TITLE: Irreducible elements for a commutative ring that is not an integral domain
QUESTION [11 upvotes]: Why does the definition of an irreducible element require us to be in an integral domain?
  Why can we not define an irreducible element exactly the same in a commutative ring that is not an integral domain?

We have that an element is irreducible if it cannot be written as a product of two non-unit elements. Unit elements are well defined and unique in a commutative ring that is not an integral domain, so I cannot see that being the problem.
I've proven a proposition of my own design (probably well known and elementary, an definitely trivial). I used irreducible elements, but otherwise nothing that requires me to move from a commutative ring to an integral domain. Do irreducible elements really require me to be in an integral domain?

REPLY [18 votes]: In an integral domain, you have the following four equivalent definitions for a nonzero nonunit $a$ to be irreducible.

$a = bc \Rightarrow (a) = (b)$ or $(a) = (c)$.
$a = bc \Rightarrow a$ is a unit multiple of $b$ or $c$.
$(a)$ is maximal among the proper principal ideals.
$a = bc \Rightarrow b$ or $c$ is a unit.

However, in commutative rings in general, we have (4) $\Rightarrow$ (3) $\Rightarrow$ (2) $\Rightarrow$ (1), and none of the implications reverse.  The literature for factorization in commutative rings with zero divisors thus has four different non-equivalent definitions of "irreducible".  (The above statements define "irreducible", "strongly irreducible", "m-irreducible", and "very strongly irreducible", respectively.)  See Factorization in Commutative Rings with Zero Divisors by Anderson and Valdes-Leon for more information.