$\DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\Supp}{Supp}$
Let $A$ be a ring, $M$ an $A$-module. The support of $M$ is defined to be the set $\Supp(M)$ of prime ideals $\frak{p}$ of $A$ such that $M_{\frak{p}}\neq 0.$ Prove that if $f: A\to B$ is a ring homomorphism and $M$ is a finitely generated $A$-module, then $\Supp(B \otimes_A M)=f^{*-1}(\Supp(M)).$
[$f^*$ is the map $\Spec(B)\to \Spec(A)$ sending $q \mapsto f^{-1}(q)$.]
I'm reading Jeffrey Carlson's solution here. They write, 
My trouble is that $1/1$ may not be the only generator of $N$, so I don't see why $(1/1) \otimes (x/1)=0$ in $N \otimes_{A_{\frak{p}}} M_{\frak{p}}$ for all $x \in M_{\frak{p}}$ implies that $N \otimes_{A_{\frak{p}}} M_{\frak{p}}=0.$ How does this follow?
I assume $N_i \subset B_{\mathfrak{q}}$, rather than $N_i \subset B$?
Any simple tensor of $N \otimes_{A_{\mathfrak{p}}} M_{\mathfrak{p}}$ is of the form $(b/t) \otimes (m/s)$ for $b \in B$, $t \in B-\mathfrak{q}$, $m \in M$ and $s \in A-\mathfrak{p}$, which is just $(b/f(s)t) \cdot (1/1)\otimes (m/1)$, considering that the $A_{\mathfrak{p}}$-action on $N$ is given by $(a/s)\cdot (b/t) = f(a)b/f(s)t$. By the proof above, any element of the form $(1/1)\otimes (m/1)$ is zero, so $N\otimes_{A_{\mathfrak{p}}} M_{\mathfrak{p}}= 0$ follows.