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I have a very basic question about the interactions of the image/kernel of module maps with tensor products.

For a ring $A$, $A$-modules $M, M'$ and $N$, and a map of $A$-modules $f: M\mapsto M'$, do the following hold?

$$\operatorname{Im}(f \otimes_A Id_N) = \operatorname{Im}(f) \otimes_A N $$ $$\operatorname{Ker}(f) \otimes_A N \subseteq \operatorname{Ker}(f \otimes_A Id_N)$$

Atticus Stonestrom
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maroxe
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1 Answers1

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Edit: What I had written here before was completely wrong; thanks to FShrike for point this out.

In fact, neither of these is true in any sense; work with $A=\mathbb{Z}$. For the first question, take $M=\mathbb{Z}$ and $M'=\mathbb{Q}$, and $f:M\to M'$ the inclusion, and $N=\mathbb{Z}/2\mathbb{Z}$. Then $M'\otimes N=0$, so in particular $\mathrm{im}(f\otimes\mathrm{id}_N)=0$, but $\mathrm{im}(f)\otimes N=\mathbb{Z}\otimes N\cong\mathbb{Z}/2\mathbb{Z}$. For the second question, take $M=\mathbb{Q}$ and $M'=\mathbb{Q}/\mathbb{Z}$, with $M\to M'$ the quotient map, and $N=\mathbb{Z}/2\mathbb{Z}$. Then $M\otimes N=0$, so in particular $\mathrm{ker}(f\otimes\mathrm{id}_N)=0$, but $\mathrm{ker}(f)\otimes N=\mathbb{Z}\otimes N\cong\mathbb{Z}/2\mathbb{Z}$.

Atticus Stonestrom
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    it is not correct that the natural map $\ker(f)\otimes M\to\ker(f\otimes1)$ is injective; it is not the case that $\ker(f)\otimes M$ is naturally a submodule of $\ker(f\otimes1)$. – FShrike Jan 11 '24 at 16:28
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    @FShrike oh dear, thanks very much for pointing this out!! was clearly not thinking clearly at the time. it is now corrected :) – Atticus Stonestrom Jan 12 '24 at 00:35