Class Number Formula

The class number formula is one of the beautiful results in number theory. It connects the arithmetic of a number field with the behavior of an analytic function at $s=1$ . On the arithmetic side stand the class number, the units, the regulator, the discriminant, and the roots of unity. On the analytic side stands the Dedekind zeta function, a Dirichlet series counting ideals by norm. The theorem says that the failure of unique factorization is visible in the residue of a zeta function.

For a number field $K$ , let $\mathcal O_K$ be its ring of integers. The Dedekind zeta function of $K$ is defined by

$\displaystyle \zeta_K(s)=\sum_{\mathfrak a\neq0}\frac{1}{N\mathfrak a^s}, \quad \text{Re}(s)>1,$

where the sum is over nonzero integral ideals of $\mathcal O_K$ . If the number of ideals of norm $n$ is $a_n$ then

$\displaystyle \zeta_K(s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}.$

Thus $\zeta_K(s)$ is an ideal-counting function. Its pole at $s=1$ measures the main term in the growth of the number of ideals of bounded norm. The analytic class number formula states that this residue is

$\displaystyle \text{Res}_{s=1}~\zeta_K(s) =~~ \frac{2^{r_1}(2\pi)^{r_2}h_KR_K}{w_K\sqrt{|D_K|}}.$

Here $r_1$ is the number of real embeddings of $K$ , $r_2$ is the number of pairs of complex embeddings, $h_K$ is the class number, $R_K$ is the regulator, $w_K$ is the number of roots of unity in $K$ , and $D_K$ is the discriminant.

The formula becomes much clearer for quadratic fields. There the Dedekind zeta function factors into the ordinary Riemann zeta function and a Dirichlet $L$ -function. The class number formula then becomes a formula for $L(1,\chi)$ , where $\chi$ is a quadratic character.

We will prove the formula in classical language.

Quadratic Fields

Let $\displaystyle K=\mathbb Q(\sqrt d),$ where $d$ is squarefree and $d\neq1$ . The ring of integers is

$\displaystyle \mathcal O_K= \begin{cases} \mathbb Z[\sqrt d], & d\equiv2,3\pmod4,\\ \mathbb Z\big [\frac{1+\sqrt d}{2}\big ], & d\equiv1\pmod4. \end{cases}$

The discriminant is

$\displaystyle D= \begin{cases} 4d, & d\equiv2,3\pmod4,\\ d, & d\equiv1\pmod4. \end{cases}$

This $D$ is a fundamental discriminant. Attached to it is the quadratic Dirichlet character

$\displaystyle \chi_D(n)=\big(\frac{D}{n}\big),$

where the symbol is the Kronecker symbol. For a rational prime $p\nmid D$ , the value of $\chi_D(p)$ determines how $p$ behaves in $\mathcal O_K$ :

$\displaystyle \chi_D(p)=1 \quad\Longleftrightarrow\quad p \text{ splits in } K,$

$\displaystyle \chi_D(p)=-1 \quad\Longleftrightarrow\quad p \text{ is inert in } K.$

If $p\mid D$ , then $p$ ramifies and $\chi_D(p)=0$ . Thus $\chi_D$ records the splitting of primes in the quadratic field.

For a quadratic field, the Dedekind zeta function factors as

$\displaystyle \zeta_K(s)=\zeta(s)L(s,\chi_D),$

where the Dirichlet L-function $L(s,\chi_D)$ is given

$\displaystyle L(s,\chi_D)=\sum_{n=1}^{\infty}\frac{\chi_D(n)}{n^s}.$

Let us prove this directly from Euler factors. For $\text{Re}(s)>1$ , the Dedekind zeta function has an Euler product over prime ideals:

$\displaystyle \zeta_K(s)=\prod_{\mathfrak p} \big(1-\frac{1}{N\mathfrak p^s}\big)^{-1}.$

Now group the prime ideals $\mathfrak p$ lying above each rational prime $p$ .

If $p$ splits, then $p\mathcal O_K=\mathfrak p_1\mathfrak p_2, \quad N\mathfrak p_1=N\mathfrak p_2=p.$ So the local Euler factor is

$\displaystyle \big(1-\frac{1}{p^s}\big)^{-2}.$

In this case $\chi_D(p)=1$ , and the local Euler factor of $\zeta(s)L(s,\chi_D)$ is

$\displaystyle \big (1-\frac{1}{p^s}\big )^{-1} \big (1-\frac{\chi_D(p)}{p^s}\big )^{-1} = \big (1-\frac{1}{p^s}\big )^{-2}.$

If $p$ is inert, then $p\mathcal O_K=\mathfrak p, \quad N\mathfrak p=p^2.$ So the local Euler factor is

