Norms of elements and principality

Dear Professor Kim,

Sorry to bombard you with these questions. I have come across a problem on your note ‘some principle ideals’. When we factorize m(x)=x^3 + x - 1 modulo 3 we get (x+1)(x^2-x-1) we then associate these factors with the ideals P_3 and P_9 respectively. When we compute the norm of x^2-x-1 we do so by calculating the determinant of the matrix L_{a^2-a-1} and find that the norm is in fact 9, so P_9 is a principle ideal. However, we could just have easily used x^2+2x-1 or x^2+2x+2 and in each case I get a different answer for the determinant. Have I made an error or is there a canonical form of sort that I should be aware of?

Thank you for your time.

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Reply:

First of all, I presume your x^2-x-1 etc. are a^2-a-1 etc. All the elements you mention do indeed belong to the ideal and can be used as generators *when used together with the element 3*. Indeed they are all all evaluations at a of polynomials that are congruent to x^2-x-1 mod 3. However, this does not mean they are generators on their own. Of course different elements in an ideal I will have different norms in general. However, an element b\in I is a generator *by itself* (making I into a principal ideal), exactly when |N(b)|=N(I). Of course such a b need not exist. I haven’t calculated the norms of the elements you mention, but if their norms come out larger than 9, it merely says they are not generators (again, by themselves), while a^2-a-1 is.

A thorny point that comes out of this discussion is that if you had initially presented the ideal as (3, a^2+2a-1), for example, then it might have been harder to see that it is principal.

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2 Comments

  1. rory
    Posted May 14, 2009 at 7:31 pm | Permalink | Reply

    Thanks for the feedback. So, if I understand you correctly, given a situation where we are checking for principality of an ideal, but we don’t find a generator with our first presentation, our only option would be to exhaust all the other possibilities?

  2. Posted May 15, 2009 at 8:59 am | Permalink | Reply

    `Exhaust other possibilities’ sounds exhausting. But recall that the equation

    N(z)=9

    for an unknown element z in our ring is not too difficult to analyze, at least in many examples. If there are no solutions at all, then a fortiori there are no solutions in our ideal. So the ideal would not be principal. However, if there is a solution, you would still need to check if there is a solution in the ideal. This can be somewhat trickier.

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