Irreducible elements for a commutative ring that is not an integral domain

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?

• The notion makes sense over any commutative ring with unity. But note that some well known results get false if the ring has zero-divisors. For example a non-trivial idempotent is never irreducible, but might be prime. – MooS Jun 24 '16 at 11:41
• See this answer and the linked papers for a starting point on factorization theory in rings with zero-divisors. – Bill Dubuque Jun 24 '16 at 20:18

In an integral domain, you have the following four equivalent definitions for a nonzero nonunit $a$ to be irreducible.
1. $a = bc \Rightarrow (a) = (b)$ or $(a) = (c)$.
2. $a = bc \Rightarrow a$ is a unit multiple of $b$ or $c$.
3. $(a)$ is maximal among the proper principal ideals.
4. $a = bc \Rightarrow b$ 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.