The Class Number Formula

Version 1.0—22nd November, 2013 klokken 11:26. A fundamental tool in the study of an algebraic number field is the Dedekind Zeta function. It is a generating function encoding a lot of the arithmetic properties of the field. We fix a number field K whose ring of integers is denoted by A, and we let d = [K : Q]. Then the Dedekind Zeta function of K is X 1 ζ (s) = (>) K N (a)s a⊆ A where a runs through all nontrivial, proper ideals of A. If we specialize K to be the field Q of rational numbers, all ideals are principal with a unique positive generator and so they are is in a one-one correspondence with the natural numbers, and we get back the Riemann Zeta function. We shall in this section establish the basic facts about the Dedekind Zeta function, and the first question that arises is: For which s is it defined? It turns out that the series in (>) converges for Re > 1, but one may show that the function can be analytically continued to the whole complex plain, except at s = 1 where the zeta function has a simple pole. For our modest needs it will be sufficient to know that the series converges for all real s > 1. Just like the Riemann Zeta function, the Dedekind Zeta function enjoys the property of having an Euler product. In the Riemann case the Euler product is a consequence of the fundamental theorem of arithmetic, stating that integers have a unique factorization in primes. Unique factorization is true for ideals in Dedkind rings, and therefore the Dedekind Zeta functions also has an Euler product: Y 1 ζ (s) = , K 1 − N (p)−s p where p runs through all prime ideals in A. The proof follows the same lines as for the Riemann Zeta function, and we shall later on give it. There are several results and many conjectures about the values of the Dedekind Zeta function at the integers, the classic result being a formula for the residue of ζk(s) at s = 1 which involves the class number of K and several other basic invariants of K. It reads 2r+tπtRh lim (s − 1)ζK (s) = p (@) s→1+ µ |∆| Where r and t are the number of real and complex conjugate pairs of embeddings of K, R is the regulator of K and µ the number of roots of unity contained in K, and of course we have the old acquaintances, the class number h and the discriminant ∆. The Class Number Formula MAT4250 — Høst 2013

This formula is a main tool in computing the class number for many fields, the other invariants occurring in the formula being more accessible, and therefore several call it the class number formula. The first instance of the class number formula was found by Dirichlet already in 1837. He proved a version for quadratic fields formulated in terms of quadratic forms. The general formula as presented here was proven by Dedekind. Let σ1, . . . , σr+t be embeddings of K, chosen like we did in section xxx. That is the first r are the real embeddings, in some order, and remaining t are chose one from each pair of complex conjugate embeddings. Dirichlet’s unit theorem gives us a set of fundamental units η1, . . . , ηr+t−1. Then the regulator is the absolute value any of the minors (they are all equal up to sign since the row-sums of the matrix vanish, see lemma3 on page 11 below) of the (r + t) × (r + t − 1)-matrix

R• = (j log |σj(ηi)|) (+) The regulator plays a more conceptual role as the volume of the fundamental parallel- lotopes of the logarithmic unit lattice in the trace-zero hyperplane H in Rr+t−1. We closely follow the presentation in chapter 5 of Borevich and Shafarevisch [BorShaf].

Introducing the class group The sum (>) above defining the Dedekind Zeta function may be split into a sum P ζK (s) = c ζc(s) of functions ζc(s), each corresponding to a class c of the class group CK . Indeed, one may take X 1 ζ (s) = c N (a)s a∈c where the ideals a over which the sum is takes, are integral and, as indicated, confined to the class c. The functions ζc(s) all have the same residue at s = 1. This explains that the class number h appears as a factor in the class group formula, and the proof of the class group formula reduces to a computation of the common residue at s = 1 of the functions ζc(s), the result being the formula 2r+tπtR lim ζc(s) = p . s→1+ µ |∆|

Now, let us fix an ideal a0 ∈ c. Any other element a of the class c is of the form (f)a0 for some f ∈ K∗. For a to be an integral ideal a necessary and sufficient condition is −1 −1 that f belongs to a0 (this is nothing else but the definition of a0 ). Introducing this in the sum defining ζc(s), we find X 1 X 1 1 X 1 ζ (s) = = = . c N (a)s N ((f)a )s N (a )s N ((f))s a∈c −1 0 0 −1 (f)⊆ a0 (f)⊆ a0

