Exercise A0.III.6.13

Proposition

Let $R$ be a ring, and let $M$ be a finitely generated module. Then any homomorphic image of $M$ is finitely generated.

Proof

Since $M$ is finitely generated, there exists a surjective morphism $\tau : R^{\oplus n} \to M$. Take any morphism $\varphi : M \to N.$ Note that $\varphi$ induces (via canonical decomposition) a surjective morphism $\overline{\varphi} : M \to \im \varphi.$ This gives a surjective morphism $(\overline{\varphi} \circ \tau) : R^{\oplus n} \to \im \varphi,$ which shows that $\im \varphi$ is indeed a finitely generated module.

$\blacksquare$