Bourbaki

Theorem 1 (Auslander-Buchsbaum)

Tag 0D4I, permanent. Cite it as [Bourbaki, Tag 0D4I].

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

Let AA be a noetherian local ring and MM a finitely generated AA-module of finite projective dimension. Then one has the equality

dpA(M)+profA(M)=prof(A) \mathrm{dp}_A(M) + \mathrm{prof}_A(M) = \mathrm{prof}(A)

Let us reason by induction on dpA(M)\mathrm{dp}_A(M).

a) If dpA(M)\mathrm{dp}_A(M) is zero, MM is a nonzero finitely generated free AA-module; its depth is equal to prof(A)\mathrm{prof}(A) (§ 1, No. 1, remark 4).

b) Suppose dpA(M)=1\mathrm{dp}_A(M) = 1 and choose a minimal projective resolution

0→L1→d1L0→d0M→0 0 \to L_1 \xrightarrow{d_1} L_0 \xrightarrow{d_0} M \to 0

of MM (No. 3, prop. 4, (iv)). The AA-modules L0L_0 and L1L_1 are nonzero finitely generated free modules, hence of depth prof(A)\mathrm{prof}(A) (§ 1, No. 1, remark 4). The mapping 1κA⊗d1:κA⊗AL1→κA⊗AL01_{\kappa_A} \otimes d_1 : \kappa_A \otimes_A L_1 \to \kappa_A \otimes_A L_0 is zero, which implies that d1d_1 belongs to mAHomA(L1,L0)m_A \mathrm{Hom}_A(L_1, L_0). By remark 5 of § 1, No. 1, one has profA(M)=prof(A)−1\mathrm{prof}_A(M) = \mathrm{prof}(A) - 1.

c) Suppose dpA(M)>1\mathrm{dp}_A(M) > 1. Choose an exact sequence

0→N→L→M→0 0 \to N \to L \to M \to 0

where LL is a finitely generated free AA-module. One then has profA(L)=prof(A)\mathrm{prof}_A(L) = \mathrm{prof}(A) (§ 1, No. 1, remark 4), dpA(N)=dpA(M)−1\mathrm{dp}_A(N) = \mathrm{dp}_A(M) - 1 (A, X, p. 135, cor. 2 c)), whence profA(N)=prof(A)−dpA(N)\mathrm{prof}_A(N) = \mathrm{prof}(A) - \mathrm{dp}_A(N) (induction hypothesis), and in particular profA(N)<profA(L)\mathrm{prof}_A(N) < \mathrm{prof}_A(L). By prop. 1 of § 1, No. 1, one then has profA(M)=profA(N)−1\mathrm{prof}_A(M) = \mathrm{prof}_A(N) - 1, which completes the proof.

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

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