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Let $M,N$ be $A$-modules and let $P$ be a prime ideal.

Can someone please explain why the following isomorphism holds?

$$(M \otimes_{A} N)_{P} \cong M_{P} \otimes_{A_{P}} N_{P}$$

Here's what I tried:

Consider the map $f: M_{P} \times N_{P} \rightarrow (M \otimes_{A} N)_{P}$ given by $$(m/s,n/s') \mapsto (m \otimes n)/(ss')$$ Since this is bilinear, the universal property induces a map $g: (M_{P} \otimes_{A_{P}} N_{P}) \rightarrow (M \otimes_{A} N)_{P}$
given by $$g(m/s \otimes m'/s') = (m \otimes n)/(ss')$$

Is it true that this map is actually an isomorphism?

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2 Answers

up vote 5 down vote accepted

Yes, the map you construct is an isomorphism. It might be easiest to verify this by first using the canonical isomorphism $A_p\otimes_A M \cong M_p$, so that one then has the following simple chain of canonical isomorphisms: $$ M_P \otimes_{A_P} N_P \cong (A_P\otimes_A M)\otimes_{A_P} (A_P\otimes_A N) \qquad$$ $$\cong (A_P\otimes_A M) \otimes_A N \cong A_P\otimes_A (M\otimes_A N) \cong (M\otimes_A N)_P.$$

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Can you please explain the step: $(A_{P} \otimes_{A} M) \otimes_{A_{P}} (A_{P} \otimes_{A} N) \cong (A_{P} \otimes_{A} M) \otimes_{A} N$? –  user6495 May 28 '11 at 1:44
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@user6495: As Amitesh Datta explains in his answer below, this is a general property of tensor product: $M\otimes_B (B\otimes_A N) = M\otimes_A N$, if $M$ is a $B$-module and $N$ an $A$-module. Regards, –  Matt E May 28 '11 at 14:50
    
thank you. –  user6495 May 31 '11 at 2:30
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@user6495 Since I cannot leave comments, I will clarify the question you asked (I hope Matt E does not mind). The point is that the tensor product is commutative and associative. Therefore, we can write $(A_p\otimes_A M)\otimes_{A_p} (A_p\otimes_A N)\cong (M\otimes_A (A_p\otimes_{A_p} A_p))\otimes_A N\cong (M\otimes_A A_p)\otimes_A N\cong (A_p\otimes_A M)\otimes_A N$. The associativity of the tensor product used here is in the general sense of bimodules; $A_p$ is an $(A,A_p)$-bimodule.

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thank you. –  user6495 May 31 '11 at 2:30
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