Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have trouble understanding an argument in the proof of the Kummer-Dedekind theorem. I am referring to a proof given in Peter Stevenhagen's notes.

This is theorem 3.1 on page 27. For the proof of the second part, the first statement says that since $p_i$ contains $pR$, it is invertible. I am not sure what the author is invoking here. For every ideal contains a principal ideal, but that is not sufficient for invertibility. Also, the fact that $p_i$ contains $pR$ should follow from the first part of the theorem since $p_i=pR+g_i(\alpha)R$. So I don't see what fact is used in proving invertibility of $p_i$ based on its containment of $pR$. Any help would be very appreciated.

share|cite|improve this question
up vote 1 down vote accepted

Principal ideals are invertible, and so too are divisors of invertible ideals, so $\mathfrak{p}\:|\:pR\ \Rightarrow\ \mathfrak{p}\ $ invertible.

Note that the ring in the theorem is not necessarily a a Dedekind domain so it need not be true that $\ I \supset J\ \Rightarrow\ I|J$

share|cite|improve this answer
Thanks for the comment, Bill. I do understand principal ideals are invertible. But given any ideal I, one can pick a non zero element in it, say a and then I contains aR. So, I am not sure how just containing a principal ideal makes the ideal invertible, since every ideal satisfies this. I am sure I am missing something trivial here. – Timothy Wagner Nov 15 '10 at 3:46
@Timothy: See my edit – Bill Dubuque Nov 15 '10 at 3:51
@Bill: I thought they were equivalent. That is the way the author defines division of ideals on page 14 of the above notes. How do you define $I|J$. I am sorry I am having a brain freeze here. – Timothy Wagner Nov 15 '10 at 3:54
OK, I think I follow why $p_i$ is invertible, since $p_i J=pR$ so $p_i (Jp^{-1}R)=R$ where $J$ is the rest of the stuff in that product of $p_i$'s. I am guessing then, that $I|J$ means that there exists an ideal $A$ such that $IA=J$. – Timothy Wagner Nov 15 '10 at 4:05
@Tim: It appears the author is inconsistent in his use of "divides". The standard definition for ideals in any ring is $\ I|J \iff I I' = J\ $ for some $ I' $. That's what is intended in the theorem. – Bill Dubuque Nov 15 '10 at 4:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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