TITLE: In a noetherian integral domain every non invertible element is a product of irreducible elements
QUESTION [5 upvotes]: I want to prove that in a noetherian ring $R$ which is also an integral domain, every non invertible element can be expressed as product of irreducible elements.

I really do not know where to start. Can someone give me hint how to prove this?

REPLY [5 votes]: Hint $\ $ Suppose not, and consider the set of principal ideals generated by all counterexamples. By ACC it contains a maximal element $(r),\,$ and $r$ is reducible $\,r = st.\,$ But by maximality, $\,s,t\,$ have irreducible factorizations, thus so too does $\,r = st,\,$ contradiction.
Remark $\ $ Interpreted positively, the proof shows if $S$ is the set of principal ideals  disjoint from a monoid $M$ then $(r)$ is maximal in $\,S \iff r\,$ is irreducible. This is a generalization of the well-known case where $\,M = \{1\},\,$ i.e. $\ (r)$ is maximal among principal ideals $\iff r$ is rreducible.   
This is a specal case of a wide class of results where elements satisfying such maximality properties are irreducible or prime, e.g. see this answer. One beautiful example along these lines is a famous theorem of Kaplansky that a domain is a UFD $\iff$ every prime ideal $\ne 0$ contains a prime $\ne 0$ (or, equivalently, every prime ideal is generated by primes).