# Are hensel valuation rings N2?

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?

• No problem Torsten, I'll edit the question.
– name
Mar 11, 2013 at 20:47
• Thanks for the reference. (by the way, you mean't Algèbre commutative ch. VI §8).
– name
Mar 11, 2013 at 21:05
• @Torsten: Where did you see the counter-examples?
– name
Mar 11, 2013 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, 2013 at 21:40
• (I deleted my earlier comments as everything relevant should now be in my answer below.) Mar 14, 2013 at 14:55

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.

• 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, 2013 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 Mar 15, 2013 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, 2013 at 2:05

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 $Frac(V_2^h) = Frac(V_1^h) \otimes_{Frac(V_1)} Frac(V_2)$.

• 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. Mar 15, 2013 at 15:34
• Thanks for the references and the correction; keeping the numbers straight in Bourbaki is impossible!
– name
Mar 16, 2013 at 6:18