Property of Flat Modules

Question

I am trying to understand a proof regarding some properties of flat modules. The place I am stuck at, essentially boils down to this situation:

Let $M'\overset{\varphi}{\rightarrow}M\overset{\psi}{\rightarrow} M''$ denote an exact sequence of $R$-modules and let $N$ denote an $R$-module. Also, assume that the diagram

has exact rows. Now the claim is, that the sequence $M'\otimes N\overset{\varphi\otimes\operatorname{id}}{\rightarrow}M\otimes_R N\overset{\psi\otimes\operatorname{id}}{\rightarrow} M''\otimes_R N$ is exact.

I tried it in the following way. Since the first row is exact we have that $\operatorname{im}(\varphi\otimes\operatorname{id})=\operatorname{im}\varphi\otimes N$. But $\operatorname{im}\varphi\otimes N=\operatorname{ker}\psi\otimes N$ and since the second row is exact the first non-zero map is an embedding and it holds $\operatorname{ker}\psi\otimes N=\operatorname{ker}(\psi\otimes \operatorname{id})$ which is the claim.

But I have the feeling, that I am missing something. If the above approach was correct, then the third row in the diagram would be obsolete, which makes me skeptical.

Answer 1

You need the third row in order to know that $\ker (\psi \otimes \mathrm{id}) \subset \ker \psi \otimes N$. The second row allows you to conclude that $\ker \psi \otimes N \subset \ker (\psi \otimes \mathrm{id})$ since $\ker \psi \otimes N$ maps to $0$ in $\operatorname{im} \psi \otimes N$, and you have a homomorphism from $\operatorname{im} \psi \otimes N$ to $M'' \otimes N$, but if the map $\operatorname{im} \psi \otimes N \to M'' \otimes N$ had a nontrivial kernel, then $\ker(\psi \otimes \mathrm{id}_N)$ could be larger than $\ker \psi \otimes N$.

Explicitly, call the map in the second row $\rho_1 : M \otimes_R N \to \operatorname{im} \psi \otimes N$, and the map in the third row $\rho_2 : \operatorname{im} \psi \otimes N \to M'' \otimes N$. Then $\psi \otimes \mathrm{id}_N = \rho_2 \circ \rho_1$, so $$\begin{align*} \ker(\psi \otimes \mathrm{id}_N) & = \rho_1^{-1}(\ker \rho_2) \\ & = \rho_1^{-1}(0) && \text{since the last row is exact} \\ & = \ker(\rho_1) \\ & = \ker \psi \otimes N && \text{since the second row is exact.} \end{align*}$$

Answer 2

You basically want to show that if ${-}\otimes_RN$ preserves short exact sequences, then it preserves all exact sequences. Indeed exactness of the bottom row for any $\psi$ is equivalent to ${-}\otimes_RN$ being left exact, so it preserves short exact sequences (because it is right exact for every $N$).

You just need to prove that $\ker(\psi\otimes_R\mathrm{id}_N)\subseteq\operatorname{im}(\varphi\otimes_R\mathrm{id}_N)$

Let's simplify the notation setting $\hat{\alpha}=\alpha\otimes_R\mathrm{id}_N$, so you want to show that $$ \ker\hat{\psi}\subseteq\operatorname{im}\hat{\varphi} $$ Factor $\hat{\psi}=gf$, where $f\colon M\otimes_RN\to\operatorname{im}\psi\otimes_RN$ and $g\colon\operatorname{im}\psi\otimes_RN\to $M''\otimes_RN$ are the maps in the diagram.

If $t\in\ker\hat{\psi}$, then $gf(t)=0$ and, since $g$ is injective, we get $f(t)=0$. Thus $t=h(u)$ for some $u\in\ker\psi\otimes_RN$ and $h\colon\ker\psi\otimes_RN\to M\otimes_RN$ is the map in the diagram.

Now the top map is surjective, so we get $v\in M'\otimes_RN$ such that $\hat{\varphi}(v)=t$.

As you see, specific properties of the tensor product are not used. Indeed, the same technique can be used to show that if $\operatorname{Hom}_R(P,{-})$ preserves short exact sequences (that is, $P$ is projective), then it preserves all exact sequences.

More generally, if an additive functor between abelian categories preserves short exact sequences, then it preserves all exact sequences.