MathOverflow is a question and answer site for professional mathematicians. 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 I have an affine subvariety $A \subset {\mathbb C}^N$ of dimension $n \geq 3$ which has an isolated singularity at $0$ (lets say for the sake of simplicity that it is non-singular everywhere else). Suppose that this variety $A$ is normal. In order to study singularities it often seems like a good idea to study hyperplane sections or intersections with lower dimensional linear subspaces passing through that singularity.

Let $L$ be a generic linear subspace of ${\mathbb C}^N$ of dimension $N - n + 2$ passing through the origin. If $L$ is generic enough then $A \cap L$ has dimension $2$ and has an isolated singularity at the origin. Is it true that $A \cap L$ is normal for $L$ generic enough?

share|cite|improve this question
up vote 9 down vote accepted

I don't think so. There are examples of isolated normal threefold singularities that are not Cohen-Macaulay. A hyperplane section is not Cohen-Macaulay, hence it can not be normal, because a normal surface is Cohen-Macaulay.

share|cite|improve this answer
E.g., affine cone over a general embedding of an Abelian surface. – Jason Starr May 20 '13 at 20:34
Excellent thanks! – Mark McLean May 20 '13 at 21:09

I'm going to assume your singularity is dimension $\geq 3$. Angelo beat me to the answer but he is right, this is not true. But it is true sometimes (including the Cohen-Macaulay case as he implied).

A singularity is normal if it is $R1$ and $S2$. In your case, an isolated singularity is normal if the depth at the singular point is at least 2.

Now, a general hyperplane section will be $R1$ by Bertini. So we just need to check that the general hyperplane is $S2$. Well, for this we just need the depth to be at least 2 again, and hence we just need the original singularity to have depth $\geq 3$.

Conclusion: If your singularity is $S3$ (in your case just $\text{depth} \geq 3$), then what you want holds after cutting down by ONE hyperplane

EDIT: As Angelo pointed out, the actual question didn't cut down by just one hyperplane. In that case you can't just have depth $\geq 3$, you need $X$ to be Cohen-Macaulay.

Of course, not all singularities satisfy this, for example a cone over an Abelian surface.

You might also look at this preprint which seems to have some related results: Tadashi Ochiai, Kazuma Shimomoto

share|cite|improve this answer
Actually, the way the question was formulated so that the intersection has dimension 2. In this case, to get a positive answer you need the singularity to be Cohen-Macaulay. – Angelo May 21 '13 at 13:18
Angelo, of course you are right, I misread the question. – Karl Schwede May 21 '13 at 13:46

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.