Thread: Ideal norms in a pid

Is it true that if two ideals in a pid have the same norm then they are in fact the same ideal (even if being generated by different elements)?

For example, I have a problem asking me to find all ideal in a certain integer ring with a given norm. But if the above is correct, then I only have to find ONE ideal with that norm, since all others will in fact be the same ideal. Correct?

Is it true that if two ideals in a pid have the same norm then they are in fact the same ideal

No. Consider the Gaussian integers Z[i]. This is a PID (indeed it is a Euclidean domain under the square of complex absolute value). Now 2+i and 2-i generate different ideals both of norm 5.

So that means if I have to find all ideals with norm 18 in Z[sqrt(2)] I need to find all solutions to a^2-2b^2=18?

Attempting to solve this the long way, I found the first five or so values in this ring with norm +/- 18, and they all generate the same ideal. So what's up with that? Why would that be true in this specific case?

Two elements generate the same principal ideal iff one is a unit times the other. In this case Z[sqrt 2] has infinitely many units, generated by 1+sqrt 2. In an algebraic number ring there are only finitely many ideals with a given norm, even though there may be infinitely many elements with that norm.

In this case there is just one ideal of norm 18, generated by 3sqrt(2).

I finally figured it out ... i factored the ideal <18> into prime ideals, and realized that any ideal with norm 18 had to divide this ideal and thus be some arrangement of the prime factors. In this case there was only one arrangement that had norm 18, the infamous 3sqrt(2).

Thanks for your help.

There is a theorem that the maximal order (the ring of integers) of an algebraic number field is a Dedekind domain, that is, every ideal can be expressed uniquely as a product of prime ideals.