−1 The two last sums are taken over all integral and principal ideals (f) contained in a0 . Now, two principal ideals (f) and (f 0) are equal if and only if the elements f and f 0 are

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associate, that is, f = uf 0 for a unit u. So, letting S be a set of representatives of the −1 associates, i.e., S contains exactly one element from each class of associates in a0 —or phrased in a slightly different manner, S is a fundamental domain for the action of the ∗ −1 unit group U = A on a0 —one has

X 1 ζc(s) = s , NK/ (f) f∈S Q where we as well has replaced the counting norm N ((f)) by the the norm NK/Q(f) (they are equal!). The proof of this formula has two main ingredients. First we establish the formula lims→1+ ζc(s) = γ/Γ where Γ is the volume of the Minkowski type lattice L −1 in a0 r t −1 K ⊗Q R = R ⊕ C associated to a0 , and γ is—with a friendly interpretation— the volume of the part of the quotient a0/U where the norm is at most one in absolute value (the absolute value of the norm is a well defined function on the quotient since N(u) = ±1 for units u). This is a special case of more general statement about counting lattice elements belonging to subsets of a certain type.

We know the volume of the lattice L −1 . It was computed in xxx: a0

−1 −tp Γ = N (a0) 2 |∆K |

The second ingredient of the proof is the computation of γ, and we shall find:

γ = 2rπtR/µ.

All together, this gives:

−1 X lim (s − 1)ζK (s) =N (a0) lim (s − 1)ζc(s) = s→1+ s→1+ c∈CK r t r+t t −1 γ −1 2 π R/µ 2 π RH =N (a0) h = N (a0) h = . −1 −tp p Γ N (a0) 2 |∆K | µ |∆K |

The Minkowski setting We recall the setting from the section on Minkowski’s geometry of numbers. Let r t V = K ⊗Q R ' R ⊕ C ; the isomorphism being an isomorphism of algebras with multiplication in Rr ⊕ Ct being defined componentwise. There is the embedding Σ of K into V which is given by

Σ(α) = (σj(α)) where the σi-s are r + t chosen embeddings, chosen according to the usual rule that the r first be the real embeddings and among the t next there should be one from each pair of complex conjugate embeddings.

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r+t The logarithmic map. Recall the logarithmic map l : V0 → R given by

l(x) = (j log |xj|) where x = (xj) and the j are the weights corresponding to the embeddings σj; with j = 1 when σj is real, and j = 2 when σj is complex. The set V0 where the logarithm is defined, consists of the elements in V = Rr ⊕Ct all of whose coordinates are non-zero. Q i Norm and trace. The norm map N(x) on V is just the product xi of the weighted coordinates. It is multiplicative, and if ρx is the endomorphism of V given by multi- r+t plication by x, then det ρx = N(x).The trace map tr(y) on R is just the sum of the coordinates of y. One has log |N(x)| = tr(l(x)). The group of units. The group U of units is completely described by Dirichlet’s r+t−1 ∗ unit theorem. It decomposes in a product U ' µK × Z where µK ⊆ K is the group of roots of unity in K. We choose a basis η1, . . . , ηr+t−1 for the free part, or as ones says, the units η1, . . . , ηr+t−1 form a fundamental set of units. We have shown (and that was the hard part of proof of Dirichlet’s unit theorem) that the vectors r+t e1 = l(η1), . . . , er+t−1 = l(ηr+t−1) are linearly independent in R ; indeed, we showed they form a basis for the trace-zero hyperplane H. Introducing the vector e0 = (j) = (1,..., 1, 2,..., 2) we thus have a basis e0, . . . , er+t−1 r+t for R . For any x ∈ V0 the coordinates of l(x) with respect to this will be denoted by ξ0, . . . , ξr+t−1; that is one has X l(x) = ξ0e0 + ξjej. 1≤j≤r+t−1

The action of the unit group The group of units U = A∗ acts on K by multiplication, of course, and when this action is transported to V —where it shows up as multiplication by Σ(u)—it will still be denoted by ux (where u ∈ u and x ∈ V ). This action is isometric since det ρu =

