On $\text{depth}_S\left(\dfrac {S}{JS+xS}\right)$, when $\text{depth}_R(R/J)=0$, and $R\to S$ is a certain flat map of local rings

Question

Let $(R, \mathfrak m) \xrightarrow{\phi} (S,\mathfrak n) $ be a flat homomorphism of local rings such that $\mathfrak n=\mathfrak m S +xS$ for some $x\in \mathfrak n \setminus \mathfrak n^2$. Let $J$ be an ideal of $R$ such that $\text{depth}_R(R/J)=0$. Then, is it true that $\text{depth}_S\left(\dfrac {S}{JS+xS}\right)=0$ ? (If needed, I am willing to assume $\phi$ is injective).

My thoughts: Using local-Cohomology, We know that for an ideal $J$ of a local ring $(R,\mathfrak m)$, $\text{depth}_R(R/J)=0$ if and only if $\mathfrak m^ny\subseteq J$ for some $n$ and $y\in R \setminus J$. Now we have to come up with an element, say $z$, of $S$ that is not in $JS+xS$, and $\mathfrak n^n z \subseteq JS+xS$. Now I have two natural chives of $z$, namely $\phi(y)$, and $\phi(y)+x$, and for both I can see that $\mathfrak n^n$ times the element is in $JS+xS$. But, I am not sure if for either of these two choices, the element belongs to $JS+xS$ or not.

Answer 1

Let $R=k[[u,v]]/(u^2,uv)$ and $S=R[x]/(x^2-u)$. Let $J=0$. You can check that $S/xS$ has depth one but $R$ has depth zero.