Depth and associated primes

Question

Suppose $R$ is a polynomial ring and $I$ is a square free monomial ideal of positive grade. Is it true that $\operatorname{depth}R/I = \min\{\operatorname{depth} R/P \mid P\in\operatorname{Ass}(R/I)\}$? Or any condition when the above is true?

Thanks in advance.

Answer 1

Ans:NO

Let $R=k[x_1,\ldots,x_4]$ be a polynomial ring and $I=\langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle$ be a square free monomial ideal.

Primary decomposition of $I$ is $I=\langle x_1,x_3\rangle \cap \langle x_2,x_4\rangle $ (since $I$ is radical). Therefore $\operatorname{Ass}R/I=\{\langle x_1,x_3\rangle,\langle x_2,x_4\rangle\}$.

$\operatorname{depth}R/\langle x_1,x_3\rangle=\operatorname{depth} R/\langle x_2,x_4\rangle=2$. But $\operatorname{depth} R/I=1$.