Bourbaki

Example

Tag 0D3Y, permanent. Cite it as [Bourbaki, Tag 0D3Y].

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

Let J be an ideal of A generated by an A-regular sequence x=(x1,…,xr)x = (x_1, \ldots, x_r). One has dp_A(A/J) ≤ r. In fact, this is clear if A is zero; in the contrary case, the Koszul complex K∙(x,A)K_\bullet(x, A) is a free resolution of A/J of length r (loc. cit., p. 159, Remark 3). Moreover, for every A-module N, the A-modules Ext_A^r(A/J, N) and N/JN are isomorphic (loc. cit.); consequently, in order that one have dp_A(A/J) = r, it is necessary and sufficient that J be distinct from A (loc. cit., p. 134, Prop. 1).

One defines analogously the injective dimension of M, denoted by di_A(M), as the greatest lower bound of the set of lengths of injective resolutions of M. One has di_A(0) = -∞, and di_A(M) ≥ 0 if M ≠ 0. In order that M be injective, it is necessary and sufficient that one have di_A(M) ≤ 0.

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

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