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This seems like the kind of thing an expert should be able to answer off the top of their head:

Recall that a valuation ring is an integral domain $A$ such that for every $a \in Frac(A)$ we have either $a \in A$ or $a^{-1} \in A$. One of the many equivalent definitions of a Hensel valuation ring, is: a valuation ring $A$ is called Henselian if for every finite field extension $Frac(A) \subseteq L$, there exists a unique valuation ring $B \subseteq L$ such that $L = Frac(B)$ and $A = Frac(A) \cap B$.

Nowhere do I assume that the rings are Noetherian. I.e., the totally ordered set of prime ideals of a valuation ring is allowed to have any order type.

Let $A$ be a Hensel valuation ring and $Frac(A) \to L$ a finite field extension of its field of fractions. Let $B$ be the normalization of $A$ in $L$. A consequence of Chevalley's extension theorem is that $B$ is the valuation ring of the unique extension of the valuation of $Frac(A)$ to $L$ (see Corollary 3.1.4 in the Engler-Prestel book "Valued fields" for example).

Question: is $B$ a finite $A$-module in general?

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No problem Torsten, I'll edit the question. –  name Mar 11 '13 at 20:47
Thanks for the reference. (by the way, you mean't Algèbre commutative ch. VI §8). –  name Mar 11 '13 at 21:05
@Torsten: Where did you see the counter-examples? –  name Mar 11 '13 at 21:11
Ok thanks. It gives me a starting place anyway. Maybe taking henselisations in exercise 3b gives a counter-example... –  name Mar 11 '13 at 21:40
(I deleted my earlier comments as everything relevant should now be in my answer below.) –  Torsten Schoeneberg Mar 14 '13 at 14:55
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up vote 2 down vote accepted

The following counterexample is very similar to exercise 2 to Bourbaki's Algèbre commutative ch. VI §8 and works as well. Actually it is example 4.9 in Scholze's Perfectoid Spaces; there, the main point is that $B$ is an "almost finitely generated" $A$-module in the sense of "almost ring theory", so there is a way around the fact that it is not finitely generated. This fact, however, makes up our counterexample.

Let $p$ be an odd prime and for all $n \ge 1$, adjoin a $p^n$-th root of $p$, call it $\pi_n$, to the field of $p$-adic numbers $\mathbb{Q}_p$. The $p$-adic valuation uniquely extends to this field; take its completion with respect to this value. This is our field $K = Frac(A)$ with its local valuation ring $(A, \mathfrak{m})$. Note that the valuation $v: K^\times \rightarrow \mathbb{R}$ is of rank 1, but non-discrete; in fact, its value group is $\bigcup_{n \ge 1} \frac{1}{p^n} \mathbb{Z} = \mathbb{Z} [\frac{1}{p}]$. Because the rank is 1 and the field is complete, the valuation is Henselian (exercise 6b in Bourbaki, loc. cit.; first sentence of paragraph 4.1 in Engler-Prestel).

Now adjoin a square root of $p$ to get the quadratic field extension $L|K$, and extend $v$ to $L$. Convince yourself that for each $n \ge 0$, there is an element in $L$, call it $\rho_n$, which is a $2p^n$-th root of $p$ and thus is a primitive element for $L|K$. The value group of $L$ is $\frac{1}{2} \mathbb{Z}[\frac{1}{p}]$ and it is not hard to see that the valuation ring $B = \lbrace x \in L: v(x) \ge 0 \rbrace$ can be written as

$$B = A \oplus \bigcup_{n \ge 0} \rho_n A$$

as $A$-module. But because $v(\rho_n) = \frac{1}{2p^n} > \frac{1}{p^{n+1}} = v(\pi_{n+1})$ and $\pi_{n+1} \in \mathfrak{m}$, we have $\rho_n \in \mathfrak{m}B$ for all $n$, hence whole union on the right is contained in $\mathfrak{m}B$. So as $A$-modules,

$$B = A + \mathfrak{m}B .$$

Now if $B$ were finitely generated over $A$, Nakayama's lemma would imply $A = B$, which is absurd since $\rho_n \in B \setminus A$.