$\displaystyle \big(1-\frac{1}{p^{2s}}\big)^{-1}.$

In this case $\chi_D(p)=-1$ , and

$\displaystyle \big (1-\frac{1}{p^s}\big )^{-1} \big (1+\frac{1}{p^s}\big )^{-1} = \big (1-\frac{1}{p^{2s}}\big )^{-1}.$

If $p$ ramifies, then $p\mathcal O_K=\mathfrak p^2, \quad N\mathfrak p=p.$ So the local Euler factor is

$\displaystyle \big(1-\frac{1}{p^s}\big)^{-1}.$

In this case $\chi_D(p)=0$ , and the local factor of $\zeta(s)L(s,\chi_D)$ is again

$\displaystyle \big(1-\frac{1}{p^s}\big)^{-1}.$

Thus every Euler factor agrees, and therefore

$\displaystyle \zeta_K(s)=\zeta(s)L(s,\chi_D).$

Since $\zeta(s)$ has residue $1$ at $s=1$ , it follows that

$\displaystyle \text{Res}_{s=1}\zeta_K(s) =L(1,\chi_D).$

So in a quadratic field, the class number formula is equivalent to a formula for the value $L(1,\chi_D)$ .

Imaginary Quadratic Fields

Suppose first that $D<0.$ Then $K$ is imaginary quadratic. It has no real embeddings and one pair of complex embeddings. That is $r_1=0, r_2=1.$ The unit group is finite. The number of roots of unity is

$\displaystyle w= \begin{cases} 2, & D<-4,\\ 4, & D=-4,\\ 6, & D=-3. \end{cases}$

The regulator is $R=1$ by convention.

The general class number formula becomes

$\displaystyle \text{Res}_{s=1}\zeta_K(s) = \frac{2\pi h}{w\sqrt{|D|}}.$

Since $\text{Res}_{s=1}\zeta_K(s)=L(1,\chi_D)$ , we get Dirichlet’s class number formula for imaginary quadratic fields:

$\displaystyle L(1,\chi_D) = \frac{2\pi h}{w\sqrt{|D|}}.$

Equivalently,

$\displaystyle h = \frac{w\sqrt{|D|}}{2\pi}L(1,\chi_D).$

The same formula appears in classical treatments of Dirichlet’s formula for imaginary quadratic fields. We now prove this formula directly using partial zeta functions for the ideal classes.

Let $\text{Cl}(K)$ be the ideal class group. Its order is the class number: $h=|\text{Cl}(K)|.$ For an ideal class $C$ , define the partial zeta function

$\displaystyle \zeta_C(s)= \sum_{\mathfrak b\in C}\frac{1}{N\mathfrak b^s},$

where the sum is over integral ideals $\mathfrak b$ in the class $C$ .

Then

$\displaystyle \zeta_K(s)=\sum_{C\in\text{Cl}(K)}\zeta_C(s).$

The proof of the class number formula reduces to proving that every partial zeta function has the same residue at $s=1$ .

In the imaginary quadratic case, the claim is

$\displaystyle \text{Res}_{s=1}\zeta_C(s) = \frac{2\pi}{w\sqrt{|D|}}$

for every ideal class $C$ . Summing over $h$ classes gives

$\displaystyle \text{Res}_{s=1}\zeta_K(s) = \frac{2\pi h}{w\sqrt{|D|}}.$

So the essential point is the residue of one partial zeta function.

Fix an ideal class $C$ . Choose an integral ideal $\mathfrak a$ representing the inverse class $C^{-1}$ . If $\mathfrak b$ is an integral ideal in the class $C$ , then $\mathfrak a\mathfrak b$ is principal, because $[\mathfrak a][\mathfrak b]=C^{-1}C=1$ . Thus

$\displaystyle \mathfrak a\mathfrak b=(\alpha)$

for some nonzero $\alpha\in\mathfrak a$ . Conversely, if $0\neq\alpha\in\mathfrak a$ , then $(\alpha)\mathfrak a^{-1}$ is an integral ideal in the class $C$ . The norm relation is

$\displaystyle N\mathfrak b=\frac{|N_{K/\mathbb Q}(\alpha)|}{N\mathfrak a}.$

However, the same ideal $\mathfrak b$ is obtained from all unit multiples of $\alpha$ . Since the imaginary quadratic unit group is finite of size $w$ , each ideal is represented exactly $w$ times.

