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?

  • 1
    $\begingroup$ 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. $\endgroup$ Mar 11, 2015 at 14:52

1 Answer 1


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 $\,\ldots\, d_3\mid d_2\mid d_1$).

  • $\begingroup$ Could you explain again how you can conclude that $I$ is generated by $a$? $\endgroup$
    – Ducky
    Mar 11, 2015 at 15:20
  • $\begingroup$ @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)\ $ $\endgroup$ Mar 11, 2015 at 15:32
  • $\begingroup$ 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). $\endgroup$
    – Ducky
    Mar 11, 2015 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.