The Analytic Class Number Formula for Orders in Products of Number Fields

Home , Arithmetic zeta function, Class number formula

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Remark 1.3. Replacing a finite-type Z-scheme X by its associated reduced subscheme Xred does not change ζX (s). On the other hand, if the scheme X = Spec in Theorem 1.1 is not assumed to be reduced, then × need not be finitely generated, soO defining the regulator O 2 × R( ) is problematic: for example, if := F2[t, ǫ]/(ǫ ), then is isomorphic to the additive groupO of F [t] via the homomorphismO 1 + fǫ f. O 2 7→ Remark 1.4. Various authors defined other zeta functions attached to a singular curve over a finite field and computed their leading terms at s = 0 or s = 1 [Gal73, Gre89, ZG97, ZG97b, Stö98], but these zeta functions are different from the usual zeta function in [Ser65] in general.

1.1. Outline of the article. Section 2 proves Theorem 1.1 for a ring of S-integers. Because we want a formula involving the S-class group and S-regulator instead of the usual class group and regulator, it is not convenient to deduce this from the classical formula for the ring of integers. Instead we redo the calculations of Tate’s thesis for S-integers. Sections 3 and 4 characterize the rings such that Spec is a reduced affine finite-type O O Z-scheme of pure dimension 1. In particular, the normalization of such a ring is a product of rings of S-integers. O O Section 5 defines all the quantities that appear in (2). e Section 6 proves Theorem 1.1. The formula for follows from the case proved in Section 2, O so our strategy is to determine how each term in (2) changes when is replaced by . In particular, we use the Leray spectral sequence toe determine how theO unit group and PicardO group change. e Finally Sections 7 and 8 illustrate (2) in examples exhibiting the phenomena that can arise in our context: both number fields and function fields, S-integers instead of just integers, multiple irreducible components, and non-maximal orders (and hence singular points of the scheme).

2. Tate’s thesis for S-integers Let K be a global field. Let µ be the torsion subgroup of K×, and let w = #µ. For each place v of K, let K be the completion of K at v. If K R, equip it with Lebesgue v v ≃ measure. If Kv C, equip it with 2 times Lebesgue measure. For each nonarchimedean v, let be the valuation≃ ring in K , let q be the size of its residue field, and equip K with Ov v v v the Haar measure dx for which vol( v) = 1; here we follow [Wei67, p. 95] instead of taking the self-dual measure as in [Tat67, p.O 310]. We write vol(T ) for the measure of a set T with respect to a measure that is implied by context. If a Kv for some v, let a v R≥0 be the factor by which multiplication-by-a scales the ∈ | | ∈ × Haar measure on Kv. The measure we use on Kv is not the restriction of the measure dx on × Kv. If v is archimedean, equip Kv with the Haar measure dx/ x v. If v is nonarchimedean, × × | | equip Kv with the Haar measure for which vol( v ) = 1. × × O × × Let Kv,1 := x Kv : x v =1 . If v is nonarchimedean, Kv,1 = v , which has volume 1 { ∈ | |× } × O for the Haar measure on Kv . If v is archimedean, then equip Kv,1 with the Haar measure 2 × compatible with the Haar measure on Kv and Lebesgue measure on R in the exact sequence log || 1 K× K× v R 0 −→ v,1 −→ v −→ −→ (for the notion of compatibility, see, e.g., [DE14, Theorem 1.53]; the notion can be extended to exact sequences of arbitrary finite length by breaking them into short exact sequences). ′ Define the adèle ring A as the restricted product v(Kv, v) with the product measure. × ′ × × O Equip the idèle group A = v(Kv , v ) with the product of the multiplicative Haar measures. The field K embeds diagonallyO in A, and K×Qembeds diagonally in A×. Equip the discrete groups K and K× andQ their subgroups with the counting measure, in order to equip A/K and A×/K× with measures. Let S be a finite nonempty set of places of K containing all the archimedean places. Let Snonarch be the set of nonarchimedean places in S. Define the ring of S-integers by := x K : v(x) 0 for all v / S . O { ∈ ≥ ∈ } Lemma 2.1. We have a measure-compatible isomorphism K A v∈S v × v∈ /S Ov ∼ −→ K Q Q and a measure-compatible short exact sequenceO K K 0 v∈S v × v∈ /S Ov v∈S v 0. −→ Ov −→ −→ −→ v∈ /S Q Q Q Y O O Proof. Since S is nonempty, strong approximation [Cas67, 15] implies that any x A can § ∈ be written as y + ǫ with y K and ǫ = (ǫv) A such that ǫv v for all v / S. In other words, the homomorphism ∈ ∈ ∈ O ∈ A K v × Ov −→ K v∈S Y vY∈ /S is surjective. Its kernel is K v∈S Kv v∈ /S v = , so we obtain the claimed isomorphism. By definition of the measures,∩ the× upperO three homomorphisO ms in the diagram Q Q K / A v∈S v × v∈ /S Ov Q Q K A v∈S v × v∈ /S Ov ∼ / K Q O Q respect the measures, so the induced isomorphism at the bottom does too. The exact sequence is obtained from the measure-compatible split exact sequence 0 K K 0 −→ Ov −→ v × Ov −→ v −→ vY∈ /S vY∈S vY∈ /S vY∈S by dividing each of the last two terms by the image of with the counting measure. O Define the S-Arakelov divisor group (cf. the usual Arakelov divisor group in [Neu99, III.1.8]) by Div := R (Z log q ), O × v v∈S v∈ /S M 3 M d where each R has Lebesgue measure and each Z log qv has the counting measure. We have a 0 homomorphism deg: Div R that sums the components, and its kernel is denoted Div , O→ 0 O which acquires a measure compatible with the sequence 0 Div Div R 0. d× → O → O → d→ The homomorphism A Div sending (xv) to (log xv v) restricts to a homomorphism 0 → O | | 0 K× Div whose cokernel is called the S-Arakelov class groupdPic . LetdA× be the kernel → O O 1 × d deg S of the composition A Div R. Equip R := v∈S R and R with Lebesgue measure. d → O → sum c Equip the sum-zero hyperplane RS := ker RS R with the measure compatible with 0 −→ L d S those (equivalently, identify R0 with its projection under the forget-one-coordinate map and use Lebesgue measure on the image). Lemma 2.2. We have a measure-preserving exact sequence

