Let $R=K[x_1,...,x_n]$ be the polynomial ring over a field $K$ and $I[x_1,...,x_n]=(u_1,...,u_t)$ be a Cohen-Macaulay monomial ideal of $R$. If $m<n$, could we say that $I[x_1,...,x_m,0,0,...,0]$ is Cohen-Macaulay in $K[x_1,...,x_m]$?
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$\begingroup$ Not all of us are ring-theory specialists. Would you provide the C-M ideal definition? $\endgroup$– Wlod AACommented Jan 15, 2020 at 7:15
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1$\begingroup$ @WlodAA An ideal $I$ of $R$ is Cohen-Macaulay if $R/I$ is a Cohen-Macaulay ring. $\endgroup$– Stephen McKeanCommented Jan 16, 2020 at 16:01
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$\begingroup$ Oh, well, a reader here has to be either young or a specialist, $\endgroup$– Wlod AACommented Jan 17, 2020 at 10:28
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No, we can't say that. For a counterexample, take $n = 3$, $m = 2$, and let $I = (x_2^2, x_1x_2, x_1x_3)$.
To see that $I$ is Cohen-Macaulay, note that the Krull dimension of $R/I$ is 1. Cohen-Macaulayness is then equivalent to $I$ having no associated prime $P$ with dimension $R/P$ equal to 0. Since $I$ is homogeneous, its associated primes are homogeneous, and so the only possible example of such an associated prime is $P = (x_1,x_2,x_3)$, which is easily seen not to annihilate any element of $R/I$, hence is not associated with $I$.
On the other hand the restricted ideal $I' = (x_2^2, x_1x_2) \subset K[x_1,x_2]$ is not Cohen-Macaulay, since $x_2$ is annihilated by $(x_1,x_2)$ in $K[x_1,x_2]/I'$, and so $(x_1,x_2)$ is an associated prime of $I'$.
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$\begingroup$ Thank you so much for your nice example. $\endgroup$ Commented Jan 18, 2020 at 4:59
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$\begingroup$ For another example, take $I=(x_1x_3,x_1x_4,x_2x_4)$. This is Cohen-Macaulay since it is the face ring (aka "Stanley-Reisner ring") of a connected graph. (It's also easy to check Cohen-Macaulayness directly.) But $I=(x_1x_3,x_1x_4)$ is not Cohen-Macaulay since it's the face ring of a nonpure simplicial complex. (The face ring of any nonpure simplicial complex has depth 1.) $\endgroup$ Commented Dec 6, 2020 at 14:03