Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ be a noetherian integral local ring. Let $K$ be its fraction field, $L$ an algebraic field extension of $K$, and $B$ the integral closure of $A$ in $L$. If $A$ is supposed to be regular, is the ring $B$ regular too?

share|improve this question

3 Answers 3

No. Take $B = \mathbb{C}[[x,y,z]]/(z^2 - xy)$ and $A = \mathbb{C}[[x,y]]$.

share|improve this answer
Ok, so what kind of hypothesis on $A$ or on $L$ could I add in order to have a positive answer ? –  user9255 Sep 15 '10 at 16:18
Dear Igor: Presumably you meant for $L/K$ to be finite separable, so $B$ is at least $A$-finite. In general it is very restrictive to expect regularity, but one available condition is given by Abhyankar's Lemma, which comes in several flavors involving "tame ramification". See A.I.11 in Freitag-Kiehl book for a version with $L/K$ Galois. –  BCnrd Sep 15 '10 at 16:26

Angelo exactly is right. In general, essentially any singularity is possible. For example, by Noether normalization, given any normal domain which is a $k$-algebra $R$, there is a subalgebra $S = k[x_1, \dots, x_n]$ such that $S \subseteq R$ is an integral extension (and thus is the normalization).

With regards to your second question

Not much can be said in general. If one knows a lot about the ramification of $A$ in $L$, then control on the singularities of $B$ can be made (eg, they can sometimes be shown to be rational, see for example, MathOverFlowQuestion).

See an old Theorem of Kunz ("Remark on the purity of the branch locus") and generalized by Griffith ("Normal extensions of regular local rings") if $A$ is regular then regular + unramified in condimension 1 implies that $A \subset B$ is unramified in general. If you can then show that the extension is flat, then it's \'etale...

share|improve this answer

Dear Igor,

Just to add some context to the answers that you have received: what you are considering is (a local version of) the following geometric situation: a map $X \to Y,$ where $Y$ is smooth, $X$ is normal, and the fibres are finite. Now by Noether normalization (which is just an algebraic way of describing projection to a hyperplane) any variety can be made finite over a smooth one, and in particular, this is true of any normal variety.

Thus your question is (essentially) ``is every normal variety smooth?'', to which the answer is no as soon as the dimension is $> 1$. This is where Angelo's answer comes from: he wrote down the simplest isolated singularity on a surface in 3-space (the vertex of a cone), and this singularity is normal but singular. (Normal implies that the singularities are in codimension 2, and for a hypersurface in an affine or projective space, the converse holds.)

To have a positive answer in some situation, you need to control the branching of your finite map in some way (it is along the branch locus that singularities can appear), and the references that BCnrd mentions deal with this kind of question.

One more remark, which you probably know (but just to be sure): if $A$ is a DVR (i.e. if you are in the one dimensional case), then $B$ will be regular (assuming some reasonable background hypotheses, so that $B$ is finite over $A$, for example); geometrically, for a curve, being normal is the same as being smooth.

share|improve this answer

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.