## Every quotient of a finitely generated module is finitely generated

Let $R$ be a ring with 1 and let $M$ be a left $R$-module. Suppose $M$ is generated by $n$ elements, say $\{a_i\}_{i=1}^n$. Show that if $N \subseteq M$ is a submodule then $M/N$ has a generating set containing at most $n$ elements. Conclude that every quotient of a cyclic module is cyclic.

We claim that $M/N = (a_i + N \ |\ i \in [1,n])$. The $(\supseteq)$ direction is clear. To see the $(\subseteq)$ direction, suppose $x + N \in M/N$ with $x = \sum r_i \cdot a_i$. Then $x + n = \sum r_i \cdot (a_i + N)$ as desired. Thus $\{a_i + N\}_{i=1}^n$ is an $R$-module generating set for $M/N$. (Which may contain redundant elements.)

In particular, If $M$ is cyclic, then it has a singleton generating set. Every quotient of $M$ is then generated by at most one element; since every module is generated by at least one element, every quotient of $M$ is also cyclic.