Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question is related to the one here: A question related to krull akizuki

In the answers to that question, some examples are given of a discrete valuation ring $A$ and a finite (necessarily inseparable) extension of its field of fractions in which the integral closure of $A$ is not finitely generated as an $A$-module. My question is whether there are similar counterexamples with $A$ Henselian (it's possible that the examples given there are Henselian, but I do not understand them well enough to know for sure).

If $A$ is a complete discrete valuation ring, then the integral closure of $A$ in any finite extension of its field of fractions is finite over $A$. For $A$ just Henselian, one can at least say that the integral closure of $A$ in a finite extension of $\mathrm{Frac}(A)$ is a Henselian discrete valuation ring. In the separable case, one knows the integral closure is a principal ideal domain which is finite over $A$, and, since $A$ is Henselian, this implies that the integral closure is a finite product of local rings, hence is itself local. For the inseparable case, the reference I know is Neukirch's book on algebraic number theory. He proves that the valuation on $\mathrm{Frac}(A)$ extends uniquely to any finite extension (with valuation ring the integral closure of $A$), and it is visible from the formula for the extended valuation that it is again discrete.

Maybe an example can be obtained by taking the Henselization of an example as in the question referenced above.

EDIT: I just noticed that Exercise 1 of section II.4 of Serre's Local Fields states that, if every finite purely inseparable extension of $\mathrm{Frac}(A)$ has integral closure finite over $A$, then the completion of $\mathrm{Frac}(A)$ is separable over $\mathrm{Frac}(A)$. So I guess a counterexample would have to be a non-excellent Henselian DVR.

share|cite|improve this question
up vote 1 down vote accepted

Let $A$ be a DVR with field of fractions $K$ such that for some purely inseparable finite extension $L$, the integral closure $B$ of $A$ in $L$ is not finite over $A$ (so $A$ is non-excellent). Then as in your guess, the henselization $A^h$ of $A$ is not excellent.

First $A^h\otimes_A B$ is the inductive limit of $A'\otimes_A B$ with $A'$ étale over $A$ so that $A'\otimes_A B$ is normal, thus $A^h\otimes_A B$ itself is normal. As $L$ is purely inseparable over $K$ and $K^h:=\mathrm{Frac}(A^h)$ is algebraic separable over $K$, $K^h\otimes_K L$ is a field and it is the field of fractions of $A^h\otimes_A B$. Therefore $A^h\otimes_A B$ is the integral closure of $K^h$ in $K^h\otimes_K L$.

It remains to show that $A^h\otimes_A B$ is not finite over $A^h$. Suppose the contrary. A system of generators involves finitely many elements of $B$. So there exists a finitely generated sub-$A$-module $M$ of $B$ such that $A^h\otimes_A M\to A^h\otimes_A B$ is surjective. So $A^h\otimes_A (B/M)=0$. By the faithful flatness of $A^h$ over $A$, we get $B=M$ is finite over $A$. Contradiction.

share|cite|improve this answer
Thank you @Qil! – Keenan Kidwell Jul 7 '12 at 22:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.