MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As we known, the integer ring R of 256-th cyclotomic field is not a Principal Ideal Domain.

And rational prime 257 is split completely in R. Suppose prime ideal P of R is an arbitrary ideal lying over 257.

My question: Is P a principal ideal in R? If it is, how to find one of its generator? Or, is there an algorithm to obtain its generator?

Thank you very very much!

share|cite|improve this question
@Dietrich: The problem with $P=(257)$ is that it is very far from being prime... – Filippo Alberto Edoardo Jun 28 '13 at 16:43
@Filippo: oh, sorry, I know that $P$ splits completely. – Dietrich Burde Jun 28 '13 at 17:47
up vote 9 down vote accepted

Let $K/\mathbf{Q}$ be the degree $32$ field of $64$th roots of unity, and let $L/K$ be the field of $256$th roots of unity. The class number of $K$ is $17$, and the degree of $K$ is just small enough to be able to compute the unit group, etc. According to pari, if $\mathfrak{p}$ is a prime of norm $257$ in $K$, then $\mathfrak{p}$ is not principal. On the other hand, if $\mathfrak{P}$ above $\mathfrak{p}$ of norm $257$ in $L$ was principal, then so would $N_{L/K}(\mathfrak{P}) = \mathfrak{p}^4$, which is a contradiction because $\mathfrak{p}$ has order $17$ in the class group.

This gives the answer to this particular problem, but if you want to understand how the computation actually works, there is an (extensive) theory of computing units and class groups in the literature: a good starting point is Henri Cohen's book A course in computational algebraic number theory.

There are general algorithms to solve generalizations of your problems, but unfortunately, they are known to be computationally difficult, and one was just fortunate that there was a trick to reduce to a smaller field in this case. We don't even know the class number of $\mathbf{Q}(\zeta_{71})$ without using GRH (we do, of course, know the plus part). Also, if $\mathfrak{p}$ had turned out to be principal, that wouldn't have said anything about $\mathfrak{P}$. There's actually a subfield $E = \mathbf{Q}(\zeta - \zeta^{-1})$ of degree $16$ which also has class number $17$, but the primes $\mathfrak{q}$ above $257$ in $E$ turn out to be principal.

share|cite|improve this answer
very helpful anwser, thank you! – huangdd Jun 30 '13 at 2:43
By the way, I do with a small correction in the first paragraph of the answer above. Here N_{L/K}(\mathfrak{P})=\mathfrak{p} because of the relative degree of \mathfrak{P} over \mathfrak{p} is 1. – huangdd Jul 1 '13 at 8:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.