** A flat module is faithfully flat if and only if it has nonzero fibers. **

Lemma 10.38.15. Let $M$ be a flat $R$-module. The following are equivalent:

$M$ is faithfully flat,

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. Since $R \to \kappa ({\mathfrak p})$ is not zero we deduce that $M \to M \otimes _ R \kappa ({\mathfrak p})$ is not zero, see Lemma 10.38.14.

Conversely assume that $M$ is flat and that $M/{\mathfrak m}M$ is never zero. 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 $H = \mathop{\mathrm{Ker}}(N_2 \to N_3)/\mathop{\mathrm{Im}}(N_1 \to N_2)$. 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 (1)

Comment #892 by Kestutis Cesnavicius on

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