Nonunits in a Noetherian Domain have an Irreducible Factor I think I've proven the following statement without using the fact that it is a domain:

Prove every nonunit in a Noetherian domain has an irreducible factor.

Proof: Suppose we have a ring which has an element with no irreducible factor. Then we can repeatedly factor it (first factoring it, then its factors, etc.) so we can form an ascending chain which doesn't stabilize, that is the ring isn't Noetherian. This proves the contrapositive of the statement.
Have I implicitly made a mistake, or is the hypothesis that we're in a domain simply extra? (This is a multi-part problem, this only being one part. In a later part, I do use the fact that we're in a domain, although all of the parts of the problem are written independently of each other.)
If I have made a mistake, and I probably have since this is a prelim. question, will you help point me in the right direction? Not necessarily a full answer, though that wouldn't be so bad, either.
 A: You haven't made a mistake. In most sources the definition of an irreducible element includes being in a domain. You may like to read the second page of this as well.
A: There isn't anything wrong with your reasoning, although some might take issue about whether or not irreducible elements are really studied outside of domains. The usual definition of irreducible is: "$a$ is irreducible if $a=bc$ implies one of $b$ or $c$ is a unit of $R$." If you aren't in a domain, it's possible to have an element $a$ which can't be "broken down" any further, and yet $a=br$ for nonunits $b$ and $r$.
We could rephrase your idea by saying "An element $b$ in the commutative ring $R$ can't be broken down anymore if $bR$ is maximal in the poset of proper principal ideals of $R$". The Noetherian condition allows you to confidently say "Yeah, that poset has maximal elements. In fact I can find a maximal element containing any $aR$ you want."
The reason it's good to have $R$ be a domain is that you can determine the generator of a principal ideal up to multiplication by a unit. If you start with $a\in bR$ and $b\in aR$, you get $a=bu$ and $b=av$, and putting them together $a=auv$. If you can cancel $a$ as you can in a domain, then you get that $1=uv$ and $u$ and $v$ are both units. Since we ignore units in factorizations, we effectively think of $a$ and $b$ as being the same thing.
But if you aren't in a domain, maybe you find that your element $c\in aR=bR$ for two elements $a, b$ which can't be broken down anymore, such that $a=br$ and $b=as$, none of $a,b,r,s$ units. So while we'd like to call both $a$ and $b$ irreducible, they have unfortunate factorizations which break the normal irreducibility definition. The elements $a$ and $b$ no longer have to differ by a negligable unit factor.
So you can see talking about factorizations outside of domains becomes harder. And don't even get me started on noncommutative factorization: that's even harder!
