## Finitely generated modules over R are precisely the module-homomorphic images of Rⁿ

Let $R$ be a ring. Prove that $M$ is a finitely generated module over $R$ if and only if $M$ is a module-homomorphic image of $R^n$.

Suppose first that $M$ is a finitely generated $R$-module – say by $A = \{a_1,\ldots,a_n\}$. Let $e_i$ denote the $i$th standard basis vector in $R^n$, and define $\varphi : R^n \rightarrow M$ by $e_i \mapsto a_i$, and extend linearly. Certainly then $\varphi$ is a surjective module homomorphism.

Conversely, suppose $\varphi : R^n \rightarrow M$ is a surjective module homomorphism, and for each $e_i$ let $a_i = \varphi(e_i)$. Since $\varphi$ is surjective, and since the $e_i$ generate $R^n$, the set $A = \{a_1, \ldots, a_n\}$ generates $M$ over $R$ as desired.