$\operatorname{Ext}^1(M,R/m)=0$ iff $M$ is projective In this answer it is suggested that over a commutative local ring, a module $M$ is projective iff $\operatorname{Ext}^1_R(M,R/m)=0$. A similar result holds for flatness and Tor. 
In the case of Tor, I can find a proof in Robert Ash's Commutative Algebra, which involves specific isomorphisms with tensor. But I cannot prove the case of Ext. Can anyone show me the proof? Thank you.
FYI, the "proof" in the next answer in that link is quite vague to me, so it would be ok if you guys can clear things out here.
 A: This is answered in the question you link to. It is clear that if $M$ is projective then $\text{Ext}_R^1(M,k)$ vanishes where $k= R/m$ is the residue field of $R$. To show that if $\text{Ext}_R^1(M,k)$ vanishes then $M$ is in fact free, you use the fact
modules over local rings admit minimal free resolutions, i.e. there is
a resolution $F\to M$ such that for each $i\geqslant 0$ the $R$-module $F_i$ is free, and such that for each $i\geqslant 1$ it holds that $dF_i \subseteq mF_{i-1}$. Now the complex $\hom_R(F,k)$ has zero differential since the
resolution is minimal. Indeed, the differential acts by $df(x) = f(dx)$, but since $dx\in mF$ and $f$ is linear, $f(dx) \in mk = 0$. Then the homology at $\hom_R(F_1,k)$, which is  $\text{Ext}_R^1(M,k)$, is zero. This means $\hom_R(F_1,k) = 0$, so $F_1=0$, which gives an exact sequence $0\to F_0\to M\to 0$, showing that $M$ is free, as desired. 
One way to think about the proof is the following: first, one shows that over
a local ring $\hom_R(F,k) = 0 $ implies $F=0$ for $F$ free. Then, by dimension
shifting and minimality, one has that $\text{Ext}_R^1(M,k) =\hom_R(F_1,k)=0$, showing that $M$ is free if and only if such Ext group vanishes.  
