## Tag `00HL`

Chapter 10: Commutative Algebra > Section 10.38: Flat modules and flat ring maps

Lemma 10.38.12. 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 from the snake lemma applied to the following diagram $$ \begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ 0 & \to & (M'')^{(I)} & \to & (M')^{(I)} & \to & M^{(I)} & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R K & \to & M' \otimes_R K & \to & M \otimes_R K & \to & 0 \\ & & & & & & \uparrow & & \\ & & & & & & 0 & & \end{matrix} $$ with exact rows and columns. The middle row is exact because tensoring with the free module $R^{(I)}$ is exact. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 8716–8726 (see updates for more information).

```
\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
from the snake lemma applied to the following diagram
$$
\begin{matrix}
& & 0 & & 0 & & 0 & & \\
& & \uparrow & & \uparrow & & \uparrow & & \\
& & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\
& & \uparrow & & \uparrow & & \uparrow & & \\
0 & \to & (M'')^{(I)} & \to & (M')^{(I)} & \to & M^{(I)} & \to & 0 \\
& & \uparrow & & \uparrow & & \uparrow & & \\
& & M''\otimes_R K & \to & M' \otimes_R K & \to & M \otimes_R K & \to & 0 \\
& & & & & & \uparrow & & \\
& & & & & & 0 & &
\end{matrix}
$$
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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