Proposition 4

Let S be a multiplicative subset of A and M an A-module. For m|s = 0 (m \in M, s \in S), it is necessary and sufficient that there exist s' \in S such that s'm = 0.

If s' \in S is such that s'm = 0, clearly m|s = (s'm)/(ss') = 0. Conversely, suppose that m|s = 0. As 1/s is invertible in S^{-1}A, m/1 = 0. For every sub-A-module P of S^{-1}A containing 1, we denote by \beta(P, m) the image of (m, 1) under the canonical mapping of M \times P to M \otimes_A P; then \beta(S^{-1}A, m) = 0. We know (Algebra, Chapter II, § 6, no. 3, Corollary 4 to Proposition 7) that there exists a finitely generated submodule P of S^{-1}A containing 1 and such that \beta(P, m) = 0. For all t \in S we denote by A_t the set of a/t, where a \in A; as P is finitely generated, there exists t \in S such that P \subset A_t (no. 1, Remark 2), whence \beta(A_t, m) = 0. The mapping a \mapsto a/t from A to A_t is surjective; let B be its kernel. It defines a surjective mapping h : M \otimes_A A \to M \otimes_A A_t, whose kernel is BM (M being identified with M \otimes A); then \beta(A_t, m) = h(tm) and consequently tm can be expressed in the form $\sum_{i=1}^r b_i m_i$, where b_i \in B, m_i \in M (1 \leq i \leq r). As b_i/t = 0 for 1 \leq i \leq r, there exists t' \in S such that t'b_i = 0 for 1 \leq i \leq r, whence t'tm = 0, which proves Proposition 4.