3 Transcendental computation of the Class number

We first need to introduce the Dedekind zeta function for a number field K, and its Euler product expansion

$$\zeta_{K}^{}$$(s) = $$\sum_{I}^{}$$$${\frac{1}{N(I)^s}}$$ = $$\prod_{Q}^{}$$$${\frac{1}{(1-{\frac{1}{N(Q)^s}})}}$$

where the sum runs over all ideals I of R and the product runs over all prime ideals Q of R. The two expressions give us two ways of computing $\lim_{s\to 1}^{}$(s - 1)$\zeta_{K}^{}$(s). The left-hand side is expressed in terms of ``arithmetic'' invariants and the right-hand side in terms of invariants for the Galois group. The resulting identity gives a way for computing the Class number h of K. The left-hand limit can be computed to be

$$\lim_{s\to 1}^{}$$(s - 1)$$\sum_{I}^{}$$$${\frac{1}{N(I)^s}}$$ = $$\lim_{r\to\infty}^{}$$$${\frac{\char93 \{I\mid N(I)\leq r\}}{r}}$$

The set {I | N(I) $\leq$ r} can be split according to ideal classes. We try to compute for each ideal class C,

z(C) = $$\lim_{r\to\infty}^{}$$$${\frac{\char93 \{I\in C\mid N(I)\leq r\}}{r}}$$.

Fixing an ideal I₀ $\in$ C, this latter set is bijective to the set {aR $\subset$ I₀^-1 | N(a) $\leq$ r ^. N(I₀)^-1}. (Here N(a) denotes the modulus of the norm of a.) We have a natural embedding K $\hookrightarrow$ K $\otimes_{{\Bbb Q}}^{}$ $\Bbb$R. The image of J = I₀^-1 is a lattice in K $\otimes_{{\Bbb Q}}^{}$ $\Bbb$R. Let $\Lambda$ denote the image of J - {0} in the quotient $\cal {S}$ = (K $\otimes_{{\Bbb Q}}^{}$ $\Bbb$R)^*/U where U is the image of the group of units in R under the above embedding. There is a natural homomorphism N : $\cal {S}$$\to$$\Bbb$R^* which restricts to the modulus of the norm on the image of K. We obtain a natural bijection between {aR $\subset$ I₀^-1 | N(a) $\leq$ r} and {l $\in$ $\Lambda$ | N(l ) $\leq$ r}. Let $\Lambda_{r}^{}$ denote the image of (1/r)J - {0} in $\cal {S}$, then we have a natural bijection between {l $\in$ $\Lambda$ | N(l ) $\leq$ r^d} and {l $\in$ $\Lambda_{r}^{}$ | N(l ) $\leq$ 1}, where d denotes the degree of K oover $\Bbb$Q. Let $\cal {S}$_{$\leq$ 1} denote locus of l $\in$ $\cal {S}$ such that N(l ) $\leq$ 1. Let $\mu$ denote the Haar measure on K $\otimes_{{\Bbb Q}}^{}$ $\Bbb$R. This is invariant under the action of U and thus gives a measure also denoted by $\mu$ on $\cal {S}$. Since J is a lattice in K $\otimes_{{\Bbb Q}}^{}$ $\Bbb$R we have

$$\lim_{r\to\infty}^{}$$$${\frac{\char93 \{l\in\Lambda_r\mid N(l)\leq 1\} }{r^d}}$$ = $${\frac{\mu({\cal S}_{\leq 1})}{\mu(K\otimes _{{\Bbb Q}}{\Bbb R}/J)}}$$

Moreover, the denominator can be re-written

$$\mu$$(K $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R/J) = N(J)$$\mu$$(K $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R/R).

In particular, we see that the limit z(C) is independent of the class C. Let (K $\otimes_{{\Bbb Q}}^{}$ $\Bbb$R)^*₁ denote the kernel of the norm map. This is a group and thus we have a Haar measure $\nu$ on it. One shows that

$$\mu$$($$\cal {S}$$_{$\leq$ 1}) = $$\nu$$((K $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R)^*₁/U)

Combining the above calculations one obtains

$$\lim_{s\to 1}^{}$$(s - 1) ^. $$\zeta_{K}^{}$$(s) = h ^. $${\frac{\nu((K\otimes _{{\Bbb Q}}{{\Bbb R}})^{*}_1 / U)}{\mu(K\otimes _{{\Bbb Q}}{\Bbb R}/R)}}$$

This often called the ``Class number formula'' for K. Note that the denominator can be computed in closed form in terms of the discriminant D of the field K and the number of pairs of conjugate complex embeddings r₂ of K.

$$\mu$$(K $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R/R) = $${\frac{1}{2^{r_2}}}$$ ^. $$\sqrt{\vert D\vert}$$

However, the numerator is in general more complicated since it involves computing the group of units of K. To expand the right-hand term we restrict our attention to abelian extensions K of $\Bbb$Q. The product term on the left can be first grouped according to rational primes

$$\prod_{Q}^{}$$$${\frac{1}{(1-{\frac{1}{N(Q)^s}})}}$$ = $$\prod_{q}^{}$$$$\prod_{Q\mid q}^{}$$$${\frac{1}{(1-{\frac{1}{N(Q)^s}})}}$$

