# Cohen-Macaulay monomial ideal

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 $$m, could we say that $$I[x_1,...,x_m,0,0,...,0]$$ is Cohen-Macaulay in $$K[x_1,...,x_m]$$?

• Not all of us are ring-theory specialists. Would you provide the C-M ideal definition? – Wlod AA Jan 15 '20 at 7:15
• @WlodAA An ideal $I$ of $R$ is Cohen-Macaulay if $R/I$ is a Cohen-Macaulay ring. – Stephen McKean Jan 16 '20 at 16:01
• Oh, well, a reader here has to be either young or a specialist, – 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 $$I'$$.
• 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.) – Richard Stanley Dec 6 '20 at 14:03