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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 $\alpha\otimes B(L\otimes f(y))=0$.

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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.

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