A Noetherian integral domain is a UFD iff $(f):(g)$ is principal

Let $R$ be a Noetherian integral domain. For $f, g \in R$, define $(f):(g)=\{h \in R \mid hg \in (f) \}$. Sow that $R$ is a UFD if and only if $(f):(g)$ is principal for all $f,g \in R$.

It is easy to show that $(f):(g)$ is an ideal. For the forward direction, I suspect that I need to use the fact that $R$ is Noetherian to show that $(f):(g)$ is principal. For the reverse direction, I know that if every $(f):(g)$ is principal then the ring $R$ is a PID. But I don't know how to proceed. Any ideas?

Solution 1:

I don't really understand what you mean by "I know that if every $(f):(g)$ is principal then the ring $R$ is a PID" : you seem to imply that every noetherian UFD is a PID, which is false.

The first implication (assuming $R$ is a UFD) is actually true even if $R$ is not noetherian. Indeed, if $R$ is a UFD, then write $f = u\prod p_i^{a_i}$ and $g=v\prod p_i^{b_i}$. Then for any $h = w\prod p_i^{c_i}$, you get $hg\in (f)$ iff $\forall i,$ $b_i + c_i \geqslant a_i$. Then putting $d_i = \max(0 ; a_i-b_i)$ you get $(f):(g) = (\prod p_i^{d_i})$.

As for the second implication, you only need that $R$ admits an irreducible factor decomposition (which of course is always true when $R$ is noetherian). In this case it is well-known that $R$ is a UFD (meaning that the decomposition is unique) if and only if irreducible elements satisfy the Gauss (or Euclid, or whatever) lemma : $p\mid ab$ implies $p\mid a$ or $p\mid b$.

First observe that if $(f):(g) = (x)$ then since $fg \in (f)$ you get $f\in (f):(g) = (x)$ and hence $x\mid f$. So if $p$ is irreducible, and $(p):(a) = (x)$, you get $x\mid p$, and thus $(x) = (1)$ or $(x) = (p)$. The first case happens if and only if $p\mid a$ by definition.

Now assume $p\mid ab$ and $p\nmid b$. Then $ab\in (p)$ so by definition $a\in (p):(b)$, but $(p):(b)=(p)$ by the previous observation, so $p\mid a$. This is the Gauss lemma.