Therefore

$\displaystyle \zeta_C(s) = \frac{(N\mathfrak a)^s}{w} \sum_{0\neq\alpha\in\mathfrak a} \frac{1}{|N_{K/\mathbb Q}(\alpha)|^s}.$

Now choose a $\mathbb Z$ -basis $\omega_1,\omega_2$ of $\mathfrak a$ . Every $\alpha\in\mathfrak a$ has the form

$\displaystyle \alpha=x\omega_1+y\omega_2, \quad x,y\in\mathbb Z.$

Define the quadratic form

$\displaystyle Q_{\mathfrak a}(x,y) = \frac{|N_{K/\mathbb Q}(x\omega_1+y\omega_2)|}{N\mathfrak a}.$

In the imaginary quadratic case, $Q_{\mathfrak a}$ is a positive definite integral binary quadratic form of discriminant $D$ . Therefore

$\displaystyle \zeta_C(s)= \frac{1}{w} \sum_{(x,y)\neq(0,0)} \frac{1}{Q_{\mathfrak a}(x,y)^s}.$

Thus the partial zeta function is, up to the factor $1/w$ , the Epstein zeta function of a positive definite binary quadratic form. So the residue calculation becomes a lattice point problem for positive definite quadratic forms.

Let $Q(x,y)=Ax^2+Bxy+Cy^2$ be positive definite, with discriminant $B^2-4AC=D<0.$ We need the behavior near $s=1$ of

$\displaystyle Z_Q(s)= \sum_{(x,y)\neq(0,0)} \frac{1}{Q(x,y)^s}.$

Let $A_Q(X) = | \{(x,y)\in\mathbb Z^2:Q(x,y)\leq X\}|.$ The region $Q(x,y)\leq X$ is an ellipse. Its area is

$\displaystyle \text{area}(Q(x,y)\leq X) =\frac{2\pi X}{\sqrt{|D|}}.$

This follows from diagonalizing the quadratic form. The matrix of $Q$ has determinant $AC-\frac{B^2}{4} = \frac{|D|}{4}.$ The area of $Q(x,y)\leq1$ is therefore

$\displaystyle \frac{\pi}{\sqrt{|D|/4}}=\frac{2\pi}{\sqrt{|D|}},$

and scaling gives the factor $X$ .

By elementary lattice point counting,

$\displaystyle A_Q(X) =\frac{2\pi}{\sqrt{|D|}}X + O(X^{1/2}).$

The exact error term is not important for the residue. What matters is the main term which equals the area of the region, the error term is bounded by the perimeter of the region.

Now use partial summation. If $A_Q(X)\sim cX,$ then $Z_Q(s)$ has a simple pole at $s=1$ with residue $c$ . Indeed,

$\displaystyle Z_Q(s) =\int_{1^-}^{\infty}X^{-s} dA_Q(X),$

and integration by parts gives

$\displaystyle Z_Q(s) = s\int_1^\infty A_Q(X)X^{-s-1} dX$

up to an entire or harmless bounded initial term. Substituting $A_Q(X)=cX+o(X)$ shows

$\displaystyle Z_Q(s) = s\int_1^\infty (cX+o(X))X^{-s-1} dX,$

so the singular part is

$\displaystyle c s\int_1^\infty X^{-s} dX = c \frac{s}{s-1}.$

Therefore

$\displaystyle \text{Res}_{s=1}Z_Q(s) = c = \frac{2\pi}{\sqrt{|D|}}.$

Since $\zeta_C(s)=\frac{1}{w}Z_{Q_{\mathfrak a}}(s),$ we get

$\displaystyle \text{Res}_{s=1}\zeta_C(s) = \frac{2\pi}{w\sqrt{|D|}}.$

This proves the imaginary quadratic class number formula:

$\displaystyle L(1,\chi_D) = \frac{2\pi h}{w\sqrt{|D|}}.$

Norm form on the ideals gives positive definite quadratic forms which define ellipses, lattice point counting gives area as the main terms, and this area essentially gives the residue.

The formula $L(1,\chi_D)=\frac{2\pi h}{w\sqrt{|D|}}$ is analytic, but we can make it very explicit. Let $\chi$ be a primitive character modulo $q$ . For $0<x<1$ we have the Fourier series

$\displaystyle \sum_{n=1}^{\infty}\frac{\sin(2\pi nx)}{n} = \pi\big(\frac12-x\big).$

Suppose $\chi$ is real and odd, so $\chi(-1)=-1$ . This is the case for $\chi_D$ when $D<0$ . The Gauss sum is

$\displaystyle \tau(\chi)= \sum_{a=1}^q\chi(a)e^{2\pi ia/q}.$

For primitive $\chi$ ,

$\displaystyle \sum_{a=1}^q\chi(a)e^{2\pi ian/q} = \chi(n)\tau(\chi).$

Thus

$\displaystyle \chi(n)= \frac{1}{\tau(\chi)} \sum_{a=1}^q\chi(a)e^{2\pi ian/q}.$

Therefore

$\displaystyle L(1,\chi) = \sum_{n=1}^{\infty}\frac{\chi(n)}{n} = \frac{1}{\tau(\chi)} \sum_{a=1}^q\chi(a) \sum_{n=1}^{\infty}\frac{e^{2\pi ian/q}}{n}.$

