In Keith Conrad's text on Tensor Products, he states on Page 8 of the text that for a fixed ring $R$ and $R$-modules $M$, $M'$, $N$ and $N'$, and maps $f:M\to M'$, $g:N\to N'$ that $\text{ker}(f)\otimes_R N$, strictly speaking is not a submodule of $M\otimes_R N$ (and similarly for $M\otimes_R\text{ker}(g)$).
The question that I have in mind is:
Given a fixed ring $R$ and $R$-modules $M$, $N$ with submodules $M'$ and $N'$ respectively, with respective inclusion maps $i:M'\to M$ and $j:N'\to N$, is it possible for (1) $M'\otimes_R N'$ to not be a submodule of $M\otimes_R N$, and (2) $i\otimes j$ to be not injective?
Best Answer
It might be good to record several examples of this, so here is another:
Take $R =\mathbb Z$ (so modules are just abelian groups), and let $M' = 2 \mathbb Z \subset \mathbb Z = M$, and $N' = N = \mathbb Z/2\mathbb Z$ (so the inclusion map $j$ is the identity). Then each of $M\otimes N$ and $M'\otimes N' = M'\otimes N$ is isomorphic to $N$, but $i\otimes j$ is the zero map.