Let $R \subseteq S$ be a faithfully flat extension of Noetherian local rings. Let $M$ be a finitely generated $R$-module such that $\operatorname{Supp}_R(M)=\operatorname{Spec}(R)$. Then, is it true that $\operatorname{Supp}_S(M\otimes_R S)=\operatorname{Spec}(S)?$
Of course, it is enough to prove that localization of $M\otimes_R S$ is non-zero at all minimal primes of $S$, but I am not sure how to achieve this.
Please help.
Yes.
Here are some results we'll need:
Suppose $p\in\operatorname{Spec} S$ is minimal with image $q\in\operatorname{Spec} R$. Then if $q$ were not minimal, it would generalize to some minimal $q'$, and by 1+2 we would have a $p'$ which was a generalization of $p$ and also mapped to $q'$. But $p$ has no nontrivial generalizations because it's a minimal prime, so $q$ is also a minimal prime.
Now let's look at $M\otimes_R S \otimes_S S_p$. This is exactly $M_q\otimes_{R_q} S_p$, and since $R_q\to S_p$ is faithfully flat, $M_q$ is nonzero iff $M_q\otimes_{R_q} S_p$ is nonzero.