Since $\chi$ is odd, the real cosine parts cancel and the sine parts remain. Hence

$\displaystyle L(1,\chi) = \frac{i}{\tau(\chi)} \sum_{a=1}^q\chi(a) \sum_{n=1}^{\infty}\frac{\sin(2\pi an/q)}{n}.$

Using the Fourier series,

$\displaystyle L(1,\chi) = \frac{i\pi}{\tau(\chi)} \sum_{a=1}^q \chi(a)\big(\frac12-\frac{a}{q}\big).$

Since $\sum_{a=1}^q\chi(a)=0$ , this becomes

$\displaystyle L(1,\chi) = -\frac{i\pi}{q\tau(\chi)} \sum_{a=1}^q\chi(a)a.$

For the quadratic character $\chi_D$ with $D<0$ , one has

$\displaystyle \tau(\chi_D)=i\sqrt{|D|}$

with the usual choice of the square root. Thus

$\displaystyle L(1,\chi_D) = -\frac{\pi}{|D|^{3/2}} \sum_{a=1}^{|D|}\chi_D(a)a.$

Combining this with

$\displaystyle L(1,\chi_D)=\frac{2\pi h}{w\sqrt{|D|}}$

gives the finite sum formula

$\displaystyle h = -\frac{w}{2|D|} \sum_{a=1}^{|D|}\chi_D(a)a.$

This is a very concrete form of Dirichlet’s class number formula. It says that the class number is a weighted imbalance of quadratic residues and nonresidues modulo $|D|$ .

Example:

Let $K=\mathbb Q(i).$ Then $D=-4,\quad w=4.$ The character $\chi_{-4}$ is

$\displaystyle \chi_{-4}(n)= \begin{cases} 0, & n\equiv0,2\pmod4,\\ 1, & n\equiv1\pmod4,\\ -1, & n\equiv3\pmod4. \end{cases}$

The $L$ -value is

$\displaystyle L(1,\chi_{-4}) = 1-\frac13+\frac15-\frac17+\cdots = \frac{\pi}{4}.$

The class number formula gives

$\displaystyle h = \frac{w\sqrt{|D|}}{2\pi}L(1,\chi_{-4}) = \frac{4\cdot2}{2\pi}\cdot\frac{\pi}{4}=1$

Thus $\mathbb Z[i]$ has class number $1$ . This is a analytic proof of unique factorization in the Gaussian integers.

Let $K=\mathbb Q(\sqrt{-5}).$ Then $D=-20,\quad w=2.$ The character $\chi_{-20}$ modulo $20$ has the values

$\displaystyle \chi_{-20}(r)= \begin{cases} 1 & \quad\text{for}\quad r \equiv1,3,7,9\pmod{20},\\ -1 & \quad\text{for}\quad r\equiv11,13,17,19\pmod{20}, \end{cases}$

with value $0$ when $(r,20)>1$ .

Compute

$\displaystyle \sum_{a=1}^{20}\chi_{-20}(a)a = 1+3+7+9-11-13-17-19==40.$

The finite class number formula gives

$\displaystyle h = -\frac{w}{2|D|} \sum_{a=1}^{|D|}\chi_D(a)a =-\frac{2}{40}(-40) =2$

Therefore

$\displaystyle h(\mathbb Q(\sqrt{-5}))=2.$

This agrees with the classical failure of unique factorization:

$\displaystyle 6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5}).$

The analytic formula measures exactly this defect. The failure of unique factorization is encoded in a finite weighted character sum.

Real quadratic fields

Now suppose $D>0.$ Then $K$ is real quadratic. There are two real embeddings and no complex embeddings: $r_1=2, r_2=0.$

The unit group is infinite. By Dirichlet’s unit theorem, it has the form

$\displaystyle \mathcal O_K^\times=\{\pm\varepsilon^n:n\in\mathbb Z\},$

where $\varepsilon>1$ is a fundamental unit. The regulator is

$\displaystyle R=\log\varepsilon.$

Also $w=2$ , because the only roots of unity in a real quadratic field are $\pm1$ .

The general class number formula becomes

$\displaystyle \text{Res}_{s=1}~\zeta_K(s) = \frac{2^2h\log\varepsilon}{2\sqrt D}=\frac{2h\log\varepsilon}{\sqrt D}$

Since $\zeta_K(s)=\zeta(s)L(s,\chi_D)$ , we get

$\displaystyle L(1,\chi_D) = \frac{2h\log\varepsilon}{\sqrt D}.$

Equivalently,

$\displaystyle h = \frac{\sqrt D}{2\log\varepsilon}L(1,\chi_D).$

The new feature is the factor $\log\varepsilon$ . In the imaginary case, the unit group was finite, so no logarithmic correction appeared. In the real case, infinitely many units generate infinitely many representatives of the same principal ideal. The regulator measures the spacing of these repetitions.

