# Any $R$-module is a quotient of some free module. Do we need “finitely generated” condition?

I'm reading Lang's Algebra on the section of semisimple ring. There is one proposition that says

If $R$ is semi simple, then every $R$-module is semi-simple.

Here we assume $R$ is unital but not necessarily commutative. In the proof, we are using that "every $R$-module is a quotient of a free module. My question is: do we need the finitely generated condition here?

Recall in Atiyah-Macdonalds, Proposition 2.3, we had that every finitely generated $A$-module $M$ is a quotient of a free module of rank $n$ where $n$ is finite.

This leads me to think that why did Atiyah-Macdonalds only restricted to the finitely generated case in commutative rings? Would there be an example where this fails for non-f.g. modules? Or is it true that this holds for all modules, no matter f.g. or not?

• It holds for all modules, simply the free module in question will not have finite rank if the module is not finitely generated – Max May 5 '18 at 12:19

do not need finite generated.if $M$ is arbitrary $R$ module.$\forall x\in M$ there is an $R$ module morphism $f_x:R\rightarrow M$ which sent 1 to $x$.by definition of direct sum,there is a $R$ module morphism$\coprod_{f_x}:\coprod_{x\in M}R\rightarrow M$.It is clear this is epimorphism.