# Projectivity and faithfully flatness (module theory) [closed]

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

Thanks!

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

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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 '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

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