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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 $\text{ker}(f)\otimes_R M'$, 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?

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2 Answers 2

Let $R = k[x]/x^2$, let $M = N = R$, let $M' = N' = (x)$, and let $i, j\colon(x) \to R$ be inclusion. Then $M' \otimes_R N' = (x) \otimes_R (x)$ is nonzero (see if you can prove this!) but under $i \otimes j$ the image of $M' \otimes_R N'$ in $M \otimes_R N = R \otimes_R R \simeq R$ is $(x)(x) = (x^2) = 0$. Thus $i \otimes j$ is the zero map.

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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.

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