Lemma 10.39.12 (00HL)—The Stacks project

Lemma 10.39.12. 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. $\square$

Comments (7)

Comment #470 by JuanPablo on March 18, 2014 at 16:38

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

Comment #485 by Johan on March 24, 2014 at 14:00

Thanks! Fixed here.

Comment #782 by Anfang Zhou on July 02, 2014 at 15:43

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

Comment #3506 by Manuel Hoff on August 08, 2018 at 12:06

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 August 08, 2018 at 16:07

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

Comment #9823 by Erhard Neher on October 24, 2024 at 09:52

Typo in the second line of the lemma. The assumption should read: If $N$ is flat, then ...

Comment #10385 by Stacks Project on July 09, 2025 at 10:30

@#9823 No, the assumption is that $M$ is flat.

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3 comment(s) on Section 10.39: Flat modules and flat ring maps

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