integral closures in separable extensions are finitely generated

The theorem below generalizes to arbitrary integral ring extensions (under certain conditions) the fact that the ring of integers of a number field is finitely generated over $\mathbb{Z}$. The proof parallels the proof of the number field result.

Theorem 1.

Let $B$ be an integrally closed Noetherian domain with field of fractions $K$. Let $L$ be a finite separable extension of $K$, and let $A$ be the integral closure of $B$ in $L$. Then $A$ is a finitely generated $B$-module.

Proof.

We first show that the trace (http://planetmath.org/Trace2) $T{r}_K^L$ maps $A$ to $B$. Choose $u\in A$ and let $f=Irr(u,K)\in K[x]$ be the minimal polynomial for $u$ over $K$; assume $f$ is of degree $d$. Let the conjugates of $u$ in some splitting field be $u=a_1,\mathrm{\dots },a_d$. Then the $a_i$ are all integral over $B$ since they satisfy $u$’s monic polynomial in $B[x]$. Since the coefficients of $F$ are polynomials in the $a_i$, they too are integral over $B$. But the coefficients are in $K$, and $B$ is integrally closed (in $K$), so the coefficients are in $B$. But $T{r}_K^L(u)$ is just the coefficient of $x^{d-1}$ in $f$, and thus $T{r}_K^L(u)\in B$. This proves the claim.

Now, choose a basis ${\omega }_1,\mathrm{\dots },{\omega }_d$ of $L/K$. We may assume ${\omega }_i\in A$ by multiplying each by an appropriate element of $B$. (To see this, let $Irr({\omega }_i,K)\in K[x]=x^d+k_1{x}^{d-1}+\mathrm{\dots }+k_d$. Choose $b\in B$ such that $b{k}_i\in B\forall i$. Then ${(b\omega )}^d+b{k}_1{(b\omega )}^{d-1}+\mathrm{\dots }+b^d{k}_d=0$ and thus $b\omega \in A$). Define a linear map $\phi :L\to K^d:a\mapsto (T{r}_K^L(a{\omega }_1),\mathrm{\dots },T{r}_K^L(a{\omega }_d))$.

$\phi$ is 1-1, since if $u\in \ker \phi ,u\ne 0$, then $Tr(uL)=0$. But $uL=L$, so $T{r}_K^L$ is identically zero, which cannot be since $L$ is separable over $K$ (it is a standard result that separability is equivalent to nonvanishing of the trace map; see for example [1], Chapter 8).

But $T{r}_K^L:A\to B$ by the above, so $\phi :A\hookrightarrow B^d$. Since $B$ is Noetherian, any submodule of a finitely generated module is also finitely generated, so $A$ is finitely generated as a $B$-module.∎