Proposition 2

*Let $A$ be a ring, $u : M \to N$ an $A$-module homomorphism and $p$ a prime ideal of $A$.

(i) Suppose that $u_p : M_p \to N_p$ is surjective and that $N$ is finitely generated. Then there exists $f \in A - p$ such that $u_f : M \to N$, is surjective.

(ii) Suppose that $u_p$ is bijective, that $M$ is finitely generated and that $N$ is finitely presented. Then there exists $f \in A - p$ such that $u_f$ is bijective.*

Let $R$ and $Q$ be the kernel and cokernel of $u$; if $g \in A$, the kernel and cokernel of $u_g$ (resp. $u_p$) are $R_g$ and $Q_g$, (resp. $R_p$ and $Q_p$) (§ 2, no. 4, Theorem 1). Then $Q_p = 0$; as $N$ is finitely generated, so is $Q$ and there exists $g' \in A - p$ such that $g'Q = 0$ (§ 2, no. 2, Corollary 2 to Proposition 4), whence $Q_{g'} = 0$. Under the hypotheses of (ii), the sequence $0 \to R_{g'} \to M \to N \to 0$ is exact, hence $R_{g'}$ is finitely generated (Chapter I, § 2, no. 8, Lemma 9). Now,

hence there exists $g_1 \in A, - pA_{g'}$ such that $g_1 R_{g'} = 0$ (§ 2, no. 2, Corollary 2 to Proposition 4). Then $g_1 = g''/{g'}^h$, where $g'' \in A - p$; as $g'/1$ is invertible in $R_{g'}$, $(g''/1)R_{g'} = 0$, whence $R_{g'g''} = (R_{g'})_{g''/1} = 0$. If $f = g'g''$, $f \in A - p$, $Q_f = 0$ and $R_f = 0$, so that $u_f$ is bijective.