Why are irreducible elements non-units? 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?
 A: 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").
A: 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 $2 = 1\cdot 2 = 1\cdot 1 \cdot 2 = \cdots$.)
