- #1

Kreizhn

- 743

- 1

My question is, is this statement redundant? It seems to me that all finitely generated R-modules are necessarily free as R-modules. In particular, if M is an finitely generated R-module with minimal generating set [itex] A \subseteq M [/itex], then isn't the free R-module on A also M? Or am I missing a technical point?