NK/Q(u) = ±1. The action is free on the set V0 where N(x) 6= 0, that is the set of elements all whose coordinates are non-zero. Indeed if u is unit, clearly all the σj(u) 6= 0, and for x ∈ V0 the equality ux = x therefore implies that u = 1. The fundamental domain. We are in need of an explicit description of a funda- mental domain for this action of U on V0, that is a subset X⊆ V0 such each orbit { uv | u ∈ U } contains exactly one point from X. The subset X we are looking for is defined by the following three constraints:

 N(x) 6= 0 P  l(x) = ξ0e0 + 1≤j≤r+t−1 ξjej where 0 ≤ ξj < 1 for 1 ≤ j ≤ r + t − 1

 0 ≤ arg x1 < 2π/m. There are two remarks to be made. First, in the second condition there is no constraint on the coordinate ξ0, Secondly, in case K has a real embedding, the last

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condition reduces to the requirement that x1 be positive. Indeed, in that case µK = {±1}, and m = 2.

Lemma 1 The set X described above is a fundamental domain for the action of U on V0.

Proof: Let x ∈ V0 and write X X l(x) = ξ0e0 + ξjej = ξ0e0 + (nj + γj)ej j j where nj = [ξj] is the integral part of ξj and γj = ξj − nj satisfies 0 ≤ γj < 1. Letting Q nj u = j ηj , one sees that

−1 X X X l(u x) = l(x) − l(u) = ξ0eo + ξjej − njej = ξ0eo + γjej. j j j

Hence u−1x satisfies the second request above. The first is trivially satisfied since −1 −1 N(u x) = N(u) N(x) = N(x). Since µm = µK ⊆ K, we clearly may find a root of 0 0 −1 unity ζ in K such that 0 ≤ arg ζx1 < 2π/m where x1 is the first coordinate of u x. Hence ζu−1x satisfies the third request, and it faithfully continues to satisfy the second since ζ being in the kernel of the logarithmic map l, gives l(ζu−1x) = l(u−1x). We have established that every U-orbit hits X. To see that it hits just in one point, 0 0 Q nj assume that ux = x for x, x ∈ X and u = ζ j η . Writing out the coordinates of l(x0) and l(ux) we get

X 0 X ξ0e0 + ξjej = ξ0eo + (nj + ξj)ej j j

0 It follows that nj = 0 since all of the ξi’s and the ξi’s lie between 0 and 1. To finish off the proof, if both x1 and ζx1 has an argument in the interval [0, 2π/m), the same holds for ζ, and being a m-th root of unity, it follows that ζ = 1. J

The fundamental domain is a cone. We shall at a certain point need this fact. So let t be a positive real number and let x ∈ X. Multiplication by t only changes the component of l(x) along e0 and does not affect the other components. Indeed, if P l(x) = ξ0e0 + j ξjej one finds

X l(tx) = (j log |txj|) = (log |t| j) + l(x) = (ξ0 + log |t|)e0 + ξjej, j

so we see that tx continues to satisfy the second request. For the third request, since t is real and positive it certainly holds true that arg tx1 = arg x1, and of course N(tx) = tdN(x), so N(tx) 6= 0 whenever N(x) 6= 0, and the first request is satisfied as well.

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Riemann’s Zeta function As everybody knows Riemann’s Zeta function is defined by

X 1 ζ(s) = (3) ns n≥1

For our modest needs we assume that s is a real number lying in the interval (0, ∞), although one may consider complex s in the half space Re s > 1. We shall not use ζ(s) extensively, and only the following property:

Proposition 1 The sum (3) converges for s > 1, and lims→0+ (s − 1)ζ(s) = 1.

Proof: Elementary calculus gives

Z n+1 1 Z n dx ≤ s ≤ s n n n−1 x where the first inequality holds for n ≥ 1 and the second for n ≥ 2. Hence, summing over n one obtains for s > 1 1 Z ∞ dx Z ∞ dx 1 = s ≤ ζ(s) ≤ 1 + s = 1 + . s − 1 1 x 1 x s − 1 J The proof is divide into a few parts:

Volumes vs the residue of the zeta function This paragraph explains the connection between the Dedekind Zeta function and cer- r s tain volumes in the space ⊕ . To be precise, the volume of the lattice L −1 and the R C a0 volume of the part of fundamental domain X where the norm takes values no greater than one. We shall work in more general situation, thus setting the points of the computation in a clearer relief. The setting is a real vector space V of dimension d equipped with an inner product and a full lattice L⊆ V . The volume of L is denoted by Γ (recall, this is the volume of a fundamental parallelotope of L). Further we are given a cone X in V , i.e., a subset such that if x ∈ X and t > 0 a real number, then tx ∈ X; and the last player in the plot, is a non-negative real-valued function F (x) defined on V , taking positive values on X and satisfying the following two conditions

d  F is homogeneous of degree d, i.e., F (tx) = t F (x)

 The set T = { x ∈ V | F (x) ≤ 1 } is bounded and has a finite volume γ.

