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.
Best Answer
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 $\sum_p e_p^*(x)e_p =x$