Let $$0\to L\xrightarrow{\alpha} M \xrightarrow{\beta} N\to 0$$ be a short exact sequence of right $R$-modules and homomorphisms. Show that if $L$ and $N$ are flat then $M$ is flat.

Let $f:A\hookrightarrow B$ be a injective left $R$-module homomorphism. My attempt is to show that $M\otimes f$ is injective.

As $L$ and $N$ are flat, $L\otimes f$ and $N\otimes f$ are inective. And one has the following commutative diagram:

$$\begin{array}{ccccccccc} 0 & \to & L\otimes A & \xrightarrow{\alpha\otimes A} & M\otimes A & \xrightarrow{\beta\otimes A} & N\otimes A & \to & 0 \\ & &\downarrow_{L\otimes f} & & \downarrow_{M\otimes f} & & \downarrow_{N\otimes f} & &\\ 0 & \to & L\otimes B & \xrightarrow{\alpha\otimes B} & M\otimes B & \xrightarrow{\beta\otimes B} & N\otimes B & \to & 0 \end{array}$$ And, as $\_\otimes A$ and $\_\otimes B$ are right-exact functors, $\beta\otimes A$ and $\beta\otimes B$ are surjective.

Take then $x\in M\otimes A$ such that $M\otimes f(x)=0$. Thus $$\beta\otimes B(M\otimes f(x))=0.$$ As the diagram commutes, $$N\otimes f(\beta\otimes A(x))=0.$$ But $N\otimes f$ is injective, given the flatness of $N$.

Thus, $\beta\otimes A(x)=0$. I.e., $x\in \ker(\beta\otimes A)=\operatorname{Im}(\alpha\otimes A)$. This way, there is a $y\in L\otimes A$ s.t. $$\alpha\otimes A(y)=x.$$

Then $M\otimes f(\alpha\otimes A(y))=0$. As the diagram is commutative, $\alpha\otimes B( L\otimes f(y))=0$.

If $x\neq 0$, then $y\neq 0$. Then $L\otimes f(y)\neq 0$. But as $\alpha \otimes B$ is not necessarily injective there is no problem with $\alpha\otimes B(L\otimes f(y))=0$.


Let $B$ be an $R$-module. The short exact sequence $$0 \to L \xrightarrow{\alpha} M \xrightarrow{\beta} N\to 0$$ induces a long Tor exact sequence which ends with

${\displaystyle {\begin{aligned}\cdots \to \mathrm {Tor}_{2}^{R}(N,B)\to \mathrm {Tor} _{1}^{R}(L,B)&\to \mathrm {Tor} _{1}^{R}(M,B)\to \mathrm {Tor} _{1}^{R}(N,B) \to \\[4pt]&\to L\otimes B\to M\otimes B\to N\otimes B\to 0 \end{aligned}}}$

By hypothesis we have $ \ \mathrm{Tor}^R_1(L,B) =0 \ $ and $ \ \mathrm{Tor}^R_1(N,B) =0 \ $. It follows that the sequence $$0 \to \mathrm {Tor} _{1}^{R}(M,B)\to 0$$ is exact. Thus $ \ \mathrm{Tor}^R_1(M,B) =0 \ $. Since $B$ is arbitrary, we have that $M$ is flat.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.