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Then we have

P 1 Proposition 2 For s > 1 the series f(s) = x∈L∩X F (x)s converges, and one has γ lim (s − 1)f(x) = . s→1+ Γ Proof: There are three points in the proof. Lattice points vs volumes. The first point of the proof is built on the intuitively obvious statement, that the number of lattice points in a bounded set W , is approx- imately equal to the volume of the set divided by the volume of the fundamental parallelotope of the lattice. To be precise, let r > 0. The scaled down lattice r−1L has a fundamental domain of volume Γ/rd. We let N(r) be the number of lattice points from r−1L lying in T , that is N(r) = #T ∩ r−1L. We may then cover T with N(r) translates of the fundamental domain of scaled lattice r−1L, and in that way obtain “tiling of T ” of total volume N(r)Γ/rd. By letting r grow, one finds (that this holds is in fact the precise content of the second assumption above)

Γ N(r) γ = lim N(r) = Γ lim , r→∞ rd r→∞ rd or N(r) γ lim = . (5) r→∞ rd Γ

Volumes vs the function F . The second point of the proof is to link the volume fraction γ/Γ to the behavior of F (x) for big values of x. Instead of making the lattice small, one may make the set T big. Clearly N(r) = rT ∩L, and since F is a homogenous function of x and X a cone, one sees that

rT = { rx | x ∈ X and F (x) ≤ 1 } = { y | y ∈ X and F (y) ≤ rd }.

Hence N(r) = #{ y ∈ L ∩ X | F (y) ≤ rd }. The set X ∩ L is countable, and we may order its the points according to increasing values of F , that is X ∩ L = {x1, x2,... } with

0 < F (x1) ≤ F (x2) ≤ F (x3) ≤,...... ,

n and we let rk be the real numbers such that rk = F (xk). The second point of the proof is to establish the equality: k γ lim = . (>) k→∞ F (xk) Γ

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To that end, we have for all  > 0

N(rk − ) ≤ k ≤ N(rk),

d hence as rk = F (xk) this gives

d N(rk − ) (rk − ) k N(rk) d d ≤ ≤ d (rk − ) rk F (rk) rk Letting k → ∞ one obtains (>) in view of (5). The function f(s) vs Riemann zeta. The last point of the proof is to compare P 1 P 1 the series n≥1 ns with the series f(s) = s∈L∩X F (x)s . For k >> 0, say k > k0, one has γ 1 1 γ 1 ( − ) < < ( − ) , Γ k F (xk) Γ k

hence γ s 1 1 γ s 1 ( − ) s < s < ( − ) s . Γ k F (xk) Γ k

Hence summing for k ≥ k0 we get

γ s X 1 X 1 γ s X 1 ( − ) s ≤ s ≤ ( − ) s Γ k F (xk) Γ k k≥k0 k≥k0 k≥k0

P 1 this shows that thew series k≥1 ks converges, and multiplying by s − 1 and letting s → 0+, we obtain γ γ ( − ) ≤ lim inf(s − 1)f(s) ≤ lim sup(s − 1)f(s) ≤ ( − ), Γ s→1+ s→1+ Γ from which it follows that lims→1+ (s − 1)f(s) = γ/Γ since lims→1+ (s − 1)ζ(s) = 1. J

The final volume computation The only missing ingredient of our computation now, is the evaluation of the volume of T . Recall that T is the subset of the fundamental domain X consisting of elements of norm no greater than one. It is described by the following constraints:

 0 < |N(x)| ≤ 1 P  l(x) = ξ0e0 + 1≤j≤r+t−1 ξjej where 0 ≤ ξj < 1 for 1 ≤ j ≤ r + t − 1

 0 ≤ arg x1 < 2π/m.

