# A non-zero and non-invertible element in a noetherian integral domain has a decomposition into irreducible elements

Let $R$ be a noetherian integral domain. I want to show that any non-zero and non-invertible element $a$ can be written as a finite product of irreducible elements.

my ideas: I should argue by contradiction and consider the set $M$ of the ideals generated by elements, which cannot be written as a product of irreducibles. Since R noetherian, we find a maximal element $(b)$ of $M$. Moreover we find a minimal overlying prime ideal $I$ of $(b)$. But now I'm stuck. I wanted to show that all minimal overlying prime ideals of $(b)$ are principal ideals, but that doesn't hold generally (I think).

I don't think we need to go far as to find prime ideals. Given a maximal element $(b)$ of $M$, observe that since $(b) \in M$, the element $b$ is not irreducible. So we can write it as a product of two non-units $b = b_1 b_2$. But then $(b_1) \supsetneq (b)$ and $(b_2) \supsetneq (b)$, so by the maximality of $(b)$ in $M$, we see that $b_1$ and $b_2$ can both be written as a product of irreducibles. So $b = b_1 b_2$ can be written as a product of irreducibles, contradicting $(b) \in M$.
• Well, I'm not sure what irreducible means outside of integral domains. Also, one place where we use that $R$ is a domain is to assert that $(b_1) \supset (b)$ is strict. To show this, suppose for sake of contradiction that $b_1 = (b)$, so $b_1 = bb_3$. Then we have $b = b_1 b_2 = (b b_3)b_2$. Using the fact that $R$ is an integral domain, we can cancel the $b$ from both sides to get $b_3 b_2 = 1$, whence $b_2$ is a unit, contradiction.