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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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  • $\begingroup$ The Minkovski bound is quite high... Are you expecting to do it by hand ? $\endgroup$ – user10676 Nov 15 '11 at 14:35
  • $\begingroup$ I know and that's what's made me post it here of course :D. Maybe there's some slick way without much casework... $\endgroup$ – Anna Nov 15 '11 at 15:13
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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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    $\begingroup$ Many thanks for the very informative answer! $\endgroup$ – Anna Dec 10 '11 at 21:26
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Take the cyclotomic integer $g(\zeta)=-1-\zeta^{3}-\zeta^{5}+\zeta^{6}-\zeta^{14}-\zeta^{17}$. The ideal $\langle g(\zeta)\rangle$ factorizes to $\langle g(\zeta)\rangle=\langle47, \psi_{47}(\zeta)\rangle^{3}$. The number $\psi_{47}(\zeta)$ is the second generator of the prime ideal that divides the prime integer $47$. Edwards proves in his marvelous book Fermat’s last theorem in chapter 4.4 that the prime ideal $\langle47, \psi_{47}(\zeta)\rangle$ is not principal. Because the class order of the prime ideal $\langle47, \psi_{47}(\zeta)\rangle$ divides the exponent n of any ideal for which the ideal $\langle47, \psi_{47}(\zeta)\rangle^n$ is principal the class order of the prime ideal must be $3$. Because the class order divides the order of the finite class group the integer $3$ divides the order of the class group.

Now take the homomorphism $\sigma: \zeta\to\zeta^5$ with the primitive root $5$ modulo $23$. The cyclotomic integer $h(\zeta)=-1-\zeta-\zeta^{7}-\zeta^{15}+\zeta^{17}-\zeta^{19}$ factorizes to $\langle h(\zeta)\rangle=\langle47, \psi_{47}(\zeta)\rangle^{2}\cdot\langle47, \sigma^{2}\psi_{47}(\zeta)\rangle$. Therefore the prime ideals $\langle47, \psi_{47}(\zeta)\rangle$ and $\langle47, \sigma^{2}\psi_{47}(\zeta)\rangle$ are in the same class or $\langle47, \psi_{47}(\zeta)\rangle\sim\langle47, \sigma^{2}\psi_{47}(\zeta)\rangle$. Consecutively applying the homomorphism $\sigma^2$ gives $\langle47, \psi_{47}(\zeta)\rangle\sim\langle47, \sigma^{2w}\psi_{47}(\zeta)\rangle$. The cyclotomic integer $k(\zeta)=\zeta^{5}+\zeta^{6}+\zeta^{9}+\zeta^{14}+\zeta^{17}+\zeta^{18}$ factorizes to $\langle k(\zeta)\rangle=\langle47, \psi_{47}(\zeta)\rangle\cdot\langle47, \sigma^{11}\psi_{47}(\zeta)\rangle$ and we get $\langle47, \psi_{47}(\zeta)\rangle^2\sim\langle47, \sigma^{11}\psi_{47}(\zeta)\rangle$. Therefore the prime ideals $\langle47, \sigma^{u}\psi_{47}(\zeta)\rangle$ divide into the two classes $\langle47, \sigma^{2w}\psi_{47}(\zeta)\rangle$ and $\langle47, \sigma^{2w+1}\psi_{47}(\zeta)\rangle$. Together with the principal class we have these three classes and a clear outline of the class group of the $23^{\text{rd}}$ cyclotomic ring of integers. The calculation with classes is explained in chapter $5$ of Edwards.

Neither the factorization nor the the generator $\psi_{47}(\zeta)$ come out of the blue. We now leave Dedekind’s theory of factorizing into prime ideals and return to Kummer. Kummer paved Dedekind the way to his theory of uniquely factorizing ideals to prime ideals by finding out that the $p^{\text{th}}$ cyclotomic ring of integers, p a prime, can uniquely be factorized to (Edwards’s term of) divisors. Divisors are principally the same as prime ideals though they are closely connected to cyclotomic integers. Edwards introduces divisors in chapter $4.12$ to the greatest common divisor of the two cyclotomic integers $(47, \psi_{47}(\zeta))$. If and only if the prime ideal $\langle47, \psi_{47}(\zeta)\rangle$ is principal a cyclotomic integer exists that has the divisor $(47, \psi_{47}(\zeta))$.

Kummer found the number $\Psi_{47}(\zeta)$ (see chapter $4.9$, Edwards, and mind the capital letter $\Psi$, the generator above has the small letter $\psi$). This number has the quality that the cyclotomic integer $r(\zeta)$ is divisible n times by the divisor $(47, \psi_{47}(\zeta))$ if $47^n$ divides $r(\zeta)\cdot\Psi_{47}^n(\zeta)$ and $47^{n+1}$ does not divide $r(\zeta)\cdot\Psi_{47}^{n+1}(\zeta)$ (see chapter $4.10$, Edwards). In the terms above the prime ideal $\langle 47, \psi_{47}(\zeta)\rangle$ divides the principal ideal $\langle g(\zeta)\rangle$ three times because $47^3$ divides $g(\zeta)\cdot\Psi_{47}^3(\zeta)$ and $47^4$ does not divide $g(\zeta)\cdot\Psi_{47}^4(\zeta)$. The generator can be calculated to

$$\psi_{47}(\zeta)=\sigma^1\Psi_{47}(\zeta)+\sigma^2\Psi_{47}(\zeta)+\dots+\sigma^{21}\Psi_{47}(\zeta).$$

If the generator has multiplicity of more than $1$ with respect to the prime ideal $\langle 47, \psi_{47}(\zeta)\rangle$ Edwards adds the integer $47$ so that it has multiplicity $1$ (see chapter $4.12$, Edwards). The number $\Psi_{q}(\zeta)$ can easily be computed though for large p or q this calculation needs some optimization.

The procedure allows to outline the class group of any $p^{\text{th}}$ cyclotomic ring of integers for a prime p. Together with Kummer’s class number formula given in chapter $6$ of Edwards you can even determine the class group. As stated in an answer above you normally have to determine the classes of a high number of ideals in order to determine the class group of a cyclotomic rings of integers. But having found all classes according to Kummer’s class number formula you can stop pretty soon. The only obstacle that remains is therefore Kummer’s class number formula that demands thorough understanding of complex analysis in $\mathbb C^n$.

The procedure of determinating the class group of cyclotomic rings of integers by factorizing cyclotomic integers as demonstrated above has been done for a couple of cyclotomic rings of integers. The result can be taken from here. The data that was used in order to compile this document can be taken from here.

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