## 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?

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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.

$N(z)=9$
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.