Bourbaki

Proposition 10

Tag 0DXA, permanent. Cite it as [Bourbaki, Tag 0DXA].

Algebra, chapter X, § 5. Modules d’extensions

Let MM be an AA-module. The following conditions are equivalent:

(i) MM is projective. (ii) Ext⁡Ai(M,N)=0\operatorname{Ext}_A^i(M, N) = 0 for every AA-module NN and for every integer i>0i > 0. (iii) Ext⁡A1(M,N)=0\operatorname{Ext}_A^1(M, N) = 0 for every AA-module NN. (iv) There exists an exact sequence

0→K→νP→ιM→0, 0 \to K \xrightarrow{\nu} P \xrightarrow{\iota} M \to 0,

where PP is projective, and where Ext⁡A1(M,K)=0\operatorname{Ext}_A^1(M, K) = 0.

(i) ⇒\Rightarrow (ii) : this is the corollary of prop. 5 of X, p. 88. (ii) ⇒\Rightarrow (iii) : this is trivial. (iii) ⇒\Rightarrow (iv) : this is clear since MM is quotient of a free module PP. (iv) ⇒\Rightarrow (i) : since Ext⁡A1(M,K)=0\operatorname{Ext}_A^1(M, K) = 0, the canonical mapping

Hom⁡A(M,P)→Hom⁡A(M,M) \operatorname{Hom}_A(M, P) \to \operatorname{Hom}_A(M, M)

is surjective (X, p. 90, corollary); hence there exists an AA-linear section of ν\nu and MM is isomorphic to a direct factor of PP, and therefore is projective.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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