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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?

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    $\begingroup$ 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. $\endgroup$
    – MooS
    Jun 24, 2016 at 11:41
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    $\begingroup$ See this answer and the linked papers for a starting point on factorization theory in rings with zero-divisors. $\endgroup$
    – Bill Dubuque
    Jun 24, 2016 at 20:18

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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.

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    $\begingroup$ fyi: That is one of the papers linked in the comment I gave above. Both are open access so can be freely downloaded. $\endgroup$
    – Bill Dubuque
    Jun 24, 2016 at 21:05
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    $\begingroup$ This is very very cool. Thanks very much! $\endgroup$
    – Commut...Integral domain...
    Jun 27, 2016 at 7:17

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