TITLE: Atiyah-Macdonald Chapter 3 Qn 19 viii
QUESTION [5 upvotes]: I am having a problem with a question in Atiyah-Macdonald Chapter 3 qn 19(viii). The question is as follows:
If $f:A\rightarrow B$ is a ring homomorphism, and $M$ is finitely generated $A$-module, then $\mbox{Supp}(B\otimes_{A}M)=f^{*-1}(\mbox{Supp}(M))$
I tried to show $\mbox{Supp}(B\otimes_{A}M)\subseteq f^{*-1}(\mbox{Supp}(M))$. Suppose ${\frak{p}}\notin f^{*-1}(\mbox{Supp}(M))$, then $f^{-1}(\mathfrak{p})\notin \mbox{Supp}(M)$. Thus for each generator $m_{i}$ of $M$, there exists $s_{i}\in A-f^{-1}(\mathfrak{p})$ such that $s_{i}m_{i}=0$. If $f(s_{i})=0$ then since $f(s_{i})\in B-\mathfrak{p}$, thus $0\in B-\mathfrak{p}$ and we are done. Otherwise for all $f(s_{i})$ if they are not zero, we have $f(s_{1})f(s_{2})\ldots f(s_{k})(b\otimes m)=b\otimes (s_{1}s_{2}\ldots s_{k})m=0$
But I am stuck with $f^{*-1}(\mbox{Supp}(M))\subseteq \mbox{Supp}(B\otimes_{A}M)$. The problem is if I start letting $\mathfrak{p}\in LHS$, so each of $m_{i}$ is not zero, but I can't show that $1\otimes m_{i}$ is not zero. Or if I start from $\mathfrak{p}\notin RHS$, I can't show that $(A-f^{-1}\mathfrak{p})M=0$. 
How should I approach? Thanks!

REPLY [11 votes]: Since no one answered you (because it's Sunday?), let me try it. 
Let $q\in f^{*-1}(\mbox{Supp}(M))$. Then $p=f^{-1}(q)\in\mbox{Supp}(M)$, that is, $M_p\neq 0$. I want to prove that $q\in\mbox{Supp}(B\otimes_{A}M)$. Since  $(B\otimes_{A}M)_q\cong B_q\otimes_{A_p}M_p$ (why?) we can reduce the problem as follows: let $f:(A,m)\to(B,n)$ be a homomorphism of local rings with $m=f^{-1}(n)$ and $M\neq 0$ a finitely generated $A$-module. If $B\otimes_AM=0$, then tensor this by $B/n$ and get that $B/n\otimes_AM=0$. But $B/n\otimes_AM\cong B/n\otimes_{A/m}M/mM$, so we obtained a field extension $A/m=K\subset L=B/n$ and $V=M/mM$ a nonzero (why?) $K$-vectorspace such that $L\otimes_KV=0$, a contradiction. Therefore $B\otimes_AM\neq 0$.