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Let $R$ be a commutative ring with unity and let $M$ be a projective and faithful $R$-module. Then is $M$ faithfully flat ? Is it true at least if $M$ is finitely generated, or say Noetherian ?

I know that I only have to show that $M\otimes_R N\ne 0$ for every non-zero $R$-module $N$. Now if $M$ is finitely generated, then by faithful ness of $M$, I can show that $M\otimes_R N\ne 0$ for every non-zero , finitely generated $R$-module $N$, because for finitely generated $R$-modules $M$ and $N$, $ \operatorname{Supp}(M \otimes_R N)=V( \operatorname{Ann}_R(M) + \operatorname{Ann}_R(N))$.

I am unable to proceed further.

Please help.

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    $\begingroup$ As Pedro's beautiful answer shows, this is false in general. However, I believe it does hold if we add the extra condition that $r_{M}(P) > 0$ for all $P \in Spec(R)$. (Recall that every projective module over a commutative local ring is free; here, $r_{M}(P)$ denotes the rank of $M_{P}$.) Indeed, suppose this is the case, and let $N$ be a nonzero $R$-module. Then for any $P \in Spec(R)$, $(M \otimes_{R} N)_{P} \cong M_{P} \otimes_{R_{P}} N_{P} \cong N_{P}^{r_{M}(P)}$. Since $r_{M}(P) > 0$ for all $P$ and $N_{P}$ must be nonzero for some $P$, it follows that $(M \otimes_{R} N)_{P}$ must be... $\endgroup$ – Alex Wertheim Aug 3 '18 at 19:37
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    $\begingroup$ ...nonzero for this same $P$, whence $M \otimes_{R} N$ is nonzero. I don't know how strong or artificial the requirement that $r_{M}(P) > 0$ for all $P \in Spec(R)$ is, however. $\endgroup$ – Alex Wertheim Aug 3 '18 at 19:39
  • $\begingroup$ @AlexWertheim Yes, I had that proof in mind too! Well, if $R$ is connected then $r_M(P)$ is constant, for example. Of course in Bourbaki's counterexample this fails miserably. :) $\endgroup$ – Pedro Tamaroff Aug 3 '18 at 19:46
  • $\begingroup$ See this. $\endgroup$ – Pedro Tamaroff Aug 3 '18 at 19:57
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According to Bourbaki, Commutative Algebra, this is false for general rings (page 29, Example (2)). See here.

The counterexample is the following: take the direct product $A$ of the rings $\mathbb Z/p$ over all primes, and consider the ideal $I$ obtained from taking the direct sum. Then $I$ is projective and faithful, but not faithfully flat.

To see that $I$ is not faithfully flat, recall that a module $M$ is faithfully flat iff it is flat and $JM\neq M$ for every (maximal) ideal. But note that $I^2=I$, so it follows that $I$ is not faithfully flat.

On the other hand, if we take $e_p\in I$ the $p$-th coordinate element, we see that for $a\in A$, $ae_p = a_p$, so if $a$ annihilates $M$ we deduce that $a=0$, and $M$ is faithful.

Finally, $I$ is projective: I will exhibit a dual basis for it. Consider the coordinate elements $e_p$ and the $A$-linear maps $e_p^*$ that are the composition of the projection to the $p$th component with the inclusion $\mathbb Z/p\to A$. It is immediate from the definitions that if $x\in I$, then $\sum_p e_p^*(x)e_p =x$

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