# A case where a UFD is a PID

Let $R$ be a unique factorization domain (UFD). Prove that if $R$ is such that every ideal generated by two elements is principal, then $R$ is a principal ideal domain (PID).

I'm having some trouble with this proof. Unfortunately I don't really have an idea for a starting place. It seems difficult to take a statement about specific ideals in $R$ and prove a statement about every ideal in $R$. Maybe something with GCD's will help?

• Try to show that every ideal is finitely generated. If $I$ is not, then there is a strictly ascending chain $I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq \dots$ of _finitely generated_ ideals in $I$. It should follow from there (using properties of UFD's) easily. Mar 11, 2015 at 14:52

Ideals are closed under gcd: $$\,a,b\in I\,\Rightarrow\, I\supseteq (a,b) = (\gcd(a,b)).$$ Thus an ideal $$I\ne 0\,$$ is generated by any $$\,0\neq a\in I$$ with least #prime factors (else $$\,a\nmid b\in I\,\Rightarrow\, \gcd(a,b)\in I$$ and, being a proper factor of $$\,a,\,$$ it has fewer prime factors than $$\,a,\,$$ contra choice of $$a).$$
Remark $$\$$ The descent used above is a generalization of the Euclidean descent (by division algorithm) in the classical proof that Euclidean domains are PIDs. More generally we have the Dedekind-Hasse criterion: a domain $$\rm\,D\,$$ is a PID $$\iff$$ given $$\rm\:0\neq a, b \in D,\:$$ either $$\rm\:a\:|\:b\:$$ or some D-linear combination $$\rm\:a\,d+b\,c\:$$ is "smaller" than $$\rm\,a\,$$ (using $$\,\Bbb N\,$$ or any well-ordered set for "size")
It is clear that such a domain must be PID (since then a "smallest" element in an ideal must divide all others). Conversely, since a PID is UFD, an adequate "smaller" metric is the number of prime factors (if $$\rm\,a\nmid b\:$$ then their gcd $$\rm\,c\,$$ must have fewer prime factors; for if $$\rm\:(a,b) = (c)\:$$ then $$\rm\,c\:|\:a\:$$ properly, else $$\rm\,a\:|\:c\:|\:b\:$$ contra hypothesis), i.e. in a PID an ideal is generated by any minimal elt, i.e. one with fewest #prime factors (e.g. minimal polynomials in $$K[x])$$.
More generally it is easy to show a domain $$D$$ is a PID $$\iff D$$ is Bezout with ACCP (Ascending Chain Condition on Principal ideals, i.e. the divisibility relations is well-founded, i.e. there are no infinite descending divisibility chains $$\,\ldots\, d_3\mid d_2\mid d_1$$).
• Could you explain again how you can conclude that $I$ is generated by $a$? Mar 11, 2015 at 15:20
• @Ducky Because $I$ is closed under gcds, we can keep taking gcds of elements of $I$ to obtain "smaller" elements of $\,I,\,$ i.e. having fewer prime factors. Pick any $\,0\ne a\in I.\,$ If $\,I\ne (a)\,$ then there is some $\,b\in I\,$ such that $\,a\nmid b,\,$ so $\,c=\gcd(a,b)\in I\,$ and $\,c\,$ has fewer prime factors than $\,a\,$ since $\,c\mid a\,$ properly (else $\,a\mid c\,$ and $\,c\mid b\,\Rightarrow\, a\mid b)\$ Mar 11, 2015 at 15:32
• Thank you for the help. I will let you know if I have any confusion writing up my version. My original idea was to prove that $R$ is in fact Euclidean, which is much stronger; it seems we are doing something similar here (where the "Euclidean norm" is the number of primes). Mar 11, 2015 at 15:46