If $R/I$ satisfies Serre's condition $S_2$ then $R/(I,x)$ satisfies $S_1$

Question

Let $I$ be an ideal of polynomial ring $R=K[x_1,\ldots,x_n]$ and $x$ be a non-zero divisor of $R/I$. Is the following statement true?

If $R/I$ satisfies Serre's condition $S_2$ then $R/(I,x)$ satisfies $S_1$.

Background:

The condition $S_i$ on ring $R$ means that $\operatorname{depth} R_p\geq \min\{\operatorname{height} p, i\}$ for all $p\in \operatorname{Spec}(R)$.

Answer 1

Let's denote $R/I$ by $A$. Then $A$ satisfies $(S_2)$ and if $x\in A$ is a regular element we wonder whether $A/(x)$ satisfies $(S_1)$. Let $p/(x)$ be a prime ideal of $A/(x)$. Then $(A/(x))_{p/(x)}\simeq A_p/xA_p$ and $\operatorname{depth}A_p/xA_p=\operatorname{depth}A_p-1$. Since $\operatorname{depth} A_p\ge \min(2, \operatorname{ht}p)$ we get $\operatorname{depth} A_p/xA_p\ge\min(1, \operatorname{ht}p/(x))$. (Note that $\operatorname{ht}p/(x)=\operatorname{ht}p-1$.)