Exercise 5, § 4

Algebra, X, p. 185

Suppose that there exists a homomorphism (unital) of $k$ -algebras π : A → k, which gives k a structure of A-module. Consider the standard resolution B(A, k) (X, p. 58); for n ≥ 0, identify B_n(A, k) with A ⊗_k A^⊗n and denote by a[a_1, ..., a_n] the element a ⊗ a_1 ⊗ ... ⊗ a_n of A ⊗_k A^⊗n. (a) Prove that the complex B(A, k) ⊗_k B(A, k) defines a resolution of the (A ⊗_k A)-module k. Prove that the (A ⊗_k A)-linear homomorphism

g : B(A \otimes_k A, k) \to B(A, k) \otimes_k B(A, k)

such that

g([x_1 \otimes y_1, ..., x_n \otimes y_n]) = \sum_{0 \leq p \leq n} \pi(x_{p+1} ... x_n) [x_1, ..., x_p] \otimes y_1 ... y_p[y_{p+1}, ..., y_n]

for x_1, ..., x_n, y_1, ..., y_n in A, is a morphism of resolutions. (b) Suppose that A is commutative. Prove that there exists on B(A, k) a structure of graded A-algebra such that one has for x_1, ..., x_n ∈ A:

[x_1, ..., x_p] [x_{p+1}, ..., x_n] = \sum_{\sigma \in M_{p,n}} \varepsilon(\sigma) [x_{\sigma(1)}, ..., x_{\sigma(n)}]

where M_{p,n} is the subset of S_n formed by the permutations σ such that σ(1) < ... < σ(p) and σ(p + 1) < ... < σ(n). Prove that B(A, k) is then a differential graded A-algebra (X, p. 183, exercise 18).

Solution

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