8

Let $A$ be Noetherian and integrally closed in its field of fractions $K$ and $L$ a finite separable field extension of $K$. Why is the integral closure $B$ of $A$ in $L$ finitely generated over $A$?

Peter Patzt
  • 3,084

1 Answers1

15

The following proof is based on the assumption that $A$ is Noetherian and is integrally closed (in its field of fractions $K$):

(1) If $L$ is a finite separable extension of $K$, then the bilinear form $(x,y)\to \text{Tr}_{L/K}(xy)$ on $L$ is nondegenerate. (Exercise)

(2) Choose a basis of $L$ over $K$. We can assume, after possibly scaling by elements of $K$, that this basis consists entirely of elements of $B$ (the integral closure of $A$ in $L$). (Exercise) Let $(v_1,\dots,v_n)$ be this basis of $L$ (as a vector space over $K$).

(3) We know by (1) that there is a basis $(w_1,\dots,w_n)$ of $L$ over $K$ "dual" to the basis $(v_1,\dots,v_n)$ of $L$ over $K$. More precisely, there is a basis $(w_1,\dots,w_n)$ of $L$ over $K$ such that $\text{Tr}_{L/K}(v_iw_j)=\delta_{ij}$ for all $1\leq i,j\leq n$ (where $\delta_{ij}$ denotes the Kronecker delta). (Exercise)

(4) Let $x\in B$ and write $x=\Sigma_{j=1}^n x_jw_j$ where $x_j\in K$ for all $1\leq j\leq n$. Note that $xv_i\in B$ for each $1\leq i\leq n$ and thus $\text{Tr}_{L/K}(xv_i)\in A$. (Exercise) However, $\text{Tr}_{L/K}(xv_i)=\Sigma_{j=1}^n x_j\text{Tr}_{L/K}(w_jv_i)= x_i$. Therefore, $x_i\in A$ for all $1\leq i\leq n$ and $x\in \Sigma_{j=1}^n Aw_j$.

(5) Since $A$ is Noetherian, it follows that $B$ is a finitely generated $A$-module. (Exercise)

An interesting corollary that is fundamental in algebraic number theory:

Corollary Let $A$ be a Dedekind domain and let $K$ be its field of fractions. If $L$ is a finite separable extension of $K$, then the integral closure of $A$ in $L$ is also a Dedekind domain.

Proof. We know that $B$ is Noetherian since it is a finitely generated $A$-module (and $A$ is Noetherian). Also, $B$ is integrally closed in $L$. (Exercise) Therefore, all that remains to show is that every non-zero prime ideal of $B$ is maximal. (Exercise) Q.E.D.

I hope this helps!

Nischay
  • 5
Amitesh Datta
  • 21,429
  • 3
    This is nicely explained. Note that the Corollary -- unlike the finite generation result itself -- actually holds without the separability hypothesis, although it is not so easy to prove. (Keyword: Krull-Akizuki Theorem) – Pete L. Clark Jul 08 '11 at 19:06
  • @Pete: Dear Pete, thanks! I do not think that I learnt about the Krull-Akizuki theorem when I studied algebraic number theory (from Janusz's Algebraic Number Fields; or perhaps I simply do not remember). Could you please suggest a textbook which covers this result? – Amitesh Datta Jul 09 '11 at 02:02
  • (Well, it has been mysteriously absent from my commutative algebra notes for some time now.) It is definitely covered in Matsumura's Commutative Ring Theory and probably in Eisenbud's Commutative Algebra... as well. (Added: I checked. Yes, it's in Chapter 11 of Eisenbud.) – Pete L. Clark Jul 09 '11 at 02:50
  • By the way, the theorem is probably more important in commutative algebra than in algebraic number theory per se. Even when one works in global function fields (which should be part of algebraic number theory but seems still to lag behind in most textbook treatments) one can usually arrange for the field extensions to be separable. – Pete L. Clark Jul 09 '11 at 02:54
  • @Pete Thanks! It does not seem to be in Atiyah and Macdonald either. I will take a look at Matsumura's Commutative Ring Theory and also Matsumura's Commutative Algebra (which I plan to read in the next few months). Unfortunately, I cannot get my hands on Eisenbud from the local library ... But if and when I do I will take a look at chapter 11. – Amitesh Datta Jul 09 '11 at 03:25
  • Why does step (5) need that $A$ is Noetherian? (4) shows $B \subset \sum_i Av_i$, but clearly $B \supset \sum_i Av_i$ holds as well, because $v_i \in B$ by construction, so really $B = \sum_i Av_i$? – red_trumpet Jan 28 '19 at 13:05
  • 1
    @red_trumpet : I think that there is a typo at the end of step (4) : it should read $x \in A' := \sum_i A w_i$, but it doesn't affect the rest of the argument. Then the hypothesis that $A$ is noetherian is used because $B$ is an $A$-submodule of the finitely generated module $A'$. – Watson Dec 14 '19 at 10:14