Let $M$ and $N$ be $R$-modules. Let $\mathrm{Supp}(M)$ be the set of primes $P$ such that $M_P\neq 0$, and let $\mathrm{Ann}(M)$ be the ideal of elements $r\in R$ such that $rm=0$ for all $m\in M$.
It is easy to see that $\mathrm{Supp}(M\otimes N)\subseteq\mathrm{Supp}(M)\cap\mathrm{Supp}(N)$ and that $\mathrm{Ann}(M)+\mathrm{Ann}(N)\subseteq \mathrm{Ann}(M\otimes N)$.
I'm looking for examples where these inclusions are strict, and in the second case, I'm looking for finitely generated $M, N$.
I tried $\mathbb Z$-modules so far, but I don't think I can find a counterexample for the second inclusion, since finitely generated abelian groups have easily computed annihilators that make the inclusion an equality.
Best Answer
Set $R=\mathbb Z$, $M=\mathbb Q$, and $N=\mathbb Z/2\mathbb Z$. Then $M\otimes_RN=0$, so $\mathrm{Supp}(M\otimes_RN)=\emptyset$, while $\mathrm{Supp}(M)=\mathrm{Spec}(\mathbb Z)$ and $\mathrm{Supp}(N)=\{2\mathbb Z\}$.
For the second one can find an example here, and another one here.