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Proposition 3 The volume γ of T is given as

2rπtR vol(T ) = µ

The whole of this paragraph is devoted to proof of this proposition. The proof is not very deep, but kind of technical. Basically, the point is to find a good parametrization of T . One uses polar coordinates for the complex coordinates, and observes that, except for the third constraint, the angles can vary freely. One might say that the “moral” content of the proof is that up to the action of roots of unity, and forgetting the arguments of the complex part, T corresponds to a logarithmic cone over the fundamental parallelotope of the unit lattice in Rr+t. The set T itself is bounded, but its image l(T ) is not, it is stretch logarithmically in a direction away from the trace-zero hyperplane. The precise content of these somehow vague descriptions is expressed in the next lemma:

Lemma 2 For x ∈ V0, one has the following expression for the logarithmic map

1 X l(x) = log |N(x)| e + ξ e d 0 j j j

Proof: For j > 0 the ej’s live in the trace-zero hyperplane, so writing l(x) = ξe0 + P P P j ξjej gives tr(l(x)) = dξ = j j log |xj| = log |N(x)|, since tr(e0) = j j = r+2t = d. J The formula in the lemma inspire us to write down a parametrization of T , in fact is a composition of two coordinate changes. Firstly, we use polar coordinates for the complex coordinates and let ri denote the radius vector, i.e., ri = |x| . The Jacobian Q i of this coordinate change is the product J1 = r+1≤i≤r+t ri. The arguments of these complex numbers do not appear in the two first constraints defining T , and if we for a moment ignore the last constraint, they can be chosen freely. The same applies to the signs of the real coordinates, and, as well, for the moment we ignore the signs and assume the real coordinates xj for j ≤ r are positive, then ri = xi for all i. This means that we shall give a parametrization of the set T0 of points from X satisfying the first constraints above and whose real coordinates are all positive. The parametrization is—in addition to keeping track of the angles— given by the expression i X  log r = log ξ + ξ  log |σ (η )| i i d 0 j i i j j where rj is given implicitly as a function of the ξj’s. One observes that N(x) = ξ0, and hence the two first constraints defining T0 translate into 0 < ξj < 1. The domain of the parameters is therefore (0, 1)r+t × [0, 2π)t.

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The remaining step is to compute the Jacobian of this parametrization, and the partials are given by:

 i i ∂ ri  if j = 0 = dξ0 ri ∂ ξj i log |σi(ηk)| if j 6= 0 .

Hence the Jacobian determinant appears as

1 ········· . . . . 1 Q r i i J = Q ··· i log |σi(ηj)| · · · dξ0 i i . . . .

r+t ·········

Q t The factor in front of the determinant simplifies to 1/( r+1≤i≤r+t ri)2 d since ξ0 = Q i N(x) = i ri . The rightmost (r + t − 1) × (r + t)-minor is nothing else but the transposed of the regulator matrix R• as defined in formula (+) on page2, hence evoking lemma3 below we see that the Jacobian is given as R J = . t Q 2 r+1≤i≤r+t ri

Computing the integral one uses the product of the two Jacobians J1 and J, the product t of the ri’s cancels, and integrand reduces to R/2 ; hence the volume of T0 is given as

t vol T0 = π R.

Coping with the third constraint. Let T1 be the subsets of X where only the two first constraints are in force. We may divide T1 into m disjoints parts of equal volume, of which T is one (in fact, the parts will be isometrically equivalent). To see this, let ζ ∈ µK be a primitive root of unity. The multiplication map x 7→ ζx is an isometry of S k V (its determinant is one, being the norm of ζ), and clearly T1 = 0≤k

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A simple lemma on determinants The following lemma use a few times when the regulator enters the scene:

Lemma 3 Let A = (aij) be an m × m-matrix whose column sums satisfy ( X a for j = 1 aij = i 0 for 2 ≤ j ≤ m that is, except for the first, the column sums all vanish. Then letting M be any the minor of A obtained by deleting the first column and any row, we have det A = aM

Proof: In det A, one may replace the elements in k-th row by the corresponding column sum, transforming the row into one with zeros everywhere except at the first entry where there is an a. Developing the determinant along that row gives the lemma. J Hence if a (m − 1) × m-matrix has all column sums equal to zero, all the minors are up to sign the same; we just add a column with all entries equal to 1/m and use the lemma.

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[BorShaf] Z. I. Borvich and I. R. Shafarevich Number theory, Academic Press 1966.

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