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In a different question that I asked (see here), there were examples of rings in which not every element is divisible by an irreducible, so my question is, do the rings which have the property of all nonzero nonunit elements being divisible by an irreducible have a specific name or classification?

Some common examples are $\mathbb{Z}$, $\mathbb{Z}[i]$, and most other standard integral domains, and technically any field counts because there are no nonzero non-unit elements.

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    I don't know, but in practice a useful sufficient condition is that every Noetherian integral domain has this property. – Qiaochu Yuan Mar 31 '15 at 04:34
  • Also, $\mathbb{N}$ is not a ring. – Qiaochu Yuan Mar 31 '15 at 04:37
  • Haha oops, I'll change that – ASKASK Mar 31 '15 at 04:38
  • I will just talk about integral domains, although some part of my comment may apply also to general rings. If every element is divisible by only finitely many irreducibles, then you can actually factor every element into a product of irreducibles; such domain is called an atomic domain. Note that if "divisible by some irreducible" is changed to "divisible by some prime", then you actually get a UFD because atomic domain + every irreducible is prime = UFD. – Jianing Song Nov 26 '23 at 15:49
  • But there are also integral domains in which there is an element divisible by infinitely many irreducibles, for example the ring of holomorphic functions over $\mathbb{C}$. This ring also satisfies every nonunit is divisible by some irreducible (actually, primes): the reciprocal of a holomorphic function without zeros is a holomorphic function. – Jianing Song Nov 26 '23 at 15:51

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