This is an exercise I found in a class test and I was struck trying to solve it.

Let $D$ be a Prüfer domain (*) and let be $\mathfrak{q}_1,\mathfrak{q}_2$ two primary ideals of $D$. Then prove that $\mathfrak{q}_1,\mathfrak{q}_2$ are coprime or one contains the other.

I tried using the fact that every localization of a Prüfer domain is a valuation domain, but it doesn't seem to solve.

Could someone give me a hint for the solution?

(*) A Prüfer domain, for what I know, is an integral domain in which every finitely generated non-zero ideal is invertible. The only characterization result I know is the one regarding valuation rings.


1 Answer 1


If $q_1,q_2$ are not coprime then they are contained in a maximal ideal, say $m$. In $R_m$ (which is a valuation ring) we then have $q_1R_m\subset q_2R_m$ or conversely, and you are done.

  • $\begingroup$ Thank you very much! Just only a detail: if the two ideals are not coprime, the best I can say is that $q_1+q_2$ is contained in a (proper) maximal ideal. The argument you propose follow either (as maximals are primes) but I was wondering if I missed something. $\endgroup$
    – Caligula
    Commented Sep 6, 2015 at 16:23
  • $\begingroup$ @Alexander Maximal ideals are prime, but if this bothers you I can edit my answer. $\endgroup$
    – user26857
    Commented Sep 6, 2015 at 16:39
  • $\begingroup$ Obviously not, that was only a detail. But where do we use the primary hypothesis? $\endgroup$
    – Caligula
    Commented Sep 6, 2015 at 16:42
  • $\begingroup$ @Alexander If we have two ideals $I_1,I_2$ contained in a maximal ideal $m$, then from $I_1R_m\subseteq I_2R_m$ we can't conclude that $I_1\subseteq I_2$. However this holds for prime and primary ideals. (This is also a good exercise.) $\endgroup$
    – user26857
    Commented Sep 6, 2015 at 16:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .