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I was attempting to prove (one of the several versions of) Gauss' lemma, and this apparently simple question popped up:

Does every nonunit element of a ring has some irreducible factor, or better ("better" because in domains, primes are irreducible) yet, a prime factor?

Amazingly, I am at loss to even begin to answer this seemingly basic question even after having had a full course on ring theory.

Any starting point?

I know that any domain that obeys ACCP, is atomic. Thus we need to search for non-Noetherian rings.

(This is a small sigh of relief as I haven't yet studied non-Noetherian rings in any significant detail. :p)

Atom
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  • Yes for Noetherian rings, see [here](https://math.stackexchange.com/questions/462876/nonunits-in-a-noetherian-domain-have-an-irreducible-factor). – Dietrich Burde Apr 03 '23 at 19:18
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    Perhaps the simplest counterexamples are antimatter domains, i.e. rings with no atoms (= irreducibles), [such as](https://math.stackexchange.com/a/37501/242) the ring of all algebraic integers, where every $\,a = \sqrt a \sqrt a\,$ is reducible. See the links [here](https://math.stackexchange.com/a/96802/242) for further examples. – Bill Dubuque Apr 03 '23 at 19:18
  • @DietrichBurde Yes, we just require ACCP indeed. I will add this in my post. – Atom Apr 03 '23 at 19:24
  • Consider the [monoid ring](https://en.wikipedia.org/wiki/Monoid_ring) $\Bbb Z[\Bbb Q_{\ge0}],$ i.e. the quotient of the polynomial ring $\Bbb Z[\{X_r\mid r\in\Bbb Q_{>0}\}]$ by the ideal generated by $\{X_rX_s-X_{r+s}\mid r,s\in\Bbb Q_{>0}\}$. In this quotient $\Bbb Z[\{\bar X_r\mid r\in\Bbb Q_{>0}\}],$ the $\bar X_r$'s have no irreducible divisor. – Anne Bauval Apr 03 '23 at 19:26
  • @BillDubuque How do you keep track of so many (helpful!) links that you so often attach!? – Atom Apr 03 '23 at 19:30
  • It is definitely not true for all commutative rings. It is one reason we restrict a lot of work to Noetherian rings. – Thomas Andrews Apr 03 '23 at 19:30
  • For *much* more on monoid domain constructions of antimatter domains like that in @Anne's prior comment see the [first paper cited by Zafrullah](https://math.stackexchange.com/a/456199/242) in the prior thread I linked. – Bill Dubuque Apr 03 '23 at 19:31
  • @BillDubuque Can you elaborate why the algebraic integers don't form a field? (I have not dabbled in this before.) – Atom Apr 03 '23 at 19:36
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    @Atom Re; keeping track of links: knowing pertinent keywords greatly helps for searching. Here these domain properties are: atomic, antimatter, factorization domain, ACCP, etc. Searching on those will locate many helpful prior posts. – Bill Dubuque Apr 03 '23 at 19:37
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    Re: $\bar{\Bbb Z}$ = ring of all algebraic integers is not a field. Unlike localizations, an essential key property of integral extensions is that [they cannot invert elements](https://math.stackexchange.com/a/3137826/242), so a nonunit in the base ring remains a nonunit in the extension. In particular every nonunit in $\Bbb Z$ remains a nonunit in $\bar {\Bbb Z}$. More generally see the [survival property](https://math.stackexchange.com/a/57702/242), and related classical properties of integral extensions LO (lying-over), GO (going-up), INC (incomparability). – Bill Dubuque Apr 03 '23 at 19:57

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