Bourbaki

Exercise 2, § 9

Algebra, X, p. 206

Let AA be a ring, u:L→Mu : L \to M an homomorphism of AA-modules.

(a) We denote by ε\varepsilon the commutation factor on Z2\mathbf{Z}^2 defined by ε(α,β)=α2β2\varepsilon(\alpha, \beta) = \alpha_2 \beta_2 for α,β∈Z2,α=(α1,α2),β=(β1,β2)\alpha, \beta \in \mathbf{Z}^2, \alpha = (\alpha_1, \alpha_2), \beta = (\beta_1, \beta_2). Now prove that there exists a unique (A, ε\varepsilon)-derivation (III, p. 118) dd of the bigraded algebra S(M)⊗AΛ(L)S(M) \otimes_A \Lambda(L), of degree (1,−1)(1, -1), such that one has d(1⊗x)=u(x)⊗1d(1 \otimes x) = u(x) \otimes 1 for x∈Lx \in L. Now prove that d∘d=0d \circ d = 0, so that S(M)⊗AΛ(L)S(M) \otimes_A \Lambda(L), endowed with the graduation deduced from that of Λ(L)\Lambda(L) and the differential dd, defines a complex K(u)K(u).

(b) If M=LM = L and u=1dLu = 1_{d_L}, now prove that K(u)K(u) is equal to the complex defined in No. 3, p. 151.

(c) Let B=SA(M)B = S_A(M), and let u‾:B⊗AL→B\overline{u} : B \otimes_A L \to B be the BB-homomorphism deduced from uu. Now prove that K(u)K(u) identifies canonically with the complex KB(u‾)K^B(\overline{u}).

(d) Let u′:L′→M′u' : L' \to M' be a second AA-homomorphism. Now prove that the complexes K(u⊕u′)K(u \oplus u') and K(u)⊗AK(u′)K(u) \otimes_A K(u') are isomorphic.

(e) Let f:L→L′f : L \to L' and g:M→M′g : M \to M' be two AA-homomorphisms such that g∘u=u′∘fg \circ u = u' \circ f. Now prove that the homomorphism S(g)⊗Λ(f)S(g) \otimes \Lambda(f) is a morphism of complexes from K(u)K(u) into K(u′)K(u'). In particular, if v:M→Pv : M \to P is an AA-homomorphism such that v∘u=0v \circ u = 0, one obtains a morphism of complexes h:K(u)→SA(P)h : K(u) \to S_A(P).

(f) Suppose that the sequence 0→L→uM→vP→00 \to L \xrightarrow{u} M \xrightarrow{v} P \to 0 is an exact sequence of flat AA-modules. Now prove that hh is an isomorphism (reduce to the case where L,M,PL, M, P are free of finite type, and use dd). Deduce for every n≥1n \geq 1 an exact sequence

0→ΛnL→M⊗AΛn−1L→⋯→Sn−1M⊗AL→SnM→SnP→0. 0 \to \Lambda^n L \to M \otimes_A \Lambda^{n-1} L \to \cdots \to S^{n-1} M \otimes_A L \to S^n M \to S^n P \to 0 .
Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.