Change of basis in general free modules

Question

Suppose $R$ is a unital ring that is not necessarily with the IBN property, and $M$ a nontivial free left $R$-module with a basis $B$. It seems to me the following proposition is correct, which makes free modules similar to vector spaces over fields.

Proposition. Let $B'\subset M$ be a $R$-linearly-independent subset such that $|B'|=|B|$. Then $B'$ is another basis of $M$.

Proof. Just left $f: B\to B'$ be a bijection, and linearly extend it to $\hat{f}: M\to \langle B'\rangle\subset M$. By the $R$-linear-independence of both $B$ and $B'$, it is clear $\hat{f}$ is an isomorphism. Hence $M=\langle B'\rangle$. $\square$

Question: However, if we replace $b\in B$ with any $rb$, where $r\in R\setminus \{0\}$, it's not guaranteed $B$ remains a basis, unless $R$ is a division ring. And This has nothing to do with IBN. Correct?

Thanks in advance. I'm just trying to understand how free modules behave differently to vector spaces.

Update: I can see why the "proof" above is wrong now: it is correct $\hat{f}: M\to \langle B'\rangle$ is a left $R$-module isomorphism, but nothing prevents $\langle B'\rangle \subsetneq M$. So we can't expect to transform a known basis to a new basis.

Wikipedia gives the counterexample of a ring $S$ such that $S^n\cong S^m$ for any two positive integers $m,n.$ For instance, $S^2\cong S,$ so that $S$ is a free $S$-module with both a basis $B$ s.t. $|B|=1$ and a basis $C={u,v}$ s.t. $|C|=2.$ Now, $B':={u}$ is a counterexample to your proposition. — Anne Bauval, Jun 29 '23 at 14:16
@AnneBauval Can you please have a look at my update 2 and check whether that's correct?? Thanks — user760, Jun 29 '23 at 18:04
How do you prove your final proposition? You seem to think that an endomorphism of $M$ is necessarily of the form $m\mapsto rm.$ — Anne Bauval, Jun 29 '23 at 18:57
@AnneBauval By "isomorphism", it's just the usual module isomorphism: an invertible homomorphism. For update 2, you are right. I changed it to only a necessary condition. — user760, Jun 29 '23 at 19:01
Your last sentence looks like a new question. I think your whole Update $2$ should be a new post. — Anne Bauval, Jun 29 '23 at 19:07
@AnneBauval Thanks. Did that. https://math.stackexchange.com/questions/4727772/conditions-so-that-change-of-basis-is-possible-in-a-free-module — user760, Jun 29 '23 at 19:19

GreginGre · Accepted Answer · 2023-06-29T15:55:54.323

This is already false for rings with IBN property, even PIDs. For example $1$ is a $\mathbb{Z}$-basis of the free $\mathbb{Z}$-module $\mathbb{Z}$ but $2$ is not.

Since you are trying to understand the difference betwwen free modules and vector spaces, here is a non exhaustive list of differences. Unlike for vecotr spaces:

there are linearly independent families $\mathcal{L}$ which cannot be extended to basis: $R=\mathbb{Z}$, $\mathcal{L}=\{2\}$
there are generating families $\mathcal{G}$ from which you cannot extract a basis: $R=\mathbb{Z}$, $\mathcal{G}=\{2,3\}$
there are free modules $M$ with non free submodules $N$ : $R$= any non PID commutative ring, $M=R$, $N=$ a non principal ideal
there are free submodules $N$ of a free module $M$ which are not a direct factor of $M$: $M=\mathbb{Z}$,$N=2\mathbb{Z}$
even $R$ has IBN, and $M$ is free of rank $n$, a linearly independent family of size $n$ is not necessarily a basis (see the example at the beginning of this post)

Thanks! So we can't expect to linearly transform a known basis into a new basis, even if the result is indeed a linearly independent set. — user760, Jun 29 '23 at 15:57
But, for a commutative ring $R$, and a free module $M$ with basis $B$, is it correct to claim, if $\det (A)\in Inv(R)$, where $A\in End(M)$, then $A(B)$ is another basis? Still correct for noncommutative rings? In that case, it seems $\det$ is not well-defined... — user760, Jun 29 '23 at 16:24

Change of basis in general free modules

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