# 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

• Isn't there somewhere a book with the exercises of Hartshorne solved? This manual is famous enough to have that thing as an attachment. – Hvjurthuk Mar 6 at 15:57
• I don't know actually – John117 Mar 7 at 16:19
• What's the point of making a trivial editing? You have two good answers, aren't you satisfied with them? – abx Mar 7 at 17:19
• Yes of course, just to be more precise – John117 Mar 7 at 17:25

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.

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.)

• Thank you very much for your very beatiful answer. I accept the SWS answer just because is more appropriate to my question (he has solved the exercise using the hint from Hartshorne). – John117 Mar 7 at 17:45