Cohen-Macaulay ring and saturated ideal

Question

Let $A=\mathbb{C}[x_0,x_1,\dots,x_n]$ and $I$ an ideal of $A$.

Is there any connection between "$A/I$ is Cohen-Macaulay ring" and "$I$ is a saturated ideal"? Does one of them imply another?

Please give me any reply or reference.

Answer 1

If $I$ is not saturated, then $m= (x_0,\dots, x_n)A$ is associated to $A/I$. Therefore, depth $(A/I)_m = 0$. This implies that $A/I$ is not Cohen-Macaulay.

From this, one sees that if $A/I$ Cohen-Macaulay, then $I$ saturated. But the converse is not true in general. For instance, take $A = k[x,y,z]$ and $I = (x) \cap (y,z)$.