Bourbaki

Proposition 3

Tag 0E2M, permanent. Cite it as [Bourbaki, Tag 0E2M].

Algebra, chapter X, § 9. Complexes de Koszul

If the A-module L is flat, the sequences (En)(\mathscr{E}_n) above are exact for n>0n > 0.

a) Let us first note that, if pLp_L is the composite homomorphism

S(L)⊗Λ(L)→αS0(L)⊗Λ0(L)→βA, \mathbf{S}(L) \otimes \Lambda(L) \xrightarrow{\alpha} \mathbf{S}^0(L) \otimes \Lambda^0(L) \xrightarrow{\beta} A,

where α\alpha is the tensor product of the canonical projections and β\beta the canonical isomorphism, it remains to prove that H(pL)H(p_L) is bijective.

b) If L=0L = 0 or if L=AL = A, the proposition is obvious.

c) Suppose L is free of finite rank; write it as the direct sum L1⊕...⊕LnL_1 \oplus ... \oplus L_n of free A-modules of rank 1. According to the remark preceding the proposition, the complex S(L)⊗Λ(L)\mathbf{S}(L) \otimes \Lambda(L) is isomorphic to the tensor product of the nn free complexes S(Li)⊗Λ(Li)\mathbf{S}(L_i) \otimes \Lambda(L_i) whose homology is free according to b). According to X, p. 79, cor. 4, the canonical homomorphism

γ:⨁i=1nH(S(Li)⊗Λ(Li))→H(S(L)⊗Λ(L)) \gamma : \bigoplus_{i=1}^n H(\mathbf{S}(L_i) \otimes \Lambda(L_i)) \to H(\mathbf{S}(L) \otimes \Lambda(L))

is bijective. According to b) H(pLi)H(p_{L_i}) is bijective for every ii. Since ⨂i=1nH(pLi)=H(pL)∘γ\bigotimes_{i=1}^n H(p_{L_i}) = H(p_L) \circ \gamma,

H(pL) is bijective. H(p_L) \text{ is bijective}.

d) In the general case, L is an inductive limit of a filtered inductive system (Li)i∈I(L_i)_{i \in I} of free modules of finite rank (X, p. 14, th. 1). Since the canonical bijective homomorphism

lim←⁡S(Li)⊗Λ(Li)→S(L)⊗Λ(L) \varprojlim \mathbf{S}(L_i) \otimes \Lambda(L_i) \to \mathbf{S}(L) \otimes \Lambda(L)

is an isomorphism of complexes, the proposition follows from X, p. 28, prop. 1.

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

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