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

  • $\begingroup$ Isn't there somewhere a book with the exercises of Hartshorne solved? This manual is famous enough to have that thing as an attachment. $\endgroup$
    – Hvjurthuk
    Mar 6, 2021 at 15:57
  • 1
    $\begingroup$ I don't know actually $\endgroup$
    – John117
    Mar 7, 2021 at 16:19
  • $\begingroup$ What's the point of making a trivial editing? You have two good answers, aren't you satisfied with them? $\endgroup$
    – abx
    Mar 7, 2021 at 17:19
  • $\begingroup$ Yes of course, just to be more precise $\endgroup$
    – John117
    Mar 7, 2021 at 17:25

2 Answers 2


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.)

  • $\begingroup$ 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). $\endgroup$
    – John117
    Mar 7, 2021 at 17:45

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