Positive results: We have theorem 2 in Bourbaki's Algèbre commutative ch. VI §8 no. 5 which gives (in a more general setup, without the Henselian assumption) equivalent criteria for $B$ being a finitely generated $A$-module, especially the famous equality $[L:K] = \sum e_i f_i$. In Neukirch's Algebraic Number Theory ch. II §6, where the setup has a Henselian rank 1 valuation $v$, this equality is proven for $L|K$ separable and $v$ discrete; it is remarked that "both conditions are really necessary" but that for a complete field, one may drop the separability. Corresponding statements can be found in Serre's Local Fields ch. I §4 and ch. II §2, and also in Engler-Prestel, theorem 3.3.5. I guess that exercise 3 in Bourbaki loc. cit. is a counterexample with $v$ discrete and $L|K$ inseparable, I just do not see right away whether $v$ is Henselian there.

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Did you see Exercise V.1.19b in Bourbaki's Algèbre Commutative?: If $A$ is integrally closed with fraction field $K$ and $\{a \in K : a^p \in A\}$ is a finite $A$-module ($p$ is the exponential characteristic of $K$), then for any finite extension $E$ of $K$, the integral closure of $A$ in $L$ is a finite $A$-module (reduce to the case $E/K$ is normal). Your example is characteristic zero right? In the Bourbaki exercise there doesn't seem to be any assumption that $A$ is Noetherian, so there is a contradiction somewhere? (I admit, I haven't done the exercise). –  name Mar 14 '13 at 21:36
Hmm, very strange. I don't see how to conclude in ex. V.1.19b without $A$ Noetherian; if it is true in general, it contradicts my answer, as it is in char. 0. (But so are ex. VI.8.2 and VI.8.4 when $char(k) = 0$, aren't they?) Very confusing. -- Btw, regarding your original question, I have found math.stackexchange.com/questions/167993 –  Torsten Schoeneberg Mar 15 '13 at 1:37
Yes, Exercise V.1.19b must be a mistake. Thanks for the stackexchange link. The henselisation of Exercise VI.8.3b works too and is a discrete valuation ring. I'll put it as an answer because there is not enough room here. –  name Mar 15 '13 at 2:05
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Exercise VI.8.3 from Bourbaki's Algèbre Commutative gives am example of a hensel discrete valuation ring and a finite (inseparable) extension of its fraction field for which the normalisation is not finite:

Define $k = \mathbb{F}_p(X_n)_{n \in \mathbb{N}}$.

Take $\alpha = \sum_{n = 0}^\infty X_n U^n \in k[[U]]$, although it looks like any transcendantal element $\alpha$ works, I don't see why we have to choose this one in particular.

Consider the sequence of fields $k(U) \subset k(U, \alpha^p) \subset k(U, \alpha) \subset k((U))$. The (complete) valuation ring $k[[U]] \subset k((U))$ induces valuation rings in all the others. Let $V_1 \subset V_2$ denote the two middle ones. We know the outer two are discrete valuation rings with residue field $k$ and therefore the same is true of the middle two.

The extension $k(U, \alpha) / k(U, \alpha^p)$ is finite and non-trivial, but induces isomorphisms of residue fields and value groups $(e = f = 1)$. It is generically purely inseparable, and therefore $R_2$ is the unique extension of $R_1$. Therefore $V_1 \subset V_2$ is not finite (Bourbaki AC Proposition 8 page 148).

Regardless of whether $V_1, V_2$ are henselian (its probably not too hard to show that they are not), we can take the associated hensel rings $V_1^h \subset V_2^h$, which are canonically valuation rings. Then $Frac(V_1^h) / Frac(V_1)$ is a separable algebraic (probably not finite) extension and therefore $Frac(V_1^h)$ does not contain $\alpha$. Henselisation preserves residue fields and value groups (Theorem 5.2.5 in Engler-Prestel) and so $V_1^h \subset V_2^h$ is not finite as long as $Frac(V_2^h) / Frac(V_1^h)$ is still purely inseparable of degree $p$.

For this we show notice that $Frac(V_2^h) = Frac(V_1^h) \otimes_{Frac(V_1)} Frac(V_2)$.

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This example (it's actually exercise AC VI.8.3) appears to go back to F.K. Schmidt and is treated in great detail in Bosch-Güntzer-Remmert's Non-Archimedean Analysis, 1.6.2. They also go through Bourbaki's exercise VI.8.2 (close to my answer) which becomes BGR 3.6.1, with context directly applicable to the question in BGR 6.4.1. –  Torsten Schoeneberg Mar 15 '13 at 15:34
Thanks for the references and the correction; keeping the numbers straight in Bourbaki is impossible! –  name Mar 16 '13 at 6:18
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