8
$\begingroup$

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?

$\endgroup$
1
9
$\begingroup$

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: $$ M_P \otimes_{A_P} N_P \cong (A_P\otimes_A M)\otimes_{A_P} (A_P\otimes_A N) \qquad$$ $$\cong (A_P\otimes_A M) \otimes_A N \cong A_P\otimes_A (M\otimes_A N) \cong (M\otimes_A N)_P.$$

$\endgroup$
2
  • $\begingroup$ Can you please explain the step: $(A_{P} \otimes_{A} M) \otimes_{A_{P}} (A_{P} \otimes_{A} N) \cong (A_{P} \otimes_{A} M) \otimes_{A} N$? $\endgroup$ – user6495 May 28 '11 at 1:44
  • 1
    $\begingroup$ @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, $\endgroup$ – Matt E May 28 '11 at 14:50
6
$\begingroup$

@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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.