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



Characterization of projective modules in terms of Ext groups
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.


