# 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

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.

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.
• Thank you. Do you think the claim would have been true if one just takes $S=R[X]_{(\mathfrak m, X)}$ ? Nov 22, 2022 at 6:58
• Why is it trivial in that case? What would be a non-zero element in the socle of $\dfrac{S}{JS+xS}$? Nov 22, 2022 at 17:34
• @feder $S/JS+XS=R/J$. Nov 22, 2022 at 17:38