Proposition 5

Let ρ : A → B be a local homomorphism of noetherian local rings, N a finitely generated B-module, x = (x_1, ..., x_r) a sequence of elements of m_B, and u : A[T_1, ..., T_r] → B the unique homomorphism of A-algebras such that u(T_i) = x_i for i = 1, ..., r. The following conditions are equivalent:

(i) the homomorphism u makes N into a flat A[T_1, ..., T_r]-module; (ii) the A-module N is flat and, for every A-module M, the sequence x is M ⊗_A N-regular; (iii) the A-module N is flat and the sequence x is κ_A ⊗_A N-regular; (iv) the A-module N/(x_1N + ... + x_rN) is flat and the sequence x is N-regular.

(i) ⇒ (ii): let us put T = (T_1, ..., T_r). Suppose that the A[T]-module N is flat. Since A[T] is flat over A, the A-module N is flat (I, § 2, No. 7, cor. 3 to prop. 8). Let M be an A-module. The sequence T is obviously M[T]-regular, hence M[T] ⊗_{A[T]} N-regular, since N is flat over A[T]. Now the A[T]-module M[T] ⊗_{A[T]} N is identified with M ⊗_A N, the homothety of ratio T_i corresponding to the endomorphism 1_M ⊗ (x_i)_N. Condition (ii) is therefore satisfied.

(ii) ⇒ (iii): this is trivial.

(iii) ⇒ (iv): this follows from proposition 10 of § 1, No. 6.

(iv) ⇒ (i): let t denote the ideal of A[T] generated by T. The (A[T]/t)-module N/tN is flat by assumption and N is ideally separated for t (III, § 5, No. 4, prop. 2). To prove that N is flat over A[T], it is enough to prove, in view of theorem 1 of loc. cit., No. 2, that the A-module Tor_1^{A[T]}(A[T]/t, N) is zero. But since the sequence T is A[T]-regular, this module is isomorphic to H_1(T, N) (A, X, p. 159, remark 3), which is zero since the sequence T is N-regular (loc. cit., p. 157, prop. 5).