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This is a follow-up to this question: was flatness really used in this argument? (Matsumura, Theorem 7.2).

The author of the given answer says we need flatness to ensure that tensoring commutes with taking images. I still find this statement unclear. Here is why:

Let $f: N' \rightarrow N$ be a morphism of $A$-modules and let $M$ be an $A$-module. Then the functor $-\otimes_A M$ takes $f$ to $f_M : N' \otimes M \rightarrow N \otimes M$ with $f_M(x \otimes y)=f(x) \otimes y$. So $\operatorname{Im} f_M = (f \otimes 1)(N' \otimes M) \subset f(N') \otimes M$. Conversely, for every $x' \in N', y \in M$ we have that $f(x') \otimes y = f_M(x' \otimes y)$ and so $f_M(N' \otimes M) = f(N') \otimes M$. Thus tensoring commutes with taking images, even if $M$ is not flat.

Is my argument correct?

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Manos
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1 Answers1

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One has to be very careful here. It is true that every element in the image of $f_M$ is a sum of elements of the form $f(x) \otimes y$. But this is a tensor in $N \otimes M$. So we can just say that it lies in the image of $\mathrm{im}(f) \otimes M \to N \otimes M$. But this map is not injective in general (unless $M$ is flat), so we cannot view $\mathrm{im}(f) \otimes M$ as a submodule of $N \otimes M$, whereas the image of $f_M$ is a submodule of $N \otimes M$. What you have observed is basically that $\mathrm{im}(f) \otimes M \to \mathrm{im}(f_M)$ is surjective. It is not a monomorphism in general, in fact it may happen that $\mathrm{im}(f_M)=0$ i.e. $f_M=0$, although $\mathrm{im}(f) \otimes M \neq 0$. The standard example is $A=\mathbb{Z}$, $f : \mathbb{Z} \to \mathbb{Z}, z \mapsto pz$ for some $p>1$, and $M=\mathbb{Z}/p\mathbb{Z}$.

Martin Brandenburg
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