If $M \otimes N=0$ then prove that either $M=0$ or $N=0$ [duplicate]

Solution 1:

1) The functor $- \otimes_A M$ is right exact. Therefore, the exact sequence $I \to A \to A/I \to 0$ yields the exact sequence $I \otimes_A M \to A \otimes_A M \to A/I \otimes_A M \to 0$. Now use that $A \otimes_A M \cong M$ and that the image of $I \otimes_A M \to M$ is exactly $IM$. This shows $A/I \otimes_A M \cong M/IM$.

You can also show directly that $M/IM$ satisfies the universal property of $A/I \otimes_A M$: Check that $([a],m) \mapsto am$ is a universal bilinear map $|A/I| \times |M| \to |M/IM|$.

2) There is always an isomorphism $(M \otimes_A N) \otimes_A B \cong (M \otimes_A B) \otimes_B (N \otimes_A B)$ for commutative $A$-algebras $B$. Just use the universal properties to show this.