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

I know that not every local seminormal ring is Cohen-Macaulay. But are 1-dimensional local seminormal rings Cohen-Macaulay?

share|cite|improve this question
No, $\mathbb{Q}[X,Y]/(X^2,XY)$ localized at $(x,y)$ provides a counterexample. – JSpecter Sep 8 '11 at 6:19

Your two sentences present some discrepancy.

Not all 1-dimensional local rings are semi-normal, so the first sentence has nothing to do with the second.

If all 1-dimensional local rings were Cohen-Macaulay, then all local rings would be Cohen-Macaulay by virtue of the definition of the Cohen-Macaulay property.

Semi-normalization can make things worse. Three lines meeting in a point that are not contained in a plane (say the 3 coordinate axes in 3-space) is semi-normal and it is not Gorenstein. On the other hand, 3 lines meeting in a point and contained in a plane is Gorenstein, but it is not semi-normal. Its semi-normalization is exactly the above union of 3 lines meeting in a point that are not contained in a plane.

share|cite|improve this answer
Three coordinate lines meeting at a point in $\mathbb{A}^3$ is Cohen-Macaulay, but it is not Gorenstein. Three lines through the origin in A^2 is Gorenstein, and as Sandor said, its semi-normalization is not. Thus semi-normalization can make things worse from the perspective of the Gorenstein condition. – Karl Schwede Sep 8 '11 at 12:41
Yes, that's what I meant. – Sándor Kovács Sep 8 '11 at 15:06

I agree with Sándor's, I'm not quite sure what you mean. And JSpecter provides a counter-example.

However, 1 dimensional seminormal rings are Cohen-Macaulay. Indeed, 1 dimensional reduced rings are Cohen-Macaulay (you just need a single non-zero-divisor).

share|cite|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.