MathOverflow is a question and answer site for professional mathematicians. 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

Is it true that every projective module is faithfully flat, if not what is a counter example.


share|cite|improve this question

closed as too localized by Pete L. Clark, Hailong Dao, S. Carnahan Aug 8 '11 at 3:06

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

$0$ is a counterexample, but I think that you want to exclude it. Every finitely generated projective module is locally free of finite rank and thus faithfully flat (or trivial). In the general case, we may approximate our module by finitely generated projective modules, at least if the ground ring is Dedekind. – Martin Brandenburg Aug 6 '11 at 15:54
A flat (right) module is faithfully flat iff $R/mR \neq 0$ for all (left) maximal ideals of $R$. Now take $R = \mathbb Z/(6)$, $M=\mathbb Z/(2)$. – Hailong Dao Aug 6 '11 at 15:56
Should be $M/mM \neq 0$ above! – Hailong Dao Aug 6 '11 at 16:35
Thanks, I found myself similar example: $R=\mathbb{Z}_{35}$ Then $R=\mathbb{Z}_7\times\mathbb{Z}_5$, thus both $\mathbb{Z}_5$ and $\mathbb{Z}_7$ are projective $R$-modules and $\mathbb{Z}_7\otimes_R\mathbb{Z}_5=0$. The same can be doe with $\mathbb{Z}_6$ (and any $\mathbb{Z}_{pq}$ for p, q distinct prime number.s – Marcin Szamotulski Aug 7 '11 at 0:47
If you like algebraic geometry, you can consider vector bundles of non-constant rank over a disconnected space. – S. Carnahan Aug 8 '11 at 3:10
up vote 2 down vote accepted

Let $k$ be a field and consider the ring $k\times k$. There are two (indecomposable) projectives. Are they faithfully flat?

share|cite|improve this answer
Perhaps this was a rhetorical question, but no: for the same reason as in Hailong Dao's comment. The point is that when $Spec R$ is disconnected (in the commutative case), you can play this kind of game. So the answer in this case is that a projective is faithfully flat iff its support is all of $Spec R$. – Donu Arapura Aug 6 '11 at 18:49
@Donu: It is clearly (to me) rhetorical. When two professionals are talking to each other as equals, too much rhetoric has the potential to offend. For questions like this, one can ask (!) whether it is better to give a sort of Socratic answer as Mariano has or simply say directly that the question is not research-level and refer the user to math.SE (or elsewhere). In this case I would argue that Mariano's response is more directly helpful. – Pete L. Clark Aug 6 '11 at 20:57
Let me note that Marino answer is an example since it is a product of fields (though distinct). – Marcin Szamotulski Aug 7 '11 at 0:52
Donu: I know, I know :) – Mariano Suárez-Alvarez Aug 7 '11 at 1:45
Dear Mariano, I thought so. It looks like our work is done here. – Donu Arapura Aug 7 '11 at 10:31

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