Checking principality again

Hi,

I was wondering how you work out when an ideal is principal. I understand that a principal ideal is generated by a single element, but how do you work this out. For example in the remarks, you mentioned (5,2^{1/3}-3) was the principal ideal (2(2^{2/3})+2^{1/3}+2). But where does this come from? (5,2^{1/3}-3) is elements a5+b(2^{1/3}-3) with a,b \in Z[2^{1/3}]. I tried to substitute x+y2^{1/3} for a and b, but I’m not sure if this is right, and if it is, where do the specific numbers come from. Thanks for your help.

Reply:

Checking for principality is not always easy, although it is sometimes. For example, when you factorize the ideal (2) in the ring Z[\alpha] where \alpha is a root of x^3-4x+2, you’ll see that (2)=P_2^2 where P_2=(2, \alpha). But clearly, 2=4\alpha-\alpha^3, so that \alpha by itself is a generator of P_2.

One general rule that comes up often is that for a principal ideal I=(a), we have N(I)=|N(a)|. From this, it is easy to deduce that

I is principal if and only if there is an element a \in I such that |N(a)|=I.

You will find a number of examples of this flavor in the note `Some principal ideals’ . The example you mention has a misprint. The actual generator for the ideal is 2^{2/3}+1. This is not so straightforward, and is worked out on page two of `Some sample class groups’.

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