# The Stacks Project

## Tag: 00HL

This tag has label algebra-lemma-flat-tor-zero and it points to

The corresponding content:

Lemma 9.36.11. Suppose that $R$ is a ring, $0 \to M'' \to M' \to M \to 0$ a short exact sequence, and $N$ an $R$-module. If $M$ is flat then $N \otimes_R M'' \to N \otimes_R M'$ is injective, i.e., the sequence $$0 \to N \otimes_R M'' \to N \otimes_R M' \to N \otimes_R M \to 0$$ is a short exact sequence.

Proof. Let $R^{(I)} \to N$ be a surjection from a free module onto $N$ with kernel $K$. The result follows by a simple diagram chase from the following diagram $$\begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ 0 & \to & (M'')^{(I)} & \to & (M')^{(I)} & \to & M^{(I)} & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\ & & & & & & \uparrow & & \\ & & & & & & 0 & & \end{matrix}$$ with exact rows and columns. The middle row is exact because tensoring with the free module $R^{(I)}$ is exact. $\square$

\begin{lemma}
\label{lemma-flat-tor-zero}
Suppose that $R$ is a ring, $0 \to M'' \to M' \to M \to 0$
a short exact sequence, and $N$ an $R$-module. If $M$ is flat
then $N \otimes_R M'' \to N \otimes_R M'$ is injective, i.e., the
sequence
$$0 \to N \otimes_R M'' \to N \otimes_R M' \to N \otimes_R M \to 0$$
is a short exact sequence.
\end{lemma}

\begin{proof}
Let $R^{(I)} \to N$ be a surjection from a free module
onto $N$ with kernel $K$. The result follows
by a simple diagram chase from the following diagram
$$\begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ 0 & \to & (M'')^{(I)} & \to & (M')^{(I)} & \to & M^{(I)} & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\ & & & & & & \uparrow & & \\ & & & & & & 0 & & \end{matrix}$$
with exact rows and columns. The middle row is exact because tensoring
with the free module $R^{(I)}$ is exact.
\end{proof}


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