I am familiar with the statement: A spanning list of the "right length" in a finite dimensional vector space $V$ is a basis.
I am wondering if this is also true for arbitrary free $R$-modules of finite rank. I am not sure because the proof that I have read on this in vector spaces does rely on a lemma that uses the hypothesis that the ring of scalars is a field.
Thanks
This is true over commutative rings. See Surjective endomorphisms of finitely generated modules are isomorphisms (a spanning list for $R^n$ of length $n$ gives a surjective homomorphism $R^n\to R^n$, which then must be an isomorphism and therefore the list is a basis).
It is not true for arbitrary rings. For instance, if $R$ is the ring of endomorphisms of an infinite-dimensional vector space $V$, then $R\cong R^2$ as an $R$-module (using an isomorphism $V\cong V^2$). So, $R^2$ can be generated by a single element $x$, and then for any $y\in R^2$, $(x,y)$ is a spanning list that is not a basis.
In general, a ring for which this property holds is called stably finite. See this answer for some discussion and related conditions.
