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I'm having a mind blank on the definition of a UFD. Obviously I've gone horrendously wrong somewhere in my reasoning. My question is: how can we possibly have an irreducible element in a UFD?

Say $r \in R$ is irreducible, with $R$ a UFD. Then by the definition of a UFD, $r \neq 0$ can be expressed as the product of irreducible elements up to order and units, let's say two.

Then $r=xy$ where $x$ and $y$ are irreducible. But then if $r$ is irreducible, by the definition of irreducibility, either $x$ or $y$ must be a unit. Let's say it's $x$. But then how can $x$ be irreducible if it is a unit?

Many thanks.

EDIT: I was totally forgetting that a product can mean just one piece, i.e. the "product" in this case is $x$.

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  • $\begingroup$ The definition of irreducible requires that the element be a non-unit. See Wikipedia $\endgroup$ – rogerl Dec 4 '16 at 23:54
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    $\begingroup$ Also the expression as a product of irreducibles can be trivial, i..e. if $r$ is irreducible, the only expression (up to multiplication by unit) is itself. $\endgroup$ – Scott Burns Dec 4 '16 at 23:57
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"Every non-zero non-unit element can be written as a product of prime elements (or irreducible elements), uniquely up to order and units". It says that each element is a product of irreducible elements uniquely up to units, if you have $r$ irreducible, you have $x$ irreducible and $y$ an unit which fits with the definition above. And if $y$ is an unit, you have a product of irreducibles (in this case it's an empty product) and an unit.

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  • $\begingroup$ Ah yes of course - I was totally forgetting that a product can mean just one piece, i.e. the "product" in this case is $x$. Thank you. $\endgroup$ – mathphys Dec 5 '16 at 0:03
  • $\begingroup$ I've just edited my post, look it. I think it can clear out something else. $\endgroup$ – Skullgreymon Dec 5 '16 at 0:05

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