We now prove the real quadratic formula by the same partial-zeta method, but this time the quadratic forms are indefinite and the unit group is infinite.

Fix an ideal class $C$ and choose an integral ideal $\mathfrak a$ in the inverse class $C^{-1}$ . As before, ideals $\mathfrak b\in C$ correspond to elements $0\neq\alpha\in\mathfrak a$ modulo units, with

$\displaystyle \mathfrak b=(\alpha)\mathfrak a^{-1},\quad N\mathfrak b = \frac{|N_{K/\mathbb Q}(\alpha)|}{N\mathfrak a}.$

Embed $K$ into $\mathbb R^2$ by its two real embeddings:

$\displaystyle \alpha\mapsto(\sigma_1(\alpha),\sigma_2(\alpha)).$

The ideal $\mathfrak a$ becomes a lattice in $\mathbb R^2$ . Its covolume is $\text{covol}(\mathfrak a) \sqrt D N\mathfrak a.$ The norm is

$\displaystyle |N_{K/\mathbb Q}(\alpha)| = |\sigma_1(\alpha)\sigma_2(\alpha)|.$

Thus the condition $N\mathfrak b\leq X$ becomes

$\displaystyle |\sigma_1(\alpha)\sigma_2(\alpha)| \leq XN\mathfrak a.$

Let $x=\sigma_1(\alpha), y=\sigma_2(\alpha).$ The inequality is

$\displaystyle |xy|\leq XN\mathfrak a.$

If we counted all lattice points satisfying this inequality, the region would have infinite area, because the hyperbola $|xy|\leq B$ stretches infinitely along the axes. The infinite unit group is exactly what removes this infinite overcounting.

Let $\varepsilon>1$ be a fundamental unit. Under the two embeddings, multiplication by $\varepsilon$ acts as

$\displaystyle (x,y)\mapsto(\varepsilon x,\varepsilon' y),$

where

$\displaystyle |\varepsilon'|=\varepsilon^{-1}.$

Therefore multiplication by $\varepsilon$ preserves $|xy|$ . Introduce logarithmic coordinates $u=\log|x|, v=\log|y|.$ Then $\log|xy|=u+v.$ Multiplication by $\varepsilon$ sends

$\displaystyle (u,v)\mapsto(u+\log\varepsilon,\ v-\log\varepsilon).$

So it translates the difference $u-v$ by $2\log\varepsilon.$ A fundamental strip for the unit action is therefore given by requiring $u-v$ to lie in an interval of length

$\displaystyle 2\log\varepsilon.$

Now compute the volume. In one quadrant, put

$\displaystyle m=|xy|, \quad t=\log|x|-\log|y|.$

Then $dx dy=\frac12 dm dt$ in absolute value. The condition is $0<m\leq XN\mathfrak a$ , and a fundamental interval for $t$ has length $2\log\varepsilon$ . Therefore the area in one quadrant is

$\displaystyle \frac12\cdot XN\mathfrak a\cdot 2\log\varepsilon = XN\mathfrak a\log\varepsilon.$

There are four quadrants, so the total area in a fundamental region is

$\displaystyle 4XN\mathfrak a\log\varepsilon.$

Divide by the lattice covolume $\sqrt D N\mathfrak a$ . This gives the number of lattice points up to the unit action, before correcting for roots of unity:

$\displaystyle \frac{4XN\mathfrak a\log\varepsilon}{\sqrt D N\mathfrak a} =\frac{4X\log\varepsilon}{\sqrt D}.$

Finally divide by $w=2$ , because $\alpha$ and $-\alpha$ generate the same principal ideal. Thus the number of ideals in the class $C$ of norm at most $X$ is asymptotic to

$\displaystyle \frac{2\log\varepsilon}{\sqrt D}X.$

Hence the partial zeta function has residue

$\displaystyle \text{Res}_{s=1}\zeta_C(s) = \frac{2\log\varepsilon}{\sqrt D}.$

Summing over the $h$ ideal classes,

$\displaystyle \text{Res}_{s=1}\zeta_K(s) = \frac{2h\log\varepsilon}{\sqrt D}.$

So we conclude that

$\displaystyle L(1,\chi_D) = \frac{2h\log\varepsilon}{\sqrt D}.$

This proves the real quadratic class number formula.

The proof has the same structure as the imaginary case, but the geometry is different. In the imaginary case we counted lattice points in ellipses. In the real case we count lattice points in hyperbolic regions, but only after cutting by a fundamental strip for the unit group. That fundamental strip has width $2\log\varepsilon$ , and that is why the regulator appears.

