Lemma 2

Let $A$ be a noetherian ring, $M$ and $N$ finitely generated $A$-modules, $u : M \to N$ a homomorphism. Let $x$ be an element of the radical of $A$, such that the homothety $x_N$ is injective. If the homomorphism $\overline{u} : M/xM \to N/xN$ induced by $u$ is injective (resp. surjective, resp. bijective), the same is true of $u$.

The assertion concerning the surjectivity of $u$ results from Nakayama's lemma (II, § 3, No. 2, cor. 1 of prop. 4), without any assumption on $x_N$. Consider the commutative diagram with exact rows

by means of the snake lemma (I, § 1, No. 4, prop. 2), one deduces from it an exact sequence $\mathrm{Ker}\, u \xrightarrow{x} \mathrm{Ker}\, u \longrightarrow \mathrm{Ker}\, \overline{u}$. If $\overline{u}$ is injective, the homothety with ratio $x$ is surjective in $\mathrm{Ker}\, u$, which implies $\mathrm{Ker}\, u = 0$ by Nakayama's lemma.