Krull dimension of tensor product Let $f: (R,m) \rightarrow (S,n)$ be a morphism of local Noetherian rings. Let $M$ be a finite $R$-module and $N$ a finite $S$-module such that $\operatorname{Supp}M = \operatorname{Spec} R$ and $\operatorname{Supp}N = \operatorname{Spec} S$.

Question: Why is it true that $\dim_S(M \otimes_R N) = \dim S$?

Remark 1: If $Q \in \operatorname{Supp}(M \otimes_R N)$, then $0 \neq M \otimes_R N \otimes_S S_Q = M \otimes_R N_Q$ and so $Q \in \operatorname{Supp} N$. Hence $\dim_S(M \otimes_R N) \le \dim S$. I don't see how equality can be achieved.
Remark 2: This question is motivated by the proof of A.5 in Bruns and Herzog, CMR.

Let's show that $\operatorname{Supp}_S(M \otimes_R N)=\operatorname{Spec}S$. 
If $P\in\operatorname{Spec}S$ is such that $(M\otimes_RN)_P=0$ then $M \otimes_R N_P=0$. Let $p=P\cap R$. From $M \otimes_R N_P=0$ we get $M_p\otimes_{R_p}N_P=0$. Now tensor by $k(p)$ (on the left), respectively by $k(P)$ (on the right). (Here $k(p)$ and $k(P)$ denote the residue fields of $R_p$ and $S_P$, respectively.) We get $M_p/pM_p\otimes_{k(p)}N_P/PN_P=0$ hence $M_p/pM_p=0$ or $N_P/PN_P=0$ since both are $k(p)$-vector spaces. By Nakayama we get $M_p=0$ or $N_P=0$, a contradiction.
