq q_m \in \spec B \\
,\]
where $p_i = q_i \intersect A$ and e.g. with $m\leq n$, then there exist $q_{m+1}, \cdots, q_n$ with $q_j \subseteq q_{j+1}$ and $q_i \intersect A = p_i$.
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:::{.proof title="?"}
Note that it's enough to lift one stage and induct.
So given $q_1 \subseteq \cdots q_n$, it's ETF $q_{m+1} \contains q_m$ with $q_{m+1} \intersect A = p_{m+1}$.
Strategy:

- Replace $A$ with $A/p_n$ and $B$ with $B/q_n$.
- Find $\bar q \in B/q_m$ with $\bar q \intersect A/p_n$ the image of $p_{n+1}$
- Use that $\bar q$ exists by Going Up I.

:::

:::{.remark}
The geometry: $A\to B\leadsto \spec B \mapsvia{\pi} \spec A$.
Increasing chains $p_i$ means $p_{i+1}\in \cl_{\spec A} p_i$, and "going up" means sequences can be completed with points in closures in $\spec B$
I.e. $\pi$ is a closed map, i.e. closed under specialization (passing to a point in the closure).
Idea: covering map, possibly with ramification or splitting.

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:::{.example title="?"}
Consider $k[x]\injects k[x,y]/\gens{y^2-x}$ over $\characteristic k = 0$ and $k=\kbar$.
Take $p=\gens{x-2}$ and $q = \gens{y-\sqrt 2}$ or $\gens{y+\sqrt 2}$ extend $p$.

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![](figures/2022-02-24_13-21-53.png)

Similarly, take $\ZZ\injects \ZZ[i]$ and consider how primes lift (see inert, split, ramified).
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:::{.definition title="Integrally closed"}
For $A \subseteq B$, say $A$ is **integrally closed in $B$** iff $A$ contains every element of $B$ which is integral over $A$.
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:::{.example title="?"}
$k[x] \subseteq k[x,y]$ is integrally closed, but $k[x] \subseteq k[x,y]/\gens{y^2-x}$ is not.
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:::{.theorem title="Going Down"}
For $A\leq B\in \IntDomain$ with $A$ integrally closed in $\ff(A)$ and $B$ integral over $A$.
If
\[
p_1 &\contains p_2 \contains \cdots \contains p_n \in \spec A \\
q_1 &\contains q_2 \contains \cdots \contains q_m \in \spec B \\
\]
with $q_i \intersect A = p_i$, then there exist $q_m \contains q_{m+1} \contains \cdots \contains q_{n+1} \contains q_n$ with $q_i \intersect A = p_i$.
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:::{.proof title="?"}
See A&M, similar to proof of going up.

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:::{.remark}
Idea: closed under *generization* (opposite of specialization, given $x$ finding a point $y$ with $x\in \cl y$), so the geometric map is almost open.
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:::{.example title="?"}
Being integrally closed corresponds to a variety being **normal** and is a smoothness condition.
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:::{.exercise title="Challenge"}
Use $k[x,y]/\gens{y^2-x^3} \cong k[x^2, x^3] \injects k[x,y]$ to construct a counterexample to the going down theorem when $A \injects \ff(A)$ is *not* integrally closed.
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## Local Properties

:::{.remark}
On local properties: for $M\in \rmod^{\fg, \Noeth}$, a HW problem shows if $p\in \spec A$ and $M\localize{p} = 0$ then $M\localize{f} = 0$ for some $f$.
This says that the property of being zero extends:

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:::{.definition title="Local"}
A property $Q$ is **local** iff given $M\in \mods{A}$, TFAE:

- $Q$ holds for $M$
- $Q$ holds for $M\localize{f_i}$ for every $\ts{f_i}$ with $\gens{\ts{f_i}} = \gens{1}$.

:::

:::{.slogan}
Local properties can be checked on an open cover of $\spec A$, and $\mods{A}$ corresponds to $\QCoh(\spec A)$.
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:::{.remark}
One can always take the set $\ts{f_i}$ to be finite since if such a collection generates the unit ideal, there is some finite sum $\sum a_i f_i = 1$.
One can also reformulate the second condition as follows: for each $p\in \spec A$, there exists some $f_p \not\in p$ such that every $M\localize{f_p}$ satisfies $Q$.
$\implies$:
Check that $\gens{\ts{f_p}} = \gens{1}$; if not then there exists some $m\in \mspec A$ with $\ts{f_p} \subseteq m$ which is maximal and hence prime.

$\impliedby$:
The claim is that given $p$ there exist $f_i\not\in p$.
If not, $\ts{f_i} \subseteq p$ and $1\in p$.
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:::{.corollary title="?"}
If $\gens{\ts{f_i}} = 1$ then $\spec A\localize{f_i} = \spec A\sm V(f_i) \subseteq \spec A$ is an open cover of $\spec A$.
Thus $\spec A$ is quasicompact for any ring $A$.
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:::{.remark}
Some local properties:

- Being zero
- Being injective/surjective/bijective
- Being finitely generated (and projective)
- Being flat

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