Cyclic faithfully flat modules I am looking for an example of a cyclic faithfully flat $R$-module that is not projective. Could someone help me?
 A: Over a von Neumann regular ring, every (left or right) module is flat.  Furthermore, a finitely generated projective module is necessarily finitely presented.  
Using these two facts, I claim that it suffices to produce a von Neumann regular ring $R$ with a left ideal $I$ such that:


*

*$I$ is not finitely generated, and

*$R/I$ contains a submodule isomorphic to $R$.


Indeed, condition 1 (along with Schanuel's Lemma) implies that $R/I$ is not finitely presented.  And the flat module $R/I$ will be faithfully flat thanks to condition 2 because the inclusion $R \hookrightarrow R/I$ remains injective after tensoring with an arbitrary right $R$-module $M$ (by flatness of $M$), so that $M \cong M \otimes_R R \hookrightarrow M \otimes_R (R/I)$ as abelian groups.
To produce an example of such $R$ and $I$ above, we can use one of the more famous pathological examples of a von Neumann regular ring: the ring $R = \operatorname{End}_k(V)$ of all linear endomorphisms of an infinite-dimensional $k$-vector $V$ space over your favorite field $k$.  It's well known that $R$ fails to have the property of invariant basis number because $R \cong R \oplus R$ as left $R$-modules.  Using this, we may inductively write left module decompositions $$R = R_0 \oplus S_1 = R_0 \oplus R_1 \oplus S_1 = R_0 \oplus R_1 \oplus R_2 \oplus S_2 = \cdots$$ where each $R_i$ and $S_i$ is isomorphic to $R$. 
In this way we obtain an infinite direct sum $I = \bigoplus_{i=1}^\infty R_i \subseteq R$ that is clearly an infinitely generated left ideal. And because $R_0 \oplus I \subseteq R$, we obtain $R \cong R_0 \hookrightarrow R/I$ as left $R$-modules. Thus conditions 1 and 2 are satisfied.
