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 ith 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.

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