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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?

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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
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up vote 5 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.

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Many thanks for the very informative answer! –  Anna Dec 10 '11 at 21:26
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