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I'm reading Aluffi's algebra book at the moment -- specifically, I'm on the introductory rings/modules chapter.

I noticed two interesting pieces of information: in a (not necessarily commutative) ring $R$,

  • $x\mapsto ux$ is injective for $u$ a left non-zero-divisor

  • $x\mapsto ux$ is surjective for $u$ a left unit

Is there any deeper (read: probably categorical) reason for this apparent "duality"?

Edit: Specifically, since we know that injectivity and surjectivity (well, mono- and epi-ness, but we're in $\sf Set$) are "dual", is there also some duality between the properties of being a left (resp. right) non-zero-divisor and being a left (resp. right) unit?

Soham Chowdhury
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  • For a left unit, it is also injective. There is no duality here – just the fact that bijective implies injective. – Bernard Jul 10 '15 at 11:25
  • Bernard, Najib: Please recall that $R$ is not assumed to be commutative, and be careful as for the "left"s here. – Martin Brandenburg Jul 10 '15 at 11:35
  • @Soham: The answer to your question is probably "no". These are absolutely trivial reformulations of the definitions. It would be better to narrow your question to get some interesting answers. – Martin Brandenburg Jul 10 '15 at 11:39
  • so $u$ a left unit implies $x\mapsto xu$ is injective? Since left units are right non-zero divisors. – Asinomás Jul 10 '15 at 11:42
  • On the other hand $u$ not a left-zero divisor does not imply $x\mapsto xu$ is surjective. Duality seems to not be dual, checkmate. – Asinomás Jul 10 '15 at 11:43
  • @MartinBrandenburg, I've edited it. Does the question make a little more sense now? – Soham Chowdhury Jul 10 '15 at 12:08
  • Yes, I think so. But I doubt that there is an honest duality between left regular elements and left units. – Martin Brandenburg Jul 10 '15 at 12:26
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    Dear @SohamChowdhury Do you have no interest in the observation that $x\mapsto ux$ is injective if $u$ is a right unit? (Of course, right units are left-non-zero divisors.) That seemed like the more obvious counterpart, but of course you may have already discarded it. In any case, these ideas about one-sided invertibility or cancelability are interesting. You might want to lookup Hopfian/coHopfian objects. Regards – rschwieb Jul 10 '15 at 14:08
  • @rschwieb, I'll definitely look those up. Thanks. – Soham Chowdhury Jul 11 '15 at 03:49
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    Notice that the zero divisors are union of minimal primes, while $R^{\times}$ is the intersection of $R \setminus m$ where $m$ goes through the maximal ideals of $R$. – Watson Jan 24 '17 at 18:05

1 Answers1

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Here is something which comes to my mind. I don't know if it answers the question properly.

Recall that a monoid (resp. a ring) is the "same" as a category (resp. linear category) with one object. The composition of morphisms is just the multiplication of monoid (resp. ring) elements. The notions of left units ( := right invertible element, what I suspect from the question) and left regular elements ( := left cancellable elements) generalize to arbitrary categories:

A morphism $f : X \to Y$ is left invertible, or more commonly, a split monomorphism, if there is a morphism $g : Y \to X$ such that $g \circ f = \mathrm{id}_X$. A morphism is called left cancellable, or more commonly, an epimorphism, if for all morphisms $g,h : Z \to X$ with $f \circ g = f \circ h$ we have $g=h$. The dual notions are split epimorphism and monomorphism. Then, we have the following:

  • $f$ is a split monomorphism if and only if $\hom(Y,Z) \to \hom(X,Z),~ g \mapsto g \circ f$ is surjective for all objects $Z$.
  • $f$ is a split epimorphism if and only if $\hom(Z,X) \to \hom(Z,Y),~ g \mapsto f \circ g$ is surjective for all objects $Z$.
  • $f$ is a monomorphism if and only if $\hom(Z,X) \to \hom(Z,Y),~ g \mapsto f \circ g$ is injective for all objects $Z$.
  • $f$ is an epimorphism if and only if $\hom(Y,Z) \to \hom(X,Z),~ g \mapsto g \circ f$ is injective for all objects $Z$.
  • Every split monomorphism is a monomorphism.
  • Every split epimorphism is an epimorphism.

From here it becomes clear that monomorphisms and split epimorphisms (similarly, epimorphisms and split monomorphisms) have dual characterizations in terms of $\hom(Z,-)$ (resp. $\hom(-,Z)$). But I doubt that this means that the notions are dual to each other.

Martin Brandenburg
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