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?