Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Suppose $A$ is a normal local domain contained in a field $K$. Suppose $B$ is the integral closure of $A$ in $K$. Under what conditions on $A$ is $B$ local?

share|cite|improve this question
up vote 8 down vote accepted

It is enough to ask $A$ to be henselian. Recall that a local ring $R$ is henselian if any finite $R$-algebra is a finite product of local rings.

If your $A$ is henselian, its integral closure in any overfield $K$ is a direct limit $A'$ of finite product of local rings (if $K$ is finite over the field of fractions of $A$, then you don't need to take direct limit). But $A'$ is an integral domain because it is contained in $K$, so it is local. Now it is easy to see that this direct limit is also local.

As examples of henselian local rings, you have complete noetherian local rings. Any noetherian local rings is contained in a ''minimal'' henselian local ring called the henselization.

If $A$ has dimension $1$ (hence a DVR), then your requirement implies that $A$ is henselian.

Further remark. Starting with a ring $B$, a general procedure to construct a local ring is to take the localization with respect to a prime ideal (e.g. the localization of a ring of integers at a maximal ideal or the stalk of the structural sheaf of an algebraic variety at some point). But this would almost never give you a henselian ring.

Namely, let $B$ be a noetherian integral non local ring. Let $\mathfrak m$ be a maximal ideal. Then $R:=B_{\mathfrak m}$ is not henselian: let $\mathfrak m'$ be a maximal ideal of $B$, different from $\mathfrak m$. Then $\mathfrak m'\ne \mathfrak m'^2$ (otherwise Nakayama would imply that $\mathfrak m'=0$, so contained in and hence equal to $\mathfrak m$). By Chinese Reminder Theorem, there exists $b\in B$ such that $b\in \mathfrak m'\setminus \mathfrak m'^2$ and $b\equiv 1 \mod \mathfrak m$. Let $n$ be an integer $\ge 2$ and prime to the characteristic of $B/\mathfrak m$. Consider the polynomial $T ^n-b\in R[T]$. In the residue field $B/\mathfrak m$ of $R$, it is separable and has a zero. If $R$ were henselian, then $T^n-b$ would have also have a zero in $R$ (this is one of the equivalent properties of henselian local rings). But then $b$ would be an $n$th power, contradiction wtih $b\in \mathfrak m'\setminus\mathfrak m'^2$.

share|cite|improve this answer
Thanks, this is what I was looking for. – Moosa Pei Oct 30 '11 at 16:23
Very nice, QiL ! – Georges Elencwajg Oct 30 '11 at 17:14
Thanks Georges ! – user18119 Oct 30 '11 at 21:54

First of all, let us remark that even if you normalize a local domain inside its fraction field $K=Frac(A)$, the result may or may not be a local ring.
For example the normalization of the local ring $\mathbb C[[T^2, T^3]]$ in its fraction field is the local ring $C[[T]]$.
On the other hand, the normalization of the local ring $(\mathbb C[X,Y ]/(Y^2-X^2-X^3))_{(\bar X,\bar Y)}$ in its fraction field is a ring with two maximal ideals.

In general, you can say that the normalization $A^\nu$ of a noetherian local domain $A$ in a finite extension $K'$ of its field of fractions $K$ is a semi-local domain i.e. a domain with finitely many maximal ideals.
If $K'=K$, you can add that $A^\nu$ is a Krull domain.
However Nagata has shown that $A^\nu$ needn't be noetherian even if $K=K'$.

In EGA, $IV_1$, (23.2.1) the local domain $A$ is defined to be unibranch if its normalization in its fraction field $K$ is a local ring, which is a special case of what you are asking about.
So you are definitely in good company with your question!

Edit I have just checked that Bourbaki gives ( in Algèbre Commutative, Chap.5, §2, exercice 8) the following nice sufficient criterion : A noetherian local domain is unibranch if its completion is a domain. This sheds an interesting light on my examples above, which I wrote without knowing that criterion!

share|cite|improve this answer
Thanks, Georges. Is there a good characterization when $A$ is normal and $K$ is Galois over the fraction field of $A$. – Moosa Pei Oct 30 '11 at 14:27
I hadn't seen QiL's answer when I started my Edit. Of course, Bourbaki's criterion is contained in QiL's class of Henselian rings. – Georges Elencwajg Oct 30 '11 at 17:09

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.