
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. $\square$

Comment #470 by JuanPablo on

There is a typo, the bottom row should have $K$ instead of $N$.

Comment #782 by Anfang Zhou on

Hi, I think it's better to say "snake lemma" here than "diagram chasing".

Comment #3506 by Manuel Hoff on

Hi, I think it's better to say "diagram chasing" here than "snake lemma". Seriously though, can somebody explain where the snake is?

Comment #3507 by Jonas Ehrhard on

By the claimed exactness of the diagram we have $0 = \mathrm{Ker}(M \otimes K \rightarrow M^{(I)})$, and

• $M'' \otimes N = \mathrm{Coker}(M''\otimes K \rightarrow (M'')^{(I)})$
• $M' \otimes N = \mathrm{Coker}(M'\otimes K \rightarrow (M')^{(I)})$
• $M \otimes N = \mathrm{Coker}(M \otimes K \rightarrow ((M)^{(I)})$

Then the snake connects $0 = \mathrm{Ker}(M \otimes K \rightarrow M^{(I)}) \rightarrow \mathrm{Coker}(M''\otimes K \rightarrow (M'')^{(I)}) \rightarrow \mathrm{Coker}(M'\otimes K \rightarrow (M')^{(I)}) \rightarrow \mathrm{Coker}(M \otimes K \rightarrow ((M)^{(I)})\rightarrow 0$.

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