2

We work in the category $R-\mathsf{Mod}$ where $R$ is unital.

Some authors define free modules to have a finite basis. If we don't require a basis to be finite, I think it is quite obvious that a finitely generated and free module (free thus means having some basis) has a finite basis and hence is isomorphic to $R^n$, for some $n \in \mathbb N$.

I don't see why a basis, if it exists, must be finite when the module is finitely generated.

  • How one can prove my assertion, that a free and f.g. module is isomorphic to $R^n$ for some $n$. –  Jun 24 '15 at 13:55

1 Answers1

1

It seems like your question is: prove that a finitely generated free module has a finite basis.

Let $E$ be a basis for such a module $M$ over $R$. Then $E$ spans $M$. Now, $M$ is finitely generated so you can find a finite ste of elements that span $M$. Write each element as (finite) linear combinations of elements from $E$. Collect all these elements from $E$ thus obtained. This will give you a finite basis for $M$.

Thomas
  • 43,555