For $D>0$ , the character $\chi_D$ is even: $\chi_D(-1)=1.$ For a primitive even character $\chi$ modulo $q$ , one uses the Fourier identity

$\displaystyle \sum_{n=1}^{\infty}\frac{\cos(2\pi nx)}{n} = -\log(2\sin\pi x), \quad 0<x<1.$

As before,

$\displaystyle \chi(n)= \frac{1}{\tau(\chi)} \sum_{a=1}^q\chi(a)e^{2\pi ian/q}.$

When $\chi$ is even, the sine parts cancel and the cosine parts remain. Hence

$\displaystyle L(1,\chi) = -\frac{1}{\tau(\chi)} \sum_{a=1}^{q-1} \chi(a)\log\big(2\sin\frac{\pi a}{q}\big).$

For the real quadratic character $\chi_D$ , one has

$\displaystyle \tau(\chi_D)=\sqrt D.$

Therefore

$\displaystyle L(1,\chi_D) = -\frac{1}{\sqrt D} \sum_{a=1}^{D-1} \chi_D(a)\log\big(2\sin\frac{\pi a}{D}\big).$

Combining this with

$\displaystyle L(1,\chi_D) = \frac{2h\log\varepsilon}{\sqrt D}$

gives

$\displaystyle 2h\log\varepsilon = -\sum_{a=1}^{D-1} \chi_D(a)\log\big(2\sin\frac{\pi a}{D}\big).$

This is the real quadratic analogue of the finite sum formula. In the imaginary case, the formula is a weighted arithmetic sum. In the real case, it is a weighted logarithmic trigonometric sum. This is exactly what one should expect: real quadratic fields have infinite units, and units are measured by logarithms.

Example:

Let $K=\mathbb Q(\sqrt5).$ Then $D=5, \quad \mathcal O_K=\mathbb Z\big[\frac{1+\sqrt5}{2}\big].$ A fundamental unit is $\varepsilon=\frac{1+\sqrt5}{2}.$

The character modulo $5$ is

$\chi_5(1)=1, \chi_5(2)=-1, \chi_5(3)=-1, \chi_5(4)=1.$

Compute

$\displaystyle S= \sum_{a=1}^{4} \chi_5(a) \log\big(2\sin\frac{\pi a}{5}\big).$

Using $\sin\frac{4\pi}{5}=\sin\frac{\pi}{5}, \quad \sin\frac{3\pi}{5}=\sin\frac{2\pi}{5},$ we get

$\displaystyle S= 2\log\big(2\sin\frac{\pi}{5}\big) + 2\log\big(2\sin\frac{2\pi}{5}\big).$

Thus

$\displaystyle S= 2\log\big( \frac{\sin(\pi/5)}{\sin(2\pi/5)} \big).$

But

$\displaystyle \frac{\sin(\pi/5)}{\sin(2\pi/5)} = \frac{1}{2\cos(\pi/5)} =\frac{1}{\varepsilon}.$

Therefore

$\displaystyle S=-2\log\varepsilon.$

The formula gives

$\displaystyle L(1,\chi_5) = -\frac{1}{\sqrt5}S \frac{2\log\varepsilon}{\sqrt5}.$

But the class number formula says

$\displaystyle L(1,\chi_5) = \frac{2h\log\varepsilon}{\sqrt5}.$

Therefore $h=1.$ So $\mathbb Q(\sqrt5)$ has class number $1$ .

General Number Fields

We now prove the general analytic class number formula by the same method. The main idea is still simple:

Split the Dedekind zeta function into partial zeta functions, one for each ideal class.
For one ideal class, choose an inverse ideal $\mathfrak a$ .
Ideals in that class correspond to elements of $\mathfrak a$ modulo units.
Embed $\mathfrak a$ as a lattice in Euclidean space.
Count lattice points in a norm-bounded region, modulo the unit lattice.
The volume calculation gives the residue.

This is the classical proof. It is the same proof as in the quadratic cases, but in higher-dimensional language.

Let $K$ be a number field of degree $n=[K:\mathbb Q].$ Let $r_1$ be the number of real embeddings, and let $r_2$ be the number of pairs of complex embeddings. Then $n=r_1+2r_2.$ Choose embeddings

$\displaystyle \sigma_1,\dots,\sigma_{r_1}:K\hookrightarrow\mathbb R$

and one embedding from each complex conjugate pair,

$\displaystyle \tau_1,\dots,\tau_{r_2}:K\hookrightarrow\mathbb C.$

The Minkowski embedding is

$\displaystyle \Phi:K\longrightarrow \mathbb R^{r_1}\times\mathbb C^{r_2}$

given by

$\displaystyle \Phi(\alpha)= (\sigma_1(\alpha),\dots,\sigma_{r_1}(\alpha), \tau_1(\alpha),\dots,\tau_{r_2}(\alpha)).$

The absolute norm of $\alpha$ is

$\displaystyle |N_{K/\mathbb Q}(\alpha)| = \prod_{i=1}^{r_1}|\sigma_i(\alpha)| \prod_{j=1}^{r_2}|\tau_j(\alpha)|^2.$

If $\mathfrak a$ is a nonzero integral ideal, then $\Phi(\mathfrak a)$ is a lattice in $\mathbb R^{r_1}\times\mathbb C^{r_2}$ .

