# Monday April 27th

## Normalization

Theorem (14.19: Integral Closure is Local)
:   A domain $R$ is integrally closed iff $R_\mfp$ is integrally closed for all $\mfp \in \spec R$ iff $R_\mfp$ is integrally closed for all $\mfm \in \maxspec R$.

> Recall that the height of a prime ideal is the supremum of lengths of chains.
> Height one corresponds to principal.

Definition (Regular Prime Ideals)
: For $R$ a Noetherian ring, $\mfp \in \spec R$ is *regular* iff $\dim_{\RR_\mfp / \mfp R_\mfp} {\mfp R_\mfp \over \qty{\mfp R_\mfp}^2} = \height(\mfp)$.

This corresponds to a mild nonsingularity property for varieties.

Theorem (14.21: Going Down)
: For $R\subset S$ an integral extension of domains, $R$ is integrally closed

\begin{center}
\begin{tikzcd}
\exists \mfp_1 \ar[d, mapsto] & \mfp_2 \ar[d, mapsto] & \spec S \ar[d, mapsto] \\
\mfp_1 \ar[r, "\subset"] & \mfp_2 & \spec R
\end{tikzcd}
\end{center}

Theorem (14.22, Noether Normalization)
:   Let $k$ be an arbitrary field, $R$ a domain and a finitely generated $k\dash$algebra.
    Then

    a. There exists a $d\in\NN$ and algebraically independent elements $y_1, \cdots, y_d$ such that $k[y_1, \cdots, y_d] \subset R$ is a finitely generated module.

    b. $\dim R = d$ equals the transcendence degree of $ff(R) / k$.

Application: Noether normalization implies Zariski's lemma.

Remark (Separable Noether Normalization):
In the above setup, if $k$ is perfect that one can choose the $y_i$ such that $ff(R) / k(y_1, \cdots, y_d)$ is finite and *separable*.

## Invariant Theory

For $G$ a finite group and $R$ an arbitrary object of a category $\mcc$, the notion of a group action by makes sense: namely a group morphism $G \to \Aut_\mcc(R)$.

Definition (Ring of Invariants)
:   For a ring $R$ and a group acting by ring automorphisms, the *ring of invariants* is given by
    \begin{align*}
    R^G \definedas \theset{x\in R \suchthat gx = x \forall g\in G}
    .\end{align*}

Theorem (Invariant Rings are Integral)
: $R^G \subset R$ is integral.

Proof
:   For $x\in R$, define $\Phi_x(t) \definedas \prod_{g\in G} (t - gx)$.
    We can extend the action of $G$ by acting on coefficients and acting on $t$ trivially, yielding $\Phi_x \in (R[t])^G = R^G[t]$ and thus $\Phi_x(x) = 0$.

Proposition (Ring of Invariants Over Integrally Closed is Integrally Closed)
: If $R$ is integrally closed, so is $R^G$.

> Note that this doesn't hold with regularity!
> Here $R^G$ corresponds to "quotient varieties" in AG, which have mild "quotient singularities".

Important point: in dimension 1, normal coincides with regular.

Example (Important in Other Areas)
:   If $G = \theset{\pm 1}$, $R = \CC[x, y]$ and let $-1$ act by $\CC\dash$algebraic automorphisms acting trivially on $\CC$ and sending $x\mapsto -x, y\mapsto -y$.
    Then 
    $$
    R^G = \CC[x^2, xy, y^2] = \CC[X, Y, Z]/\gens{XZ - Y^2}
    $$ 
    which is a conic, and in particular an *affine quadric cone*.
    Since we started out with a UFD, which is integrally closed (and in fact regular), this is integrally closed but the maximal ideal $\gens{X, Y, Z} + (XZ - Y^2)$ is singular.

> Note that here the Picard group is trivial but the divisor class group has order 2.


Theorem (14.32, Noether 1928)
:   Suppose $R$ is a finitely generated $k\dash$algebra and $G$ acts on $R$ by $k\dash$algebra automorphisms.
    Then

    a. $R$ is finitely generated as an $R^G\dash$module.
    b. $R^G$ is finitely generated as $k\dash$algebra.

> This says one can take a quotient of an affine coordinate ring by an affine variety again yields an affine coordinate ring. 

Proof
:   \hfill
    
      a. $R$ is finitely generated as an $R^G\dash$algebra and $R^G \subset R$ is integral, which implies that $R$ is finitely generated as an $R^G\dash$module.

      b. Apply the Artin-Tate lemma for $k \subset R^G \subset R$ where $k \subset R$ is a finitely generated algebra and $R^G\subset R$ is a finitely generated module.


## Normalization Theorems

Normalization is a geometric synonym for integral closure.
Next up: theorems giving good properties of the integral closure.

Let $R$ be an integrally closed domain with fraction field $K$, $L/K$ a field extension, and set $S = I_L(R)$ the integral closure.

\begin{center}
\begin{tikzcd}
S \ar[r, "\subset"] \ar[d, dash] & L \ar[d, dash] \\
R \ar[r, "\subset"] & K
\end{tikzcd}
\end{center}

We know that $S$ is integrally closed, $ff(S) = L$, and $R\subset S$ is integral, we can conclude that $\dim R = \dim S$.

Questions:

1. Is $S$ finitely generated as an $R\dash$module?
2. Is $S$ Noetherian?

WLOG, we can assume $L/K$ is algebraic since we can pass to fraction fields, and integral over a field implies algebraic.
Also note that if $[L: K]$ is infinite, then we shouldn't expect either answer to be "yes".

Example
:   Take $R = \ZZ, k = \QQ, L = \bar \QQ, S = \bar \ZZ$. 
    Then $\bar \ZZ$ is not Noetherian and not finitely generated as a $\ZZ\dash$module.

Is $S$ finitely generated over $R$ or even Noetherian? 
Not necessarily, we need to assume $R$ Noetherian integrally closed and $[L: K]$ finite.

Theorem (18.1, First Normalization Theorem)
:   Supposing that $R$ is an integrally closed domain, $L/K$ finite and separable, then

    a. If $R$ is Noetherian, so is $S$.
    b. If $R$ is a PID, $S \cong_R R^{[L:K]}$ (free of finite rank).

Proof uses discriminant and some linear algebra.
There are (complicated) counterexamples: $R$ a DVR in characteristic 2, $L/K$ a quadratic extension.

Upcoming: two more normalization theorems.