In a noetherian integral domain every non invertible element is a product of irreducible elements 
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?
 A: 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).
A: Let $X$ be your set of nonzero, nonunits which cannot be written as a product of irreducibles. Towards a contradiction, suppose $X\neq\emptyset$, and pick $x_0\in X$. Then $x_0$ itself is not irreducible, so we may write $x_0=xy$, where $x$ and $y$ are nonzero nonunits. If both $x$ and $y$ can be written as a product of irreducibles, then $x_0$ can as well, a contradiction. So at least one of $x$ and $y$ is not a product of irreducibles, say $x$. Call it $x_1\in X$. Then $(x_0)\subset (x_1)$.
Continue this process to yield an ascending chain
$$
(x_0)\subset (x_1)\subset\cdots
$$
in $R$. Now use the fact that $R$ is Noetherian to find a contradiction.

 Since $R$ is Noetherian, $(x_n)=(x_{n+1})$ for some $n$. By our construction, $x_n=yx_{n+1}$ for $y$ a nonunit. But then $x_{n+1}\in (x_n)=(yx_{n+1})$, so for some nonzero $z\in R$, we have $x_{n+1}=zyx_{n+1}$. By cancellation, since we are in a domain, $1=yz$, so $y$ is a unit, a contradiction. So $X=\emptyset$.

