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Let $R$ be a local ring and $I \subset R$ an ideal such that $R/I$ is cohen-macaulay. Is $R/I^2$ also cohen-macaulay?

I've seen that it always works for principal ideals here: Is a quotient of ring of polynomials Cohen-Macaulay? So a counter-example has to have 2 or more generators.

bliipbluup
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    With regards to your proposed example, the square of a principal graded ideal is still principal graded, no? – C.D. Jul 26 '22 at 18:19
  • You're absolutely right. Had a brain freeze for a moment. I'll edit it – bliipbluup Jul 26 '22 at 18:24
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    I must be confused by what you mean. If you mean $R/(I^2)$, then in dimension 1 can't you take $I$ the maximal ideal of a DVR? The quotient $R/I$ is a field, but $R/(I^2)$ has nilpotents, so isn't reduced, so isn't $S_1$ (with regards to Serre's criterion) hence not CM? – C.D. Jul 26 '22 at 18:24
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    Huh, I guess in dimension zero everything is Cohen-Macaulay, which feels a bit weird to me, but I guess is true. – C.D. Jul 26 '22 at 19:20

1 Answers1

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No. It is not true even for principal ideals unless one assumes more about the local ring $R$, e.g., that $R$ is a domain. For a simple example, let $R=k[\![x,y]\!]/(xy)$ and take $I=(x)$. Then, $R/I \cong k[\![y]\!]$ which is Cohen-Macaulay but $R/I^2 \cong k[\![x,y]\!]/(x^2,xy)$ which is famously not Cohen-Macaulay.

For more exotic examples, note there is a short exact sequence $0 \to I/I^2 \to R/I^2 \to R/I \to 0$ which relates Cohen-Macaulayness of $I/I^2$, which is known as the conormal module of $I$, to that of $R/I^2$. The conormal module (and thus $R/I^2$) puts up some resistance to being Cohen-Macaulay. In several situations it is known that Cohen-Macaulayness of the conormal module forces $R/I$ to be Gorenstein. Several of these situations are detailed in this work of Mantero-Xie. One can use these criteria to construct various types of examples.

metalspringpro
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