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This is a follow-up of this 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 $B$.

Under what exact conditions, for every injective $R$-module endomorphism $h: M\to M$, the set $h(B)$ is a new basis?

For $M$ a finitely generated free left $R$-module, we have

Lemma 1. For an injective endomorphism $h:M\to M$, the set $h(B)$ is a new basis iff the map $h$ is surjective (hence invertible).

For $R$ a unital commutative ring, we further have

Lemma 2. For a finitely generated free left $R$-module $M$ over a unital commutative ring $R$, an $R$-module endomorphism $h: M\to M$ is injective iff the determinant $\det(h)$ is not a zero-divisor in $R$. (See here, here, and here for the proofs.)

Based on the two lemmas, the following might be true.

Proposition. Let $M$ be a non-zero finitely generated free left $R$-module over a unital commutative ring $R$, and $B\subset M$ a basis. Then, the image of $B$ by every injective endomorphism of $M$ is a basis of $M$ iff the ring $R$ only consists of units and zero-divisors.

Proof. For every injective endomorphism $h$, by lemma 2, $\det(h)$ is not a zero-divisor. If $R$ consists of only units and zero-divisors, then $\det(A)$ is a unit, and so $h$ is invertible. By lemma 1, $h(B)$ is then a basis.

Conversely, if $h$ is an endomorphism and $h(B)$ is a basis, then $h$ is invertible by lemma 1, and so $\det(A)$ is a unit. If this is true for every injective endomorphism $h$, then by lemma 2, every element in $R$ that is not a zero-divisor must be a unit. $\square$

But I'm not sure how to answer when $R$ is noncommutative. Thanks in advance.

Update: With a helpful comment, the proposition would clearly be false for infinitely-generated free modules.

Anne Bauval
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user760
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  • In your explicitely non-necessarily-commutative context (since you distinguish between left-zero-divisors and right-zero-divisors), your maps $h_r:m\mapsto rm$ are not endomorphisms of your left module $M.$ Moreover, I cannot see how your proposition should result from the lemmas, more precisely why non-units should be right-zero-divisors. I think you should better restrict to commutative rings.
    • – Anne Bauval Jun 30 '23 at 15:30
  • @AnneBauval Thanks. I'll try to phrase it better. I don't understand why $h_r: m\mapsto rm$ is not an endomorphism. Isn't $h_r$ a module homomorphism from $M$ to itself? – user760 Jun 30 '23 at 15:47
  • If $rs\ne sr$, $h_r(sm)\ne sh_r(m)$ in general. – Anne Bauval Jun 30 '23 at 15:48
  • @AnneBauval Oh, right. But if restricted to commutative rings, then the lemmas imply the proposition, correct? Becuase lemma 2 applies to all injective endomorphisms, and the $h_r$'s are special ones. If $M$ is also finitely generated, then $A: M\to M$ is injective iff $\det(A)$ is not a zero-devisor. So if the ring only consists of units and zero-divisors, then all the injective $A$'s automatically are invertible. That makes the condition sufficient. – user760 Jun 30 '23 at 15:59
  • @AnneBauval Perhaps I didn't express my question clearly enough. What I want to ask is: if $B$ is a basis, then under what conditions is $f(B)$ guaranteed to be also a basis, where $f: M\to M$ is an arbitrary injective endomorphism. So I shouldn't have started from equal cardinality. I'll update it. – user760 Jun 30 '23 at 16:10
  • Yes, if restricted to commutative rings, everything was ok for me (you should just clarify this from the beginning, and only speak of zero-divisors, not mentionning left/right/two-sided). As for your question, it was clear and my answer would have been the same as yours, reasoning on determinant, once you would have agreed to restrict to commutative rings, and to the case of free modules with finite basis. So there is no question left. But the new question with which you intend to replace the now solved one is too vague, imo. – Anne Bauval Jun 30 '23 at 18:00
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    @AnneBauval Thanks. I updated. Does it all make sense now? I hope there's no mistakes left. – user760 Jun 30 '23 at 18:37
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    For non-finitely generated it never happens. If $B$ is a basis, then $B$ is infinite, so there exists a non-surjective embedding $f\colon B\to B$. Then $f$ extends to an injective endomorphism of $M$ which is not surjective; in particular, $f(B)$ cannot be a basis, since it doesn't even span. – Arturo Magidin Jun 30 '23 at 20:57
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    This is exactly the reason of the advice 2) of my first comment above. @ArturoMagidin (and user760: please follow it by editing your last sentence). – Anne Bauval Jun 30 '23 at 21:38
  • @user760 Also, your proposition is not well-phrased because of the quantifiers. It is not true that "if $B\subset M$ is a basis and $h: M\to M$ an injective endomorphism, then the set $h(B)$ is also a basis iff The ring $R$ consists of only units and zero-divisors" (take $h=id$ and $R=$ any ring). Instead, write something like "For a finitely generated free left $R$-module $M$ with basis $B$ over a unital commutative ring $R$, the image of $B$ by every injective endomorphism of $M$ is a basis of $M$ iff the ring $R$ consists of only units and zero-divisors"; – Anne Bauval Jun 30 '23 at 21:49
  • Or simply (without reference to $B$): "[...] every injective endomorphism of $M$ is bijective iff the ring $R$ consists only of units and zero-divisors". Btw, $M$ must moreover be assumed to be non-zero. – Anne Bauval Jun 30 '23 at 21:54
  • @AnneBauval Thanks! I'll update. Yes, at the beginning of the post, I assumed $M$ to be "nontrivial". – user760 Jul 01 '23 at 06:01
  • Although you added the word "every", your proposition remains ambiguous at best (and false at worst). – Anne Bauval Jul 01 '23 at 06:28
  • Your proposition remains ambiguous at best (and false at worst): "$(\forall h\dots)$ [the image h(B) is also a basis iff The ring R consists...]" – Anne Bauval Jul 01 '23 at 09:18
  • @AnneBauval I did seem to find a potential mistake in the attempted proof. But I'm not sure how to phrase the (potentially false) proposition better. You are welcome to edit it. – user760 Jul 01 '23 at 09:43
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    There is no essential mistake, and every element of $R$ is obviously a $\det.$ The only "mistake" was the wording of the propostion. – Anne Bauval Jul 01 '23 at 11:33