Exercise A0.III.6.18

Proposition

Let $M$ be an $R$-module and $N$ a submodule of $M.$ Suppose $N$ and $M/N$ are finitely generated. Then $M$ is finitely generated.

Proof

Let $\pi : M \to M/N$ be the quotient map.

Let $\{g_1,\ldots g_k\}$ be the generators of $N$ and $\{q_1,\ldots q_l\}$ the generators of $M/N.$

Take some set-theoretic section $\pi' : M/N \to M$ and let $q'_n = \pi'(q_n).$

Let $m \in M$ and let $\pi(m) = \sum r_j q_j.$ Let $x = \sum r_jq'_j.$ Since $\pi(x) = \pi(m),$ $x - m \in \ker \pi = N.$ That is, $x - m = \sum u_jg_j.$

So $m = x - \sum u_jg_j = \sum r_jq'_j - \sum u_jg_j,$ and $M$ is generated by $\{g_1,\ldots g_k\} \cup \{q'_1,\ldots q'_l\},$ a finite set.

$\blacksquare$