Support of module and faithfully flat base change 
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.
 A: Yes, this is true.
Indeed, we can prove the following.

Let $\pi:R\rightarrow S$ be a faithfully flat homomorphism of commutative rings. Assume $M$ is an $R$-module such that $\text{Supp}_R(M)=\text{Spec}(R)$, then $\text{Supp}_S(M\otimes_R S)=\text{Spec}(S)$.

First, $M\otimes_R S$ is non-zero as $\pi$ is faithfully flat. For each prime ideal $\mathfrak{q}$ of $S$. We denote $\mathfrak{p}=\pi^{-1}\mathfrak{q}$. This is a prime ideal of $R$. Note that $(M\otimes_R S)_{\mathfrak{q}}\cong M\otimes_R S_{\mathfrak q}\cong M_{\mathfrak{p}}\otimes_{R_{\mathfrak p}}S_{\mathfrak q}$. This is non-zero. The point is the induced map $R_{\mathfrak p}\rightarrow S_{\mathfrak q}$ is faithfully flat and $M_{\mathfrak p}\neq 0$ as $\text{Supp}_R(M)=\text{Spec}(R)$. To see $R_{\mathfrak p}\rightarrow S_{\mathfrak q}$ is faithfully flat, we just need to note that the flat homomorphism of local rings is faithfully flat.
A: Yes.
Here are some results we'll need:

*

*Flat ring maps satisfy going down.

*A ring map satisfies going down iff generalizations lift along it.

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.