Now for each rational prime q which is unramified in K we have

$$\prod_{Q\mid q}^{}$$$${\frac{1}{(1-{\frac{1}{N(Q)^s}})}}$$ = $$\prod_{\chi}^{}$$$${\frac{1}{(1-{\frac{\chi(q)}{q^s}})}}$$

where $\chi$ runs over all characters of the Galois group and $\chi$(q) = $\chi$(Frob_q) is the value of $\chi$ on a Frobenius element associated with q. We define the Dirichlet L-series and their Euler product formulas as follows

L(s,$$\chi$$) = $$\sum_{n}^{}$$$${\frac{\chi(n)}{n^s}}$$ = $$\prod_{p}^{}$$$${\frac{1}{(1-{\frac{\chi(p)}{p^s}})}}$$

where we set $\chi$(p) = 0 when $\chi$ is ramified at p. We also define the additional factor

F(s) = $$\prod_{p \text{ ramified}}^{}$$$${\frac{1}{(1-{\frac{1}{p^{f_p}}})^{g_p}}}$$

where the product runs over all ramified primes and f_p denotes the residue field extension over p and g_p the number of distinct primes in K lying over p. The product expansion of $\zeta_{K}^{}$(s) becomes

$$\zeta_{K}^{}$$(s) = F(s) ^. $$\prod_{\chi}^{}$$L(s,$$\chi$$).

Thus the computation of the limit can be reduced to the corresponding computation for the Dirichlet L-series. For the case of the unit character we get by comparison with the zeta function

$$\lim_{s\to 1}^{}$$(s - 1)F(s)L(s, 1) = 1.

So the right-hand limit gives

$$\lim_{s\to 1}^{}$$(s - 1)$$\zeta_{K}^{}$$(s) = $$\prod_{\chi\neq 1}^{}$$L(1,$$\chi$$).

There is a positive integer m such that $\chi$ is determined on classes modulo m and $\chi$ is zero on all primes p dividing it; m is called the conductor of $\chi$. We rewrite the L-function associated with $\chi$ as follows

L(s,$$\chi$$) = $$\sum_{x\in ({\Bbb Z}/{m{\Bbb Z}})^{*}}^{}$$$$\left(\vphantom{ \chi(x) \cdot \sum_{n\cong x \pmod m} {\frac{1}{n^s}} }\right.$$$$\chi$$(x) ^. $$\sum_{n\cong x \pmod m}^{}$$$${\frac{1}{n^s}}$$$$\left.\vphantom{ \chi(x) \cdot \sum_{n\cong x \pmod m} {\frac{1}{n^s}} }\right)$$

The latter sum can be rewritten using the identity

$$\sum_{i=0}^{m-1}$$$$\omega^{xi}_{}$$ = $$\begin{cases} 0, \text{ if } x \not\cong 0 {\pmod m} \\ m, \text{ if } x \cong 0 {\pmod m} \end{cases}$$

where $\omega$ is a primitive m-th root of unity. The second sum then becomes

$$\sum_{n\cong x \pmod m}^{}$$$${\frac{1}{n^s}}$$ = $${\frac{1}{m}}$$$$\sum_{n=1}^{\infty}$$$${\frac{1}{n^s}}$$$$\sum_{i=0}^{m-1}$$$$\omega^{(x-n)i}_{}$$

Thus we obtain

L(s,$$\chi$$) = $${\frac{1}{m}}$$$$\sum_{i=0}^{m-1}$$$$\left(\vphantom{ \sum_{x\in ({\Bbb Z}/{m{\Bbb Z}})^{*}} \chi(x) \omega^{ix} }\right.$$$$\sum_{x\in ({\Bbb Z}/{m{\Bbb Z}})^{*}}^{}$$$$\chi$$(x)$$\omega^{ix}_{}$$$$\left.\vphantom{ \sum_{x\in ({\Bbb Z}/{m{\Bbb Z}})^{*}} \chi(x) \omega^{ix} }\right)$$ ^. $$\sum_{n=1}^{\infty}$$$${\frac{\omega^{-in}}{n^s}}$$

The expression

$$\tau_{i}^{}$$($$\chi$$) = $$\sum_{x\in ({\Bbb Z}/{m{\Bbb Z}})^{*}}^{}$$$$\chi$$(x)$$\omega^{ix}_{}$$

is called the Gaussian sum associated with the integer i and the character $\chi$. If $\chi$ is not the unit character then $\tau_{0}^{}$($\chi$) = 0. Moreover, if i $\neq$ 0 then we have the identity

$$\sum_{n=1}^{\infty}$$$${\frac{\omega^{-in}}{n}}$$ = - log(1 - $$\omega^{-i}_{}$$)

Hence, we obtain the formula when $\chi$ is not the unit character

L(1,$$\chi$$) = - $${\frac{1}{m}}$$$$\sum_{i=1}^{m-1}$$$$\tau_{i}^{}$$($$\chi$$) ^. log(1 - $$\omega^{-i}_{}$$)