Lemma 10.39.15 (00HP)—The Stacks project

Lemma 10.39.15. Let $M$ be a flat $R$-module. The following are equivalent:

$M$ is faithfully flat,
for every nonzero $R$-module $N$, then tensor product $M \otimes _ R N$ is nonzero,
for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$ the tensor product $M \otimes _ R \kappa (\mathfrak p)$ is nonzero, and
for all maximal ideals $\mathfrak m$ of $R$ the tensor product $M \otimes _ R \kappa (\mathfrak m) = M/{\mathfrak m}M$ is nonzero.

Proof. Assume $M$ faithfully flat and $N \not= 0$. By Lemma 10.39.14 the nonzero map $1 : N \to N$ induces a nonzero map $M \otimes _ R N \to M \otimes _ R N$, so $M \otimes _ R N \not= 0$. Thus (1) implies (2). The implications (2) $\Rightarrow $ (3) $\Rightarrow $ (4) are immediate.

Assume (4). Suppose that $N_1 \to N_2 \to N_3$ is a complex and suppose that $N_1 \otimes _ R M \to N_2\otimes _ R M \to N_3\otimes _ R M$ is exact. Let $H$ be the cohomology of the complex, so $H = \mathop{\mathrm{Ker}}(N_2 \to N_3)/\mathop{\mathrm{Im}}(N_1 \to N_2)$. To finish the proof we will show $H = 0$. By flatness we see that $H \otimes _ R M = 0$. Take $x \in H$ and let $I = \{ f \in R \mid fx = 0 \} $ be its annihilator. Since $R/I \subset H$ we get $M/IM \subset H \otimes _ R M = 0$ by flatness of $M$. If $I \not= R$ we may choose a maximal ideal $I \subset \mathfrak m \subset R$. This immediately gives a contradiction. $\square$

Comments (6)

Comment #892 by Kestutis Cesnavicius on August 09, 2014 at 12:58

Suggested slogan: A flat module is faithfully flat iff it has nonzero fibers

Comment #4585 by Bjorn Poonen on October 07, 2019 at 18:09

It would be nice to include another part between (1) and (2) (I'll call it (1.5) but of course you should renumber):

(1.5) for every nonzero $R$ -module $N$ , the tensor product $M \otimes_R N$ is nonzero,

Then you can replace the first paragraph of the proof by

Proof. Assume $M$ faithfully flat and $N \ne 0$ . By Lemma 10.39.14, the nonzero map $1\colon N \to N$ induces a nonzero map $M \otimes_R N \to M \otimes_R N$ , so $M \otimes_R N \ne 0$ .

Comment #4765 by Johan on December 09, 2019 at 13:00

Thanks and fixed here.

Comment #8097 by teovvv on February 06, 2023 at 11:18

To prove 1 implies 2 wouldn't it also be correct the following argument? $N=0 \iff 0 \rightarrow N \rightarrow 0$ exact $\iff 0 \rightarrow N\otimes M\rightarrow 0$ exact $\iff N\otimes M=0$ To me it feels very slightly cleaner. Or am I missing something?

Comment #8098 by Laurent Moret-Bailly on February 06, 2023 at 12:03

Typo one line 2 of proof: "imlpications".

Comment #8210 by Stacks Project on March 03, 2023 at 12:10

Dear teovw, i like the current argument too much but of course what you say is in some sense better... Dear Laurent, thanks. Change is here.

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

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