If the $R$-module $M$ is finitely generated, but some generating set is not linearly independent, is $M$ not free?

Question

If an $R$-module $M$ is finitely generated, then there exists a finite set of generators. But there can exist infinitely many finite sets of generators.

So, if one of these finite sets of generators is not linearly independent, does this mean that the $R$-module is not free?

Is it not possible for some other finite generating set to form a basis?

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Edit: I guess I meant to say minimal finite generating set. If we have one minimal finite generating set that is not linearly independent, does this mean $M$ is not free?

Is it not possible for some other minimal finite generating set to form a basis?

Answer 1

Just because an $R$-module $M$ has a finite set of generators that is not linearly independent does not imply that $M$ is not free. For example, take any free module $M$ of finite rank with basis $B$ and add any element of $M$ to $B$ - the resulting set is again a finite set of generators, but no longer linearly independent.

Even when you require the generating set to be minimal, this is not true. Consider the set of generators $\{3, 2\}$ for $\Bbb Z$. This set is minimal because both $2$ and $3$ generate a proper $\Bbb Z$-submodule of $\Bbb Z$, however it is also not linearly independent because $3\cdot2-2\cdot3=0$.

Answer 2

R itself is a free R-module. Let $x\neq 0$. Then {1, x} generates R and is linearly dependent.