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

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

2 Answers 2

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.

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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).

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