Let $(A,\mathfrak{m})$ a noetherian local ring. If $\mathrm{depth}(A)=0$ then all non unit elements are 0 divisors so $\mathfrak{m}$ is a subset of the union of all the prime divisor of $(0)$ (the associated primes of $A$) so $\mathfrak{m}$ is one of these associated prime. I'm intersted of the converse: if $\mathfrak{m}$ is an associated prime of $A$, is it true the $\mathrm{depth}(A)=0$?
I had the idea of taking $k[x,y]/(x^2,xy)_{(x,y)}$, here $\mathfrak{m}$ is an embedded prime ideal (so it is an associated ideal), but I don't have any idea of the depth of such an ring.
Best Answer
If $\mathfrak m$ is an associated prime, then there is $x \in \mathfrak m - \{0\}$ such that $x\mathfrak m = 0$, so every element in $\mathfrak m$ is a zero-divisor, so $\operatorname{depth}(A) = 0$.