Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm studying some results about flatness and faithful flatness and I'd like to keep in my mind some examples about faithfully flat modules.

In general, free modules are the typical examples.
Another (unusual) example of faithfully flat module is the "Zariski Covering". (Let $R$ be a ring, $(f_1,\dots,f_n)=R$, and let $R_{f_i}$ be a localization $\forall i$. Then $S:=\bigoplus_{i=1}^n R_{f_i}$ is called Zariski covering.)

Do you have any other example?

share|cite|improve this question
Your definition of Zariski covering is missing the important condition that the $f_i$ generate the unit ideal. Regards, – Matt E May 27 '11 at 12:17
Indeed, if the $f_i$ do not generate the unit ideal, there exists a maximal ideal $m$ containing all the $f_i$. In this case, the extension of $m$ to $S$ is all of $S$ and $S$ cannot be a faithfully flat $R$-module. – Amitesh Datta Jun 1 '11 at 2:53

Formal properties
The tensor product of two faithfully flat modules is faithfully flat.
If $M$ is a faithfully flat module over the faithfully flat $A$-algebra $B$, then $M$ is faithfully flat over $A$ too.
An arbitrary direct sum of flat modules is faithfully flat as soon as at least one summand is.
(But the converse is false: see caveat below.)

An $A$-algebra $B$ is faithfully flat if and only if it is flat and every prime ideal of $A $ is contracted from $B$, i.e. $\operatorname{Spec}(B) \to\operatorname{Spec}(A)$ is surjective.
If $A\to B$ is a local morphism between local rings, then $B$ is flat over $A$ iff it is faithfully flat over $A$.

Caveat fidelis flatificator
a) Projective modules are flat, but needn't be faithfully flat. For example $A=\mathbb Z/6=(2)\oplus (3)$ shows that the ideal $(2)\subset A$ is projective, but is not faithfully flat because $(2)\otimes_A \mathbb Z/2=0$.

b) A ring of fractions $S^{-1}A$ is always flat over $A$ and never faithfully flat [unless you only invert invertible elements, in which case $S^{-1}A=A$].

c) The $\mathbb Z$-module $\oplus_{{{\frak p}}\in \operatorname{Spec}(\mathbb Z}) \mathbb Z_{{\frak p}}$ is faithfully flat over $\mathbb Z$ . All summands are flat, however none is faithfully flat.

share|cite|improve this answer
Thanks everyone. – user11428 May 31 '11 at 9:36
The fact, "$M$ flat over $A$ implies that the base change $M \otimes_A B$ is flat over $B$" would fit into the "Formal properties" section nicely, I think. – Dylan Moreland Jun 20 '12 at 19:18

If $A$ is a noetherian ring, $I\subseteq A$ is an ideal, and $\widehat{A}$ is the $I$-adic completion of $A$, if $I$ is contained in the Jacobson radical of $A$ (for example, if $A$ is local) then $A\to \widehat{A}$ is faithfully flat.

share|cite|improve this answer

Let $f:A\to B$ be a flat ring homomorphism. If $q$ is a prime ideal of $B$, then $p=f^{-1}(q)$ is a prime ideal of $A$. Furthermore, we have an induced homomorphism of rings $\overline{f}:A_p\to B_q$ such that the composition $A\to B\to B_q$ (the first arrow is $f$ and the second arrow is the localization homomorphism $B\to B_q$) is equal to $A\to A_p\to B_q$ (the first arrow is the localization homomorphism $A\to A_p$ and the second arrow is $\overline{f})$. (Of course, this follows from the universal property of localization; every element of $A$ not in $p$ is mapped to an element of $B$ not in $q$.)

In this situation, $\overline{f}:A_p\to B_q$ is a faithfully flat ring homomorphism. (Proof: We use the transitivity of flatness. In particular, $B_q$ is a local ring of $B_p$ and is therefore flat over $B_p$ and $B_p$ is flat over $A_p$ since $B_p\cong B\otimes_{A} A_p$ and flatness is preserved under base change. Faithful flatness follows since $pA_p$, the unique maximal ideal of $A_p$, is mapped into $qB_q$, a proper (in fact, maximal) ideal of $B_q$, under $\overline{f}$.)

In particular, the induced map on spectra $\overline{f}^{*}:\text{Spec}(B_q)\to \text{Spec}(A_p)$ is surjective.

share|cite|improve this answer

If $A$ is a coherent ring, any ring of power series $A[[X_1,\dots,X_n]]$ is faithfully flat over $A$.

Also, the direct sum of a flat $A$-module and a faithfully flat $A$-module is faithfully flat.

The direct sum $\bigoplus\limits_{\mathfrak m\in\operatorname{Max}A} A_{\mathfrak m}$ is faithfully flat over $A$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.