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