RS 0 0 0 Pic Pic 1. −→ im × −→ O −→ O −→ O Proof. Apply the snake lemma to c 1 / × / K× / K×/ × / 1 O O 0 0 / RS / Div / Div / 0. 0 O O If char K > 0, let q be the size of thed constant ﬁeld of K. In all the formulas below, the term in square brackets involving log q should be present only if char K > 0. Each copy of R has Lebesgue measure, which induces a measure on quotients such as R/(Z log q). Lemma 2.3. We have a measure-compatible exact sequence

× × v Kv,1 A1 0 R R 1 × Pic 0. −→ µ −→ K −→ O −→ Z log qv −→ Z log q −→ Q v∈Snonarch M Proof. Since S contains all archimedeanc places, we have a commutative diagram

/ / / R / 0 / R Z log qv / R Z log qv / / 0 × × Z log qv v∈S v∈Snonarch vMarch v nonarchM M Mv∈ /S M sum sum 0 / R R / 0 with exact measure-compatible rows. The second vertical homomorphism is deg : Div 0 O → R, so it is surjective with kernel Div . The upper left group is im A× Div , so the O → O 0 d kernel of the first vertical homomorphism is im A× Div . If K is a number field, d 1 → O d then the first vertical homomorphism is surjective. If K is a function field, then its image is Z log q since a smooth projective curve over Fq has closedd points of every sufficiently large degree. Thus the snake lemma explains exactness at the last three nontrivial positions in 4 the exact sequence

× × 0 R snake R 1 Kv,1 A1 Div 0, −→ −→ −→ O −→ Z log qv −→ Z log q −→ v v∈Snonarch Y M d × × and exactness at the ﬁrst two positions follows since Kv,1 is the kernel of log v : Kv R. × × × || → The discrete subgroup K and compact subgroup v Kv,1 of A1 intersect in µ, so forming quotients yields the claimed exact sequence. By construction, each of the exact sequences above is measure-compatible. Q

