Characterization of projective modules in terms of Ext groups This is from Hartshrone exercise 6.6 part (a).
Let $A$ be a regular local ring and $M$ be a finitely generated $A$-module, prove the following

$M$ is projective $\iff$ $\operatorname{Ext}^{i}(M,A)=\{0\}$ for all $i>0$

The hint is to use the following
Proposition (6.11 A)
If $A$ is a regular local ring, then
(1) for every $M$, pd$(M)\le \dim(A)$  where pd(M) is the projective dimension and dim(A) is the Krull dimension
(2) If $K=A/m$ then $\operatorname{pd}(K)=\dim(A)$
and to use the descending induction to prove that $\operatorname{Ext}^i(M,N)=\{0\}$ for all $i>0$ and all finitely generated $A$-module $N$. Then finally show that $M$ is a direct summand of a free module.
I really don't know how to put together all this information. Any help is appreciated
 A: If $M$ is projective, then $\mathrm{Ext}^i(M,-) = 0$ for every $i>0$; this is because $\mathrm{Ext}$ can be computed by taking a projective resolution of the first argument.
For the converse, we prove a more general result: Let $R$ be a noetherian local ring and $M$ a finitely generated $R$-module with finite projective dimension. (The first hint guarantees this.) Suppose that $M$ is not projective. Then $\mathrm{pd}(M) > 0$, so a minimal projective resolution of $M$ looks like
$$
0 \to F_p \to F_{p-1} \to \cdots \to F_0 \to 0
$$
(Here $p  = \mathrm{pd}(M)$ and $p>0$.) Since this is a minimal resolution, the entries in the matrix describing the map $F_p \to F_{p-1}$ is inside the maximal ideal. Hence
$\mathrm{Ext}^p(M,R)$ is the cokernel of the dual of this map, which is nonzero, by Nakayama lemma. (The entries in the (dual) map are in the maximal ideal, so it cannot be surjective.)
A: Along the lines of Hartshorne:
by (1) for all finitely generated $\mathrm{N}$ we have $\mathrm{Ext^i(M,N)}=0$  ($i>\mathrm{dim(A)}$).
Since $\mathrm{N}$ is finitely generated, we may find an exact sequence of the form
$$0\rightarrow\mathrm{K}\rightarrow\mathrm{A}^{\oplus r}\rightarrow\mathrm{N}\rightarrow 0.$$
Taking the $\mathrm{Ext^i(M,-)}$ long exact sequence and using the vanishing $\mathrm{Ext^i(M,A)}=0$ shows that $\mathrm{Ext^{i+1}(M,-)}=0$  implies $\mathrm{Ext^{i}(M,-)}=0$, as required by descending induction.
For the last claim consider an exact sequence of the above form for $\mathrm{M}$ instead of $\mathrm{N}$; the condition $\mathrm{Ext^{1}(M,K)}=0$ means that it must be split.
