I know that not every local seminormal ring is Cohen-Macaulay. But are 1-dimensional local seminormal rings Cohen-Macaulay?
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4$\begingroup$ No, $\mathbb{Q}[X,Y]/(X^2,XY)$ localized at $(x,y)$ provides a counterexample. $\endgroup$– JSpecterCommented Sep 8, 2011 at 6:19
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2 Answers
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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).
answered Sep 8, 2011 at 12:40
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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.
answered Sep 8, 2011 at 11:18
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2$\begingroup$ 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. $\endgroup$ Commented Sep 8, 2011 at 12:41
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