Class number

Let be an algebraic number field. Then its class number is the order of the (always finite) ideal class group of . ${\displaystyle K}$ ${\displaystyle h_{K}}$ ${\displaystyle K}$

Number theoretic meaning

If you want to solve an equation over a number field, one possible strategy is to solve the equation over the ideal group and the ideal class group . 1 is the only solution to the ideal class group, so every ideal is having a principal ideal: . This number solves the original equation modulo units . ${\displaystyle F(x)=1}$ ${\displaystyle I_{K}}$ ${\displaystyle Cl_{K}}$ ${\displaystyle {\mathfrak {a}}}$ ${\displaystyle F({\mathfrak {a}})=1}$ ${\displaystyle {\mathfrak {a}}=(\alpha )}$ ${\displaystyle \alpha }$

To solve the equation above , it is sufficient to know the structure of as an Abelian group. In most cases it is sufficient to know the prime factorization of . (e.g. for , or: if .) ${\displaystyle Cl_{K}}$ ${\displaystyle Cl_{K}}$ ${\displaystyle h_{K}}$ ${\displaystyle x^{n}=1\Rightarrow x=1}$ ${\displaystyle (n,h_{K})=1}$ ${\displaystyle x^{n}=1}$ ${\displaystyle h_{K}|n}$

For this reason, the determination of the ideal class number is one of the central tasks of number theory.

Example circle division body and Fermat's conjecture

In the early attempts to prove Fermat's conjecture , it was tacitly assumed that the fields of circular division important for this problem (with the respective exponent in the Fermat equation and the -th root of unit ) had a clear prime factorization (class number 1), which was refuted by Ernst Eduard Kummer . Kummer introduced new algebraic objects, the ideals , and was thus able to save the evidence for a large class of circle-dividing fields by moving from calculating with the algebraic numbers themselves to calculating with the subsets of the numbers of the algebraic number field that made up the ideals. The cyclotomic fields, for which he was able to prove Fermat's last theorem had one that a regular prime figured they called the class number of cyclotomic not share: . ${\displaystyle \mathbb {Q} (\zeta _{p})}$ ${\displaystyle p}$ ${\displaystyle \zeta _{p}}$ ${\displaystyle p}$ ${\displaystyle p}$ ${\displaystyle p\nmid h_{\mathbb {Q} (\zeta _{p})}}$

The special case of the Fermat's conjecture was then: Let be an odd regular prime number. Then the equation has no integer solutions. ${\displaystyle p}$ ${\displaystyle x^{p}+y^{p}=z^{p},\quad (xyz,p)=1}$

Evidence sketch: The equation can be rewritten to . If one now goes over to the ideals of , one obtains the equations, since the ideals on the left are coprime . Since the mapping on the ideal class group of is injective, the equations are obtained from it with a unit that can lead to a contradiction. ${\displaystyle \prod _{i=0}^{p-1}(x+\zeta _{p}^{i}y)=z^{p}}$ ${\displaystyle \mathbb {Q} (\zeta _{p})}$ ${\displaystyle x+\zeta _{p}^{i}y={\mathfrak {a}}^{p}}$ ${\displaystyle x\mapsto x^{p}}$ ${\displaystyle \mathbb {Q} (\zeta _{p})}$ ${\displaystyle x+\zeta _{p}^{i}y=\epsilon \cdot \alpha ^{p}}$ ${\displaystyle \epsilon }$

A regular prime number can also be defined using Bernoulli numbers :

${\displaystyle p|h_{K}\Leftrightarrow p|B_{j}}$ for a ${\displaystyle j\in \{2,4,\ldots ,p-3\}}$

Be . Then: ${\displaystyle n>0}$ ${\displaystyle p|h_{\mathbb {Q} (\zeta _{p})}\Leftrightarrow p|h_{\mathbb {Q} (\zeta _{p^{n}})}}$

Example of imaginary-quadratic number fields and Gaussian class number problem

There are exactly 9 so-called Heegner numbers , for which the class number has: and . They represent the solution of the Gaussian class number problem for imaginary-quadratic number fields, the question of which imaginary-quadratic number fields have the class number 1, that is, unambiguous prime factorization. The solution comes from Kurt Heegner . ${\displaystyle d\in \mathbb {N} }$ ${\displaystyle K=\mathbb {Q} ({\sqrt {-d}})}$ ${\displaystyle h_{K}=1}$ ${\displaystyle d=1,2,3,7,11,19,43,67}$ ${\displaystyle 163}$

properties

• Class number formula : The following applies to the class number : ${\displaystyle h_{K}}$

${\displaystyle \lim _{s\to 1}(s-1)\zeta _{K}(s)={\frac {2^{r_{1}}\cdot (2\pi )^{r_{2}}\cdot h_{K}\cdot \operatorname {Reg} _{K}}{w_{K}\cdot {\sqrt {\mid D_{K}\mid }}}}}$

Here, the number of roots of unity in , the discriminant of the extension and the regulator of . ${\displaystyle w_{K}}$ ${\displaystyle K}$ ${\displaystyle D_{K}}$ ${\displaystyle K/\mathbb {Q} }$ ${\displaystyle \operatorname {Reg} _{K}}$ ${\displaystyle K}$

The class number formula is suitable for the practical calculation of the class number.

• Be an -extension, i.e. H. and . Be the fraction of the class number . Then there is of independent natural numbers , , such that for sufficiently large . (See: Iwasawa theory ) ${\displaystyle K|k}$ ${\displaystyle \mathbb {Z} _{p}}$ ${\displaystyle k=k_{0}\subset k_{1}\subset \cdots \subset K}$ ${\displaystyle G(k_{n}|k)\cong \mathbb {Z} /p^{n}\mathbb {Z} }$ ${\displaystyle p^{e_{n}}}$ ${\displaystyle p}$ ${\displaystyle h_{k_{n}}}$ ${\displaystyle n}$ ${\displaystyle \lambda }$ ${\displaystyle \mu }$ ${\displaystyle \nu }$ ${\displaystyle e_{n}=\lambda n+\mu p^{n}+\nu }$ ${\displaystyle n}$
• Vandiver's conjecture (not generally proven, forverified): ${\displaystyle p<12\cdot 10^{6}}$

Be . Then there is no divisor of . ${\displaystyle K^{+}:=\mathbb {Q} (\zeta _{p})^{+}=\mathbb {Q} (\zeta _{p})\cap \mathbb {R} }$ ${\displaystyle p}$ ${\displaystyle h_{K^{+}}}$

See also

• Relative class number

literature

• Jürgen Neukirch : Algebraic number theory . Springer, Berlin et al. 1992, ISBN 3-540-54273-6 .
• Lawrence C. Washington Introduction to Cyclotomic Fields (= Graduate Texts in Mathematics. Vol. 83). 2nd edition. Springer, New York NY 1997, ISBN 0-387-94762-0 .