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I previously posted about this here: comaximality of ideals in a commutative ring with unit

Sadly, having an unregistered account at the time, I can't edit that post. I will say thanks to Arturo Magidin, here, instead, and hope he reads it.

By Dedekind domain, I will mean an integral domain such that every non-0 proper ideal has a unique factorization into prime ideals. I would like to show that for ideals $I,J$ of a Dedekind domain $R$, "contains means divides"|that is, that $I\subseteq J$ iff $I=JK$ for some ideal $K$ of $R$|from which it will follow that non-0 prime ideals are maximal (my ultimate goal).

The approach I was taking was to attempt to show that if $I,J$ are non-0 ideals of a Dedekind domain and $IJ=I\cap J$, then $I+J=R$. From that, I believe I can show that for any non-0 proper ideals $I,J$ of $R$, we have $IJ=(I\cap J)(I+J)$|a la "product = lcm times gcd"|which should do the trick, I think.

I'm trying not to detour through equivalent definitions of Dedekind domain to obtain this result (since I'd have to prove such an equivalence, first), but all of the discussions I've been able to find either just list it as a result or use a different definition.

Can anyone give me a sketch for my approach, or give me an idea for an easier approach to take (given the definition I'm working with for Dedekind domain)?


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Accounts have been merged. Should this question also be merged as an edit? – Zev Chonoles Apr 12 '12 at 3:12
Do you know about invertible ideals, or do you not have that tool available for this problem? – Rankeya Apr 12 '12 at 3:19
@ZevChonoles: Thanks! I will go ahead and keep the questions separate, though. – Cameron Buie Apr 12 '12 at 3:20
@Rankeya: I do know about them, from looking into this online. I haven't actually proved any results about them, though. Do you think it would be easier to go that route? – Cameron Buie Apr 12 '12 at 3:21
@CameronBuie: Look at this link…. Arturo Magidin gave a complete answer here. – Rankeya Apr 12 '12 at 3:33
up vote 3 down vote accepted

Once you take a look at A sufficient condition for a domain to be Dedekind?, you will know that every fractional ideal of a Dedekind domain (where the definition of a Dedekind domain is the one you gave) is invertible.

Let $R$ be your Dedekind domain. Then note that $I \subset J$ implies that $IJ^{-1} \subset R$. Now $IJ^{-1}$ is an ideal of $R$. Then clearly, $I = (IJ^{-1})J$, and you are done. The other implication is not hard to prove.

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Fantastic! Thanks very much! – Cameron Buie Apr 12 '12 at 3:56

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