The Stacks project

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

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

  1. $M$ is faithfully flat,

  2. for every nonzero $R$-module $N$, then tensor product $M \otimes _ R N$ is nonzero,

  3. for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$ the tensor product $M \otimes _ R \kappa (\mathfrak p)$ is nonzero, and

  4. 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

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

Comment #4585 by Bjorn Poonen on

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 -module , the tensor product is nonzero,

Then you can replace the first paragraph of the proof by

Proof. Assume faithfully flat and . By Lemma 10.39.14, the nonzero map induces a nonzero map , so .

Comment #8097 by teovvv on

To prove 1 implies 2 wouldn't it also be correct the following argument? exact exact To me it feels very slightly cleaner. Or am I missing something?

Comment #8098 by Laurent Moret-Bailly on

Typo one line 2 of proof: "imlpications".

Comment #8210 by on

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.

There are also:

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

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00HP. Beware of the difference between the letter 'O' and the digit '0'.