If K is a number field, let (ei)1≤i≤n be a basis for the ring of integers of K, and define Disc K := det(Tr(e e )) Z. If K is a global function field of genus g over F , define i j 1≤i,j≤n ∈ q Disc K := q2g−2; this is so that in both cases it is Disc K s/2 times the completed zeta function below that is symmetric with respect to s | 1 |s. Define the S-class number 7→ − × S h( ) := # Pic . By the proof of the Dirichlet S-unit theorem, the image of R0 is a fullO lattice; itsO covolume is called the S-regulator R( ). O → O Lemma 2.4. We have A vol = Disc K 1/2 K | | K vol v∈S v = Disc K 1/2 | | Q O 2, if K R; v ≃ vol K× = 2π, if K C; v,1  v ≃  1, if Kv is nonarchimedean. 0 vol Pic = h( ) R( ) O O O A× 2r1 (2π)r2 h( ) R( ) [log q] vol c 1 = O O . K× w( ) log q O v∈Snonarch v (Again, the term in square brackets should be presentQ only if char K > 0.) Proof. The first formula can be found in [Wei82, 2.1.3]. It implies the second, by Lemma 2.1. × § For vol(Kv,1) for archimedean v, see Tate’s thesis [Tat67, p. 337]. For nonarchimedean v, × × the group Kv,1 = v has volume 1 by definition of the measure. Lemma 2.2 computes 0 O vol Pic . Lemma 2.3 yields the last formula. O −s/2 1−s Definec gamma factors ΓR(s) := π Γ(s/2) and ΓC(s) := (2π) Γ(s), and define the completed zeta function by

r1 r2 ζK(s) :=ΓR(s) ΓC(s) ζK(s). (Warning: We have used the definitions of [Wei67, VII. 6], but other authors use definitions of b § ΓR(s)and ΓC(s) that are nonzero constant multiples of these; cf. [Del73, 3.2] and [Den91, 2]. s/2 § § Some authors also include a factor Disc K in the definition of ζK(s) [Neu99, p. 467].) | | 5 b Lemma 2.5. We have × A1 vol K× lim(s 1)ζK(s)= . s→1 − Disc K1/2 [log q] | | Proof. See the proofs of [Wei67, Theoremsb 3 and 4]. Theorem 2.6 (Analytic class number formula for S-integers). We have

r1 r2 −1 2 (2π) h( ) R( ) v∈Snonarch ((1 qv ) / log qv) lim(s 1)ζO(s)= O O − . s→1 − w( ) Disc K 1/2 OQ| |

Proof. The functions ζK(s) and ζO(s) diﬀer only in that the former contains

gamma factors ΓR(s) and ΓC(s), which take the value 1 at s = 1, and • b −s −1 −1 −1 Euler factors (1 qv ) for each v Snonarch, which take the value (1 qv ) at • s = 1. − ∈ − The formula follows from this, the last formula of Lemma 2.4, and Lemma 2.5.

3. Characterizing rings of S-integers Lemma 3.1. Let be an integrally closed domain that is finitely generated as a Z-algebra. Let K = Frac . IfO the Krull dimension dim is 1, then K is a global field and is a ring of S-integers inO K in the sense of Section 2.O O Proof. Case 1: The image of Spec Spec Z is a closed point. Then is a 1-dimensional O→ O algebra over Fp, so Spec is a regular curve, equal to C S, where C is the smooth projective curve with function fieldOK, and S is a nonempty finite− set of places. Case 2: The morphism Spec Spec Z is dominant. Then it is of relative dimension 0 since dim = 1 = dim Z. ThusOK →is a finite extension of Q. Since is integrally closed, O −1 O O contains the integral closure K of Z in K. Thus = K [S ] for some set S of places of K. Since is finitely generated,O S is finite. O O O 4. Describing 1-dimensional schemes From now on, X is a reduced affine finite-type Z-scheme of pure dimension 1, say X = Spec . Let(Ci)i∈I be the 1-dimensional irreducible components of X. Let Ki be the function field ofO C . Let K be the total quotient ring of , obtained by inverting all non-zerodivisors, i O so K = Ki. Let X be the set of closed points