Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How does one prove that the class number of $\mathbb{Q}(\zeta_{23})$ is divisible by $3$? And afterwards how do you show that it is precisely $3$. Any help?

Thanks in advance!

//Ok, so I proved the divisibility (I was really tired to ask this for hints for that I guess). What about the equality?

share|improve this question
The Minkovski bound is quite high... Are you expecting to do it by hand ? –  user10676 Nov 15 '11 at 14:35
I know and that's what's made me post it here of course :D. Maybe there's some slick way without much casework... –  Anna Nov 15 '11 at 15:13
add comment

1 Answer

up vote 4 down vote accepted

As you likely discovered by the time of your edit, divisibility is pretty straight-forward. By class field theory, the class group of $\mathbb{Q}(\zeta_{23})$ surjects onto that of $\mathbb{Q}(\sqrt{-23})$, which has class number 3 by a (comparatively) easy calculation. So voila! Divisibility.

Finding class numbers of cyclotomic fields in in generally a very tough problem. But for $p=23$, the single smallest non-trivial case, things aren't soooo horrendous. As I describe below, the worst of the computation comes from the real cyclotomic subfield. So even though SAGE stalls at a direct attempt to find the class number of $\mathbb{Q}(\zeta_{23})$ (without assuming, say, GRH, etc.), it could eventually be pieced together as follow:

  • The Minkowski bound for $\mathbb{Q}(\zeta_{23}+\zeta_{23}^{_1})$ is a mere 900, as opposed to 9 million or so for $\mathbb{Q}(\zeta_{23})$. A brute forces factorization of primes in that range concludes that the real cyclotomic field has class number 1.

  • Kummer's formula for the relative class number: $$ h_{23}^-:=\frac{h(\mathbb{Q}(\zeta_{23})}{h(\mathbb{Q}(\zeta_{23}+\zeta_{23}^{-1}))}=-\frac{23}{2^{10}}\prod_{1\leq k\leq \frac{p-1}{2}} B_{1,\omega^{2k+1}} $$ evaluates to 3.

Neither of these could be done in under a few minutes by hand, but you could do it if you were stranded on a desert island and had to kill some time. In any case, once they're accomplished, we put them together to get $$ h_{23}=h_{23}^+h_{23}^-=3\cdot 1=3. $$ This probably isn't even the most efficient approach (though I don't think anything as slick as Odlyzko bounds will apply) -- the 1982 paper "Class Number Computations of Real Abelian Number Fields" by van der Linden establishes a lot of these small real cyclotomic class numbers with minimal computational power (but a lot of work!).

For a more up-do-date state-of-the-affairs, see Schoof's 2002 article "Class Numbers of Real Cyclotomic Fields of Prime Conductor," especially for its very clear exposition of the computational difficulties (which end up being linear-algebraic-theoretic...Jordan-Hölder factors of the groups of units modulo cyclotomic units, viewed as a module over the group ring of the real cyclotomic Galois group). Worse, it's not even an "asymptotic" problem in the sense that our algorithms become inefficient only for increasingly large $p$. As of Schoof's massive calculation in 2002 cited above, we don't know a single one of these $h(K^+)$'s for sure for $p\geq 71$, and only get up to $p=163$ under the assumption of GRH.

share|improve this answer
Many thanks for the very informative answer! –  Anna Dec 10 '11 at 21:26
add comment

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.