Let $L/K$ be a field extension.
Let $R,S$ be two local commutative $K$-algebras and let $\varphi : R \to S$ be a homomorphism of $K$-algebras, not assumed to be local. Let's call a prime ideal $\mathfrak{p} \subseteq R \otimes_K L$ *good* when $\mathfrak{p} \cap R = \mathfrak{m}_R$. Notice that good prime ideals correspond to prime ideals in the tensor product of fields $R/\mathfrak{m}_R \otimes_K L$.

I wonder if the following is true:

**Question.** Is there some good prime ideal $\mathfrak{p} \subseteq R \otimes_K L$ such that for all $f \in R \otimes_K L$ with $f \notin \mathfrak{p}$ the image of $f$ in $S \otimes_K L$ is not contained in every good prime ideal of $S \otimes_K L$? Equivalently, the image of $f$ in $S/\mathfrak{m}_S \otimes_K L$ is not nilpotent.

In terms of the local $K$-schemes $X=\mathrm{Spec}(R)$, $Y=\mathrm{Spec}(S)$ and the morphism $Y \to X$, the question is the following: Is there some point in $X_L$ over the closed point of $X$ such that every open neighborhood of it pulls back to an open subset in $Y_L$ which contains a point over the closed point of $Y$?

I have checked some special cases, but either they were trivial or too hard to understand, since tensor products of fields can be nasty. Of course it is true when $L=K$, and it is also true when $\varphi$ is local. My feeling is that the statement is false in general, but I might be wrong. Maybe counterexamples can be constructed from localizations of $R$.

The context for this question is to prove a certain result for locally ringed spaces, and for locally ringed spaces with exactly two points it comes down to the question above.