# Does $A\!\leq\!M$ and $B\!\leq\!N$ imply $A\!\otimes_R\!B\hookrightarrow M\!\otimes_R\!N$? (tensor product of modules)

Let $R$ be a commutative unital ring. What would be an example of a $R$-modules $M,N$ with submodules $A,B$, such that there does not exist an embedding of $R$-modules $$A\!\otimes_R\!B\hookrightarrow M\!\otimes_R\!N.$$

I've tried with finitely generated $\mathbb{Z}$-modules, by searching for $M$ and $N$ with $M\!\otimes\!N \cong 0$, but then every time also $A\!\otimes\!B \cong 0$ happens in my attempts.

Take $A=M=\mathbb{Z}/2$ and $B=\mathbb{Z}$ and $N=\mathbb{Q}.$ Then $A \otimes B \cong \mathbb{Z}/2$ while $M \otimes N \cong 0.$
Hint: Try $A=\mathbb{Z}$, $B=\mathbb{Z}/2\mathbb{Z}=N$, and $M=\mathbb{Q}$.