Under the Minkowski embedding, $\mathcal O_K$ becomes a lattice whose fundamental volume is controlled by the embedding determinant; this determinant is essentially $\sqrt{|D_K|}$ , with a factor $2^{-r_2}$ from replacing complex conjugate coordinates by real and imaginary parts. Since an ideal $\mathfrak a\subset\mathcal O_K$ has index $N\mathfrak a$ , its lattice is $N\mathfrak a$ times sparser, So the covolume is

$\displaystyle \text{covol}(\Phi(\mathfrak a)) = 2^{-r_2}\sqrt{|D_K|} N\mathfrak a.$

The unit group $\mathcal O_K^\times$ acts on $K$ by multiplication and therefore acts on the Minkowski space.

The logarithmic map is

$\displaystyle \ell:\mathcal O_K^\times \longrightarrow \mathbb R^{r_1+r_2}$

defined by

$\displaystyle \ell(u)= \big( \log|\sigma_1(u)|,\dots,\log|\sigma_{r_1}(u)|, 2\log|\tau_1(u)|,\dots,2\log|\tau_{r_2}(u)| \big).$

The product formula says that

$\displaystyle \sum_{i=1}^{r_1}\log|\sigma_i(u)| + \sum_{j=1}^{r_2}2\log|\tau_j(u)| =1$

Thus $\ell(u)$ lies in the hyperplane

$\displaystyle H= \big \{ (t_1,\dots,t_{r_1+r_2}): t_1+\cdots+t_{r_1+r_2}=0 \big \}.$

Dirichlet’s unit theorem says that the image of the unit group modulo roots of unity is a lattice in this hyperplane of rank $r=r_1+r_2-1.$ The covolume of this lattice is the regulator $R_K$ .

More concretely, choose fundamental units

$\displaystyle \varepsilon_1,\dots,\varepsilon_r.$

Form the matrix of logarithms

$\displaystyle \big( \log|\sigma_i(\varepsilon_j)| \big)$

with the complex embeddings counted using $2\log|\tau_i(\varepsilon_j)|$ . Delete one row, take the absolute value of the determinant, and obtain $R_K$ with the standard normalization. This is the multiplicative lattice part of the proof. The ring of integers gives an additive lattice in Minkowski space. The units give a multiplicative lattice after taking logarithms. The class number formula is obtained by comparing these two lattices.

Let $C$ be an ideal class. Define

$\displaystyle \zeta_C(s)= \sum_{\mathfrak b\in C} \frac{1}{N\mathfrak b^s}.$

Then

$\displaystyle \zeta_K(s)= \sum_{C\in\text{Cl}(K)}\zeta_C(s).$

Choose an integral ideal $\mathfrak a$ representing the inverse class $C^{-1}$ .

As before, integral ideals $\mathfrak b\in C$ correspond to nonzero elements $\alpha\in\mathfrak a$ modulo multiplication by units, through

$\displaystyle \mathfrak b=(\alpha)\mathfrak a^{-1}.$

The norm relation is

$\displaystyle N\mathfrak b = \frac{|N_{K/\mathbb Q}(\alpha)|}{N\mathfrak a}.$

Thus the counting function

$\displaystyle A_C(X)= | \{\mathfrak b\in C:N\mathfrak b\leq X\}|$

is the number of nonzero elements $\alpha\in\mathfrak a$ , modulo units, satisfying

$\displaystyle |N_{K/\mathbb Q}(\alpha)| \leq XN\mathfrak a.$

Therefore, to estimate $A_C(X)$ , we count lattice points of $\Phi(\mathfrak a)$ in the region

$\displaystyle \prod_{i=1}^{r_1}|x_i| \prod_{j=1}^{r_2}|z_j|^2 \leq XN\mathfrak a,$

but only one representative modulo the action of the infinite unit group.

