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I know this may seem trivial but I'm trying to grasp why irreducible elements are non-units. If an element p is a unit and b is its inverse, then $pb = 1, \forall p,b \in R$, R is a ring. Does this imply that b is a factor of p, thus making it reducible?

Squires McGee
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    It is a definition. The whole point is that reducible elements are supposed to have nontrivial factorizations. In an integral domain, the nonzero elements fall into three classes: units, irreducible elements, and reducible elements. For a unit, every factorization into a product of two terms has both factors equal to units (that is to be proved), for an irreducible element every factorization into a product of two elements has exactly one factor equal to a unit. For a reducible element there is some factorization into a product of two elements that are both non-units. – KCd Mar 03 '15 at 14:10

2 Answers2

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Irreducible elements are non-units by definition. That is, they are non-units because the definition explicitly requires them to be; if it didn't, they could be units. (The reason why the definition insists that they be non-units is that otherwise, the notion of a UFD would be useless: even in $\mathbb Z$, the factorization of a nonzero integer into irreducibles would not be unique since $2 = 1\cdot 2 = 1\cdot 1 \cdot 2 = \cdots$.)

darij grinberg
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Here is another example: suppose your teacher instructs you to factor:

$p(x) = x + 1$ in the rational number system.

You might reply, "it's already factored", but then your teacher says:

"No, $p(x) = 2\cdot \left(\dfrac{x}{2} + \dfrac{1}{2}\right)$".

You realize, with a sinking feeling, you'll never "be finished factoring". Intuitively, factoring ought to stop "at some basic level". Factoring "up to units" IS that basic level, an opt-out that allows us to finally be "done". (note that $2$ is a unit in $\Bbb Q$, so the two factorizations above are "really the same").

David Wheeler
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