Bourbaki

Proposition 4

Tag 0E0Z, permanent. Cite it as [Bourbaki, Tag 0E0Z].

Algebra, chapter X, § 8. Dimension homologique

Let nn be an integer ≥0\geq 0. The following conditions are equivalent:

(i) dh⁡(A)≤n\operatorname{dh}(A)\leq n,

(ii) for every AA-module M, we have dpA(M)≤n\mathrm{dp}_A(M)\leq n,

(ii′) for every finitely generated AA-module M, we have dpA(M)≤n\mathrm{dp}_A(M)\leq n,

(iii) for every exact sequence

0→K→Pn−1→Pn−2→…→P0 0 \to K \to P_{n-1} \to P_{n-2} \to \ldots \to P_0

where the PiP_i are projective, KK is projective,

(iv) for every exact sequence

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

where the IiI^i are injective, NN is injective,

(v) every AA-module possesses an injective resolution of length ≤n\leq n.

The equivalence of conditions (i), (ii) and (iii) follows from prop. 1. We have obviously (ii) ⇒\Rightarrow (ii’). It is therefore enough to proving (ii’) ⇒\Rightarrow (iv) ⇒\Rightarrow (v) ⇒\Rightarrow (i).

(ii’) ⇒\Rightarrow (iv): with the notation of (iv), let KK be the kernel of I0→I1I^0 \to I^1. By X, p. 128, cor. 4, we have for every AA-module MM an isomorphism ExtA1(M,N)→ExtAn+1(M,K)\mathrm{Ext}_A^1(M, N) \to \mathrm{Ext}_A^{n+1}(M, K); it then follows from (ii’) that ExtA1(M,N)\mathrm{Ext}_A^1(M, N) is zero for every finitely generated AA-module MM. By X, p. 93, prop. 11, this implies that NN is injective, whence (iv).

(iv) ⇒\Rightarrow (v): let MM be an AA-module. Applying (iv) to the exact sequence

0→M→I0(M)→I1(M)→…→In−1(M)→Kn−1(M)→0 0 \to M \to I^0(M) \to I^1(M) \to \ldots \to I^{n-1}(M) \to K^{n-1}(M) \to 0

of X, p. 52, we conclude that Kn−1(M)K^{n-1}(M) is injective, whence (v).

(v) ⇒\Rightarrow (i): this follows from X, p. 100, th. 1.

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

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