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In Dummit&Foote's Abstract Algebra on p.284, the authors give the definition of irreducibles(irreducible elements): "Suppose $r\in R$ is nonzero and is not a unit. Then $r$ is called irreducible in $R$ if whenever $r=ab$ with $a,b\in R$, at least one of $a$ or $b$ must be a unit in $R$. Otherwise $r$ is said to be reducible."

I am confused about the reason why the authors use the words "at least one of" in the above context: if both of $a$ and $b$ are units with $r=ab$ an irreducible element, then there must exist $a^{-1},b^{-1}\in R^{\times}$ s.t. $aa^{-1}=bb^{-1}=1$, hence $rb^{-1}a^{-1}=abb^{-1}a^{-1}=1$ implies that $r$ is also a unit, contrary to "Suppose $r\in R$ is nonzero and is not a unit".

Quay Chern
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    Technically, you are right, they can't be both units. But it makes sense to define like this. If you already know $r$ is not a unit then given a decomposition $r=ab$ you just have to check that one of them must be a unit. If you checked $a$ is a unit, you don't have to check anything about $b$. If in the definition you require "exactly one has to be a unit" then it seems like you have to always check both elements, which would be annoying. So it makes more sense to define like the authors did, and what you wrote can be given as a simple proposition, not part of the definition. – Mark Jul 28 '24 at 07:42
  • @Yathi An irreducible must not be factored into two non-units, there exists a typo in your sentence. – Quay Chern Jul 28 '24 at 07:42
  • @Mark Thanks for your replying. The words "at least one of" truly provide convenient to check whether $r=ab$ is a irreducible. If it is changed into "one of $a$ and $b$", then I should check both "one is unit" and "the other is not a unit", which is harder than "at least one of". – Quay Chern Jul 28 '24 at 07:52
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    The phrase "at least one of" is unnecessary - delete it to get the standard definition for domains (in rings with zero-divisors the notion bifurcates into a few inequivalent notions). $\ \ $ – Bill Dubuque Jul 28 '24 at 08:09

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