# Good prime ideals in tensor products of local rings

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.

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.