I just want this result for something that I am proving. I am proving if you have $f : M \rightarrow N$ be a surjective homomorphism then N has a finite rank. I showed that $N \cong M/ker(f)$, since $M/ker(f)$ is finitely generated then N is finitely generated. But I don't right away for general rings that finitely generate implies finite rank.
Recall here: Finitely generated module is defined as follows
M is finitely generated R-module iff there exists a generating set $\{x_i\}$ such that every $m = \Sigma r_ix_i$.
M has finite rank iff there exists a generating set $\{x_i\}$ that is a basis, i.e every m can be written uniquely as above.
Take a finitely generated free module $M$ over a ring $R$. The existence of a surjection $M \twoheadrightarrow N$ to an $R$-module $N$ does not imply that $N$ is free. That would imply that every finitely generated module over $R$ is free.
For example $\mathbb{Z}$ is a finitely generated free module, and the quotient map $\mathbb{Z} \rightarrow \mathbb{Z} / n \mathbb{Z}$ is a surjection, but $\mathbb{Z} / n \mathbb{Z}$ is not free.
However, a finitely generated free module $M$ over $R$ has finite rank. Suppose $M$ has basis $\{b_i \}_{i \in I}$ and let $\{ x_i \}_{1 \leq i \leq n}$ be a finite set of generators. Express each $x_i$ as $\sum_{j \in F_i} b_{j}$ for a finite set $F_i \subset B$. $\left\{ b_j | j \in \cup_{i = 1}^n F_i \right\}$ is finite, and consists of linearly independent elements. But it also spans $M$ since each $x_i$ is contained in the submodule of $M$ it generates. Thus $\left\{ b_j | j \in \cup_{i = 1}^n F_i \right\}$ testifies to the fact that $M$ has finite rank.