Bourbaki

Exercise 13, § 5

Commutative Algebra, II, 150

Bourbaki marks this one as harder than the rest.

Let AA be a (commutative) ring and BB an AA-algebra (not necessarily commutative). Suppose that the AA-module BB is faithful and generated by a finite number of elements bib_i (1≤i≤n1 \leq i \leq n); recall that B0B^0 denotes the opposite algebra of BB and that a canonical homomorphism is defined of the AA-algebra

B \otimes_A B^0 to the A-algebra End_A(B) mapping b⊗b′b \otimes b' to the endomorphism z↦bzb′z \mapsto bzb'.

(a) Show that the two following conditions are equivalent: (α) The bib_i form a basis of the A-module B and the canonical homomorphism B⊗AB0→EndA(B)B \otimes_A B^0 \to \mathrm{End}_A(B) is bijective. (β) The matrix U=(uij)U = (u_{ij}) such that uij=bjbiu_{ij} = b_j b_i is invertible in Mn(B)M_n(B).

(b) Suppose further that A is a local ring with maximal ideal mm; let βi\beta_i denote the canonical image of bib_i in B/mBB/mB. Show that conditions (α) and (β) are then equivalent to each of the conditions: (γ) The βi\beta_i form a basis of the A/m-module B/mBB/mB and B/mBB/mB is a simple central A/m-algebra (Algebra, Chapter VIII, § 5, no. 4). (δ) The matrix (βjβi)(\beta_j \beta_i) is invertible in Mn(B/mB)M_n(B/mB). (To prove that (δ) implies (β), use the fact that mBmB is contained in the Jacobson radical of B and Exercise 5 of Algebra, Chapter VIII, § 6).

Solution

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Transcribed from the printed volume. Not translated.