# Wednesday April 22nd

Recall that we had an antitone bijection

\begin{align*}
\text{prime ideals of } R & \iff \text{irreducible closed subsets of } \sec R \\
\text{maximal ideals} &\iff \text{Singletons } \theset{p}, p\text{ closed} \\
\text{minimal primes} &\iff \text{maximal irreducible closed subsets}
.\end{align*}

Note that the last item on the RHS corresponds to an irreducible component.

Definition (Noetherian Space)
: A topological space $X$ is *Noetherian* iff the ascending chain conditions holds for open sets.

Proposition (13.14: In Noetherian Spaces, Every Subset is Quasicompact)
: $X$ is a Noetherian space iff every *open* subset $U \subset X$ is quasicompact. iff *all* subsets are quasicompact.

Note that this is a strange class of spaces!

Exercise
:   If $X$ is Noetherian and Hausdorff, then $X$ is finite.
    Conversely every finite space is Noetherian, regardless of whether or not it is Hausdorff.

Proposition (Noetherian Spaces are Finite Direct Sum )f Connected Components)
: A Noetherian space has finitely many connected components and is the direct sum (disjoint union) of its connected components.

This holds e.g. for manifolds, but not every topological space.
Thus if we're considering the topology of such spaces, we may as well assume they're connected.

Recall $\spec R$ is connected when $R$ is connected (no nontrivial idempotents), and a similar condition holds here:

Theorem (Spec R is Noetherian iff ACC Holds for Radical Ideals)
:   For a ring $R$, TFAE

    1. $\spec R$ is Noetherian.
    2. The ACC holds on radical ideals of $R$.

Note that if $R$ is Noetherian, (2) holds, so this is a weaker condition.

Proof
: This follows from the Galois correspondence.

Thus if $R$ is Noetherian, $\spec R$ is Noetherian.

Next up: any Noetherian ring has finitely many minimal primes.

Proposition (Noetherian Spaces Have Finitely Many Irreducible Components)
: A Noetherian space has finitely many irreducible components,


**Step 1:**
Show that $X$ is a finite union of irreducible closed subsets.

Let $f$ be the set of counterexamples: closed subsets of $X$ that are *not* a finite union of irreducible closed subsets.
Since $X$ is Noetherian, if $f\neq \emptyset$ it has a minimal element $Y$.
Since $Y$ is not irreducible, we have $Y = Y_1 \union Y_2$ which are proper closed subsets.
By minimality, $Y_1, Y_2 \not \in f$, so they must be finite unions of irreducible closed subsets, and thus so is $Y$.

So $f$ must be empty, and $X = Y_1 \disjoint \cdots \disjoint Y_n$ with the $Y_i$ irreducible.
If there exists an $i, j$ such that $Y_i \subsetneq Y_j$, so remove $Y_i$ and similarly all redundancies.
This yields a cover $X = \union_{i=1}^n Y_i$ where $Y_i \not\subset Y_j$ for any $i, j$.

Claim
: The irreducible components of $X$ are precisely the $Y_i$.

Proof
:   Let $Z$ be irreducible and closed, $Z = \union_{i=1}^n \qty{Z\intersect Y_i}$.
    Since $Z$ is irreducible, $Z\intersect Y_i = Z$ for some $i$, so $Z \subset Y_i$.

$\qed$

Corollary
: $R$ Noetherian implies $\mathrm{MinSpec} R$ is finite.

This wraps up everything we'll say about the Zariski topology.

## Integral Extensions

Leading toward normalization theorems, potentially Krull-Akuzuki.

We'll first generalize the notion of "algebraic" for fields

Definition (Integral Elements of Rings)
: For $R\subset S$ rings, $\alpha\in S$ is *integral* over $R$ if there exists a monic $P \in R[t]$ such that $P(\alpha) = 0$.

Note that we don't require $P$ to be monic.
We say $R\subset S$ is integral iff every $\alpha \in S$ is integral over $R$.

Theorem (Characterization of Integral Elements)
:   For $R\subset T \ni \alpha$, TFAE:

    1. $\alpha$ is integral over $R$
    2. $R[\alpha]$ is finitely generated as an $R\dash$module.
    3. There exists an $S$ such that $R \subset S \subset T$ such that $\alpha\in S$ and $S$ is finitely generated as an $R\dash$module.

This says that an integral extension is "locally finite".

Example
: For $L/K$ an algebraic field of extension of infinite degree (e.g. $\bar \QQ / \QQ$), then $K \subset L$ is integral but not finitely generated as a $K\dash$module.

> Note that we've used this to prove Zariski's lemma.

Exercise
:   Take $\bar \ZZ$ to be the set of algebraic integers, i.e. $\bar \ZZ = \theset{\alpha \in \CC \suchthat \alpha \text{ is integral over } \ZZ}$, then $\ZZ \subset \bar \ZZ$ is integral but $\bar \ZZ$ is *not* finitely generated as a $\ZZ\dash$module.

    > E.g. show that $\bar \ZZ$ is not Noetherian.

If $R\subset S$ is integral, then $S$ is finitely generated as an $R\dash$module iff $S$ is finitely generated as an $R\dash$algebra.

Proposition (Integral Properties Push to Quotients and Localizations)
:   Let $R\subset S$ be integral, then

    a. For $J\normal S$, there is an injection $R/\qty{J\intersect R} \injects S/J$ and this is integral.
    b. For $T \subset R\nonzero$ multiplicatively closed, localization induces $R_T \injects S_T$ which is integral.


Corollary (Towers of Successive Integral Containments are Integral)
: For $R \subset S \subset T$, if $R\subset S$ and $S\subset T$ are integral then $R\subset T$ is integral.

Proof
:   Take $\alpha\in T$, then there exist $b_0, \cdots, b_{n-1} \in S$ such that $$\alpha^n + b_{n-1}\alpha^{n-1} + \cdots b_1 \alpha + b_0 = 0.$$
    Then $R[\theset{b_i}, \alpha]$$ is finitely generated as an $R[\theset{b_i}]\dash$module.
    Since $R\subset S$ is integral, $R[\theset{b_i}]$ is finitely as an $R\dash$module.
    Therefore $R[\theset{b_i}, \alpha]$ is finitely generated as an $R\dash$module as well.
    \
    By condition (3) in the integrality condition theorem, $\alpha$ is thus integral over $R$.