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 $A$?
The integral closure of a finite separable field extension of the fraction field is finitely generated
1 Answers
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 $x\in \Sigma_{i=1}^n Av_i$.
(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!
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3This 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**) – 2011-07-08
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0@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? – 2011-07-09
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0(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.) – 2011-07-09
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0By 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. – 2011-07-09
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0@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. – 2011-07-09