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 $u\in \mathrm{ker}\phi ,u\ne 0$, 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. ∎
References
- 1 P. Morandi, Field and Galois Theory^{}, Springer, 2006.
Title | integral closures in separable extensions are finitely generated |
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Canonical name | IntegralClosuresInSeparableExtensionsAreFinitelyGenerated |
Date of creation | 2013-03-22 17:02:12 |
Last modified on | 2013-03-22 17:02:12 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 5 |
Author | rm50 (10146) |
Entry type | Theorem |
Classification | msc 13B21 |
Classification | msc 12F05 |
Related topic | IntegralClosureIsRing |