# Is the prime ideal principal which is in 256-th cyclotomic ring lying over 257?

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!

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@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

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.
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.