I am reading this paper.
It is proven there that if $f:A\rightarrow B$ is a faithfully flat morphism of rings and $M$ an $A$-module such that the $B$-module $M\otimes_A B$ is projective, then $M$ is projective.
The proof there is long a complicated. Is there a simpler one? I mean to prove $M$ projective one needs to prove $\mathrm{Hom}_A(M,-)$ exact. But this is functorially isomorphic to $\mathrm{Hom}_B(M\otimes_A B,-)$!
Also is it true that if $M\otimes_A B$ is free, then so is $M$?
For reasonable rings, one way to do this would be to use the fact that a module is projective if and only if it is locally so and thus we may assume $A$ is local with residue field $k$. Now, $M$ is projective if and only if $\operatorname{Tor}^1_A(k,M)=0$. Then one has $0=\operatorname{Tor}^1_B(k\otimes_A B, M\otimes_A B) =\operatorname{Tor}^1_A(k,M)\otimes_A B$, since $M\otimes_A B$ is $B$ projective. Now faithful flatness proves $\operatorname{Tor}^1_A(k,M)=0$.
The last part is false in general. For example, take $A=\mathbb{R}[x,y,z]/(x^2+y^2+z^2-1)$, the coordinate ring of the real sphere and let $B=A\otimes_{\mathbb{R}}\mathbb{C}$. If $P$ is the tangent bundle of the real sphere, it is not free over $A$, but $P\otimes_A B$ is free. Of course, $A\to B$ is faithfully flat.