Let $m=r_1+r_2.$ For each real coordinate, write $y_i=|x_i|.$ For each complex coordinate, write $y_{r_1+j}=|z_j|^2.$

Then the absolute norm condition becomes

$\displaystyle y_1y_2\cdots y_m \leq XN\mathfrak a.$

The volume element contributes a factor

$\displaystyle 2^{r_1}\pi^{r_2}$

from real signs and complex angular integration. Indeed, each real coordinate has two signs, while in a complex coordinate $z$ , the area element satisfies

$\displaystyle dA(z)=\pi d(|z|^2).$

Thus after passing to the positive variables $y_1,\dots,y_m$ , the Euclidean volume is

$\displaystyle 2^{r_1}\pi^{r_2} dy_1\cdots dy_m.$

Now take logarithms:

$\displaystyle t_i=\log y_i.$

Then the product condition becomes

$\displaystyle t_1+\cdots+t_m \leq \log(XN\mathfrak a).$

The unit group acts by translation in the hyperplane

$\displaystyle H:t_1+\cdots+t_m=0.$

Choose a fundamental parallelepiped for the unit lattice in $H$ . Its volume is the regulator $R_K$ . Therefore, after quotienting by the free part of the unit group, the logarithmic directions contribute the factor $R_K.$

The remaining radial variable is

$\displaystyle M=y_1y_2\cdots y_m.$

Since the measure contains $dM$ in this radial direction, integration over

$\displaystyle 0<M\leq XN\mathfrak a$

contributes

$\displaystyle XN\mathfrak a.$

Thus the volume of a fundamental region for the unit action inside the norm-bounded region is

$\displaystyle 2^{r_1}\pi^{r_2}R_K XN\mathfrak a.$

Now divide by the lattice covolume

$\displaystyle 2^{-r_2}\sqrt{|D_K|} N\mathfrak a.$

This gives

$\displaystyle \frac{ 2^{r_1}\pi^{r_2}R_K XN\mathfrak a }{ 2^{-r_2}\sqrt{|D_K|} N\mathfrak a } = \frac{2^{r_1+r_2}\pi^{r_2}R_K}{ \sqrt{|D_K|}}X= \frac{ 2^{r_1}(2\pi)^{r_2}R_K }{\sqrt{|D_K|} }X.$

Finally, divide by $w_K$ , the number of roots of unity, because multiplying by a root of unity does not change the generated principal ideal. Therefore

$\displaystyle A_C(X) \sim \frac{ 2^{r_1}(2\pi)^{r_2}R_K }{ w_K\sqrt{|D_K|} }X.$

This is the central asymptotic estimate for one ideal class. Summing over all the ideal classes gives

$\displaystyle \text{Res}_{s=1}\zeta_K(s) = h_K \frac{2^{r_1}(2\pi)^{r_2}R_K }{w_K\sqrt{|D_K|}}.$

Therefore

$\displaystyle \text{Res}_{s=1}\zeta_K(s) = \frac{ 2^{r_1}(2\pi)^{r_2}h_KR_K}{w_K\sqrt{|D_K|}}.$

This is the analytic class number formula.

The formula is not a collection of mysterious constants. Each factor has a precise origin. The factor $h_K$ appears because the Dedekind zeta function is the sum of $h_K$ partial zeta functions, one for each ideal class. The factor $\sqrt{|D_K|}$ appears because $\mathcal O_K$ and its ideals are lattices under the Minkowski embedding, and their covolumes are governed by the discriminant. The factor $R_K$ appears because the infinite unit group causes infinitely many elements to generate the same principal ideal. Taking logarithms turns the units into a lattice, and the regulator is the volume of a fundamental region for that lattice. The factor $w_K$ appears because roots of unity give finite repetitions. If $\zeta$ is a root of unity, then $\alpha$ and $\zeta\alpha$ generate the same ideal. The factor $2^{r_1}$ comes from real signs. Each real embedding gives a coordinate that can be positive or negative. The factor $(2\pi)^{r_2}$ comes from complex embeddings. Each complex embedding contributes angular integration in the complex plane. Thus the formula is an exact accounting of all the ways elements represent ideals.

The proof can be summarized as follows.

The Dedekind zeta function counts ideals:

$\displaystyle \zeta_K(s)= \sum_{\mathfrak a\neq0} \frac{1}{N\mathfrak a^s}.$

Splitting by ideal class gives

$\displaystyle \zeta_K(s)= \sum_C\zeta_C(s).$

For one ideal class $C$ , choose an inverse ideal $\mathfrak a$ . Ideals in $C$ correspond to elements of $\mathfrak a$ modulo units. Thus the partial zeta function is controlled by counting elements in a lattice, modulo a unit lattice.

The additive lattice has covolume involving $\sqrt{|D_K|}$ . The multiplicative unit lattice has covolume $R_K$ . The roots of unity have order $w_K$ . The number of ideal classes is $h_K$ . The archimedean coordinates supply $2^{r_1}(2\pi)^{r_2}$ .

Therefore the residue must be

$\displaystyle \frac{ 2^{r_1}(2\pi)^{r_2}h_KR_K }{ w_K\sqrt{|D_K|} }.$

This is why the class number formula is so natural. It is the equality obtained by counting the same arithmetic objects in two languages: ideals through their norms; elements through lattice points modulo units.

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