How to check Cohen-Macaulayness?

Question

Let $R=k[x,y,z]$. Consider the ideal $I=(x^2z^2,xyz,y^2z^4,y^4z^3,x^3y^5,x^4y^3)$. Is $R/I$ Cohen-Macaulay ?

By definition it seems tough to solve this problem. Is there any other way to check this?

Answer 1

The ideal $I=(x^2z^2,xyz,y^2z^4,y^4z^3,x^3y^5,x^4y^3)$ is monomial, so you can find a primary decomposition of $I$. (If you don't know how to do it, then look here.) Then notice that $I$ has an embedded associated prime, so $R/I$ is not Cohen-Macaulay.

Another approach: the height of $I$ is two, so $\dim R/I=1$. On the other side, $(x,y,z)y^3z^3\in I$, but $y^3z^3\notin I$, so $\operatorname{depth}R/I=0$.

Answer 2

You can take a primary decomposition of the ideal $I$. For monomials $x,y$ and monomial ideal $J$,

$(x + J) \cap (y + J) = (xy) + J$ provided that $\gcd(x,y) \sim 1$.

Then \begin{align} I&= (x^2z^2,xyz,y^2z^4,y^4z^3,x^3y^5,x^4y^3) \\ &= (xyz,x^2z^2,y^2z^4,y^4z^3,x^3y^5,x^4y^3) \\ &= (x,x^2z^2,y^2z^4,y^4z^3,x^3y^5,x^4y^3) \cap (y, x^2z^2,y^2z^4,y^4z^3,x^3y^5,x^4y^3) \\ &\qquad \cap (z, x^2z^2,y^2z^4,y^4z^3,x^3y^5,x^4y^3) \\ &= (x,y^2z^4,y^4z^3) \cap (y, x^2z^2) \cap (z,x^3y^5,x^4y^3) \\ &= [ (x, y^2) \cap (x,z^4,y^4z^3)] \cap [(y,x^2) \cap (y,z^2)] \cap [(z,x^3) \cap (z, y^5,x^4y^3)] \\ &= [ (x, y^2) \cap (x,z^4,y^4) \cap (x,z^3)] \cap [(y,x^2) \cap (y,z^2)] \cap [(z,x^3) \cap (z, y^5,x^4) \cap (z,y^3)] \\ &= (x, y^2) \cap (x,z^3) \cap (y,x^2) \cap (y,z^2) \cap (z,x^3) \cap (z,y^3) \cap (z, y^5,x^4) \cap (x,z^4,y^4) \end{align}

Since $(x,y,z)$ is associated to $I$, $\operatorname{depth} R/I = 0$. But $\dim R/I = 1$; hence $R/I$ is not Cohen-Macaulay. In other words, $I$ is a mixed ideal.