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

Thanks!

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 $0$ 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 2011 at 15:54
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 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 2011 at 15:56
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 Should be $M/mM \neq 0$ above! – Hailong Dao Aug 6 2011 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 2011 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 2011 at 3:10

closed as too localized by Pete L. Clark, Hailong Dao, S. Carnahan♦ Aug 8 2011 at 3:06

1 Answer

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

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 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 2011 at 18:49
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 @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 2011 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 2011 at 0:52
Donu: I know, I know :) – Mariano Suárez-Alvarez Aug 7 2011 at 1:45
Dear Mariano, I thought so. It looks like our work is done here. – Donu Arapura Aug 7 2011 at 10:31

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