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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!

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@navigetor23: Thanks! Noted. – enoughsaid05 Nov 18 '12 at 10:36
up vote 4 down vote accepted

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 $B\otimes_AM\neq 0$.

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What are (A,m) and (B,n)? How does $B/n\otimes_{A}M\cong B/n\otimes_{A/m}M/mM$ leads to the field extension? Why is $M/mM$ a non-zero K-vsp, and what does it contradict? Thanks! – enoughsaid05 Nov 19 '12 at 6:20
Well, it seems that you are not used with the reduction techniques in commutative algebra. Here $A$ denotes $A_p$ and $m$ is $pA_p$ (similar for $B$). If you have an extension of rings $R\subset S$ and $P$ a prime of $S$ whose contraction to $R$ is $p$, then $R/p\subset S/P$. $M/mM=0$ leads to $M=0$ via Nakayama. (Btw, the new $M$ is $M_p$.) – user26857 Nov 19 '12 at 9:59
sorry if i ask too many questions but why $B/n$ is a field extension of $A/m$? (and I probably have other questions on some identities but i will try to figure them out first). Anyway, what is the reduction technique in commutative algebra? – enoughsaid05 Nov 19 '12 at 17:43
Because they are fields and there is a ring homomorphism from one to another (which is necessarily injective, right?). – user26857 Nov 19 '12 at 17:45
One question: to answer why it leads to a contradiction, am I right in saying that because $M/mM\cong K\otimes_{K}M/mM\cong \phi(K)\otimes_{K}M/mM\subseteq L\otimes_{K}V$ where $\phi$ is the field homomorphism from $K$ to $L$. so since if we assume that $L\otimes_{K} M/mM$ is zero, implies that so is $M/mM$ and hence $M=0$ by Nakayama's lemma. Other than that, I think they are correct! How do you manage to come up with this proof (especially on your first try!). What is a reduction technique in commutative algebra? – enoughsaid05 Nov 19 '12 at 20:11

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