Good prime ideals in tensor products of local rings

Question

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.

Answer 1

Let $x$, resp $y$ be the closed point of $X$, resp $Y$. Denote $f : Y \to X$ the given morphism. Then $f(y) \leadsto x$ (specialization). Let $x_L$ be any point of $X_L$ mapping to $x$. The morphism $X_L \to X$ is flat. Hence there is a specialization $z \leadsto x_L$ in $X_L$ such that $z$ maps to $f(y)$. Since $Y_L = Y \times_X X_L$, there is a point $w$ of $Y_L$ which maps to $y$ in $Y$ and $z$ in $X_L$. This proves what you want as $w$ will be in the inverse image of any open neighborhood of $x_L$ you pick.