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 $K[x_1,...,x_m]$?

  • $\begingroup$ Not all of us are ring-theory specialists. Would you provide the C-M ideal definition? $\endgroup$ – Wlod AA Jan 15 '20 at 7:15
  • 1
    $\begingroup$ @WlodAA An ideal $I$ of $R$ is Cohen-Macaulay if $R/I$ is a Cohen-Macaulay ring. $\endgroup$ – Stephen McKean Jan 16 '20 at 16:01
  • $\begingroup$ Oh, well, a reader here has to be either young or a specialist, $\endgroup$ – Wlod AA Jan 17 '20 at 10:28

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 $I'$.

  • $\begingroup$ Thank you so much for your nice example. $\endgroup$ – Amir Mafi Jan 18 '20 at 4:59
  • $\begingroup$ 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.) $\endgroup$ – Richard Stanley Dec 6 '20 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.