According to https://stacks.math.columbia.edu/tag/031O, Definition 10.157.1(2), a Noetherian ring $R$ has property $S_2$ if for every prime $\mathfrak{p}$ the local ring $R_{\mathfrak{p}}$ has depth at least $\min\{2, \dim R_{\mathfrak{p}}\}$.
On the other hand, in p.252 (Theorem 11.5) of Eisenbud's book on commutative algebra, the condition $S_2$ is defined as follows: every associated prime of a principal ideal generated by a nonzero divisor in $R$ is of codimension $1$, and every associated prime of $0$ is of codimension $0$. Let us say that $R$ has propert $S_2'$ if this holds.
Are these two conditions equivalent? Actually I want to know that whether $S_2'$ implies $S_2$.How can we prove this?
$S_2'\implies S_2$
If $\dim R_p=0$ there is nothing to prove.
Suppose $\dim R_p=1$. We want to show $\operatorname{depth} R_p\ge1$. If $\operatorname{depth} R_p=0$ then $p$ is an associated prime of $R$ (or of $(0)$ in Eisenbud's terminology), so the height (codimension) of $p$ equals $0$ which contradicts $\dim R_p=1$.
Suppose $\dim R_p\ge2$. We want to show that $\operatorname{depth}R_p\ge 2$. As before, $p$ is not an associated prime of $R$, and thus there is a non-zerodivisor $a\in p$. If $p$ is an associated prime of $(a)$, then its height (codimension) is $1$, a contradiction. For the same reason $p$ can't be contained in any other associated prime of $(a)$, so it contains a non-zerodivisor $b$ of $R/(a)$. Then $a,b$ is an $R$-sequence, and this implies that $\operatorname{depth} R_p\ge2$.
Remark. In fact, $S_2$ is equivalent with $S_2'$.