Bourbaki

Proposition 1

Tag 0D3Z, permanent. Cite it as [Bourbaki, Tag 0D3Z].

Commutative Algebra, chapter X, § 3. Profondeur et dimension homologique

Let A be a ring, M an A-module and n an integer ≥ 0. The following conditions are equivalent:

(i) one has di_A(M) ≤ n (in other words, M possesses an injective resolution of length ≤ n) ; (ii) for every A-module N and every integer i > n, one has Ext_A^i(N, M) = 0 ; (iii) for every ideal a of A, one has Ext_A^{n+1}(A/a, M) = 0 ; (iv) for every exact sequence of A-modules

0→M→I0→I1→…→In−1→Q→0, 0 \to M \to I^0 \to I^1 \to \ldots \to I^{n-1} \to Q \to 0,

where the IiI^i are injective, the A-module Q is injective.

(i) ⇒ (ii) : this follows from A, X, p. 100, Th. 1. (ii) ⇒ (iii) : this is clear. (iii) ⇒ (iv) : in the situation of (iv), for every A-module N one has an isomorphism of Ext_A^1(N, Q) onto Ext_A^{n+1}(N, M) (A, X, p. 128, Cor. 4) ; under assumption (iii), the A-module Ext_A^1(A/a, Q) is zero for every ideal a of A, and Q is injective (A, X, p. 93, Prop. 11). (iv) ⇒ (i) : consider the exact sequence (A, X, p. 52)

0→M→I0(M)→…→In−1(M)→Kn−1(M)→0; 0 \to M \to I^0(M) \to \ldots \to I^{n-1}(M) \to K^{n-1}(M) \to 0;

if condition (iv) is satisfied, the A-module Kn−1(M)K^{n-1}(M) is injective, whence (i).

Recall (A, X, p. 138, Def. 2) that the homological dimension of the ring AA, denoted by dh(A)\mathrm{dh}(A), is the least upper bound in Z‾\overline{\mathbf{Z}} of the set of integers nn for which there exist two AA-modules MM and NN such that ExtAn(M,N)\mathrm{Ext}^n_A(M, N) is nonzero. It is therefore also the least upper bound of the set of projective dimensions (or injective dimensions) of all the AA-modules; when AA is noetherian, it is enough to restrict oneself to finitely generated AA-modules (A, X, p. 139, Cor.).

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

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