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Is there an integral domain in which all elements are reducible?

I can't think of a counter example. Some sugestion?

Levi
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Douglas Souza
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  • What do you mean "integrity ring"? Do you mean integral domain https://en.wikipedia.org/wiki/Integral_domain? – Ben Blum-Smith Nov 04 '19 at 14:00
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    Vacuously, a field has this property. But I guess this is not what you're after :) – Levi Nov 04 '19 at 14:00
  • Yes, Integral Domain, i speak portuguese, sorry. – Douglas Souza Nov 04 '19 at 14:01
  • Consider the ring $R=\bigcup_n{k[[X^{1/n!}]]}$. – Aphelli Nov 04 '19 at 14:01
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    Hint Consider a domain, not a field, that is closed under square-roots. It has no irreducibles since every element factors $\rm\ d\ =\ \sqrt{d}\ \sqrt{d}:.\ $ For example, the domain of all algebraic integers. Such domains are known as antimatter domains so you can learn more by searching on that term. – Bill Dubuque Nov 04 '19 at 16:12

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The adjectives "reducible" and "irreducible" are typically only taken to apply to nonzero, nonunit elements of an integral domain.

So if you want an integral domain which has lots of nonzero, nonunit elements that are reducible, and no irreducible elements, then the ring of algebraic integers is an example. Every such element is the square of another element.

rschwieb
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