Welcome to our community

Be a part of something great, join today!

Free and Finitely Generated Modules


Well-known member
MHB Math Helper
Feb 5, 2012
Hi everyone, :)

Want to confirm my understanding about Free and Finitely Generated modules. I want to know whether the following ideas are correct. Thank you for all your help. :)

1) Is every free module a finitely generated module?

No. Because a free module may have an infinite basis. So we cannot say it's finitely generated. However if the free module has a finite basis it's finitely generated.

2) Is every finitely generated module a free module?

No again. If \(M\) is a \(R\)-module which is finitely generated by a set \(S=\{x_1,\,x_2,\,\cdots,\,x_n\}\subset M\) then for each element \(x\in M\) we have,

\[x=r_1 x_1+\cdots+r_n x_n\]

where \(r_1,\cdots,r_n\in R\).

However we don't know whether \(S\) is linearly independent. So generally \(M\) is not free.

Am I correct? :)


Well-known member
MHB Math Scholar
Feb 15, 2012
That looks correct.

An easy counter-example for (2) is given by the $\Bbb Z$-module $\Bbb Z_n$.

Clearly, we have $\{1\}$ as a generating set, but this is not a basis because:

$n.1 = 0$ but $n \neq 0$.

Something for you to think about:

Suppose $R$ is a principal ideal domain, and that $M$ is a finitely-generated $R$-module.

Let $T = \{m \in M: \exists r \in R^{\ast}\text{ with }r.m = 0\}$.

Is $M/T$ free? (This is a subtler question than might appear at first. Ask yourself: why do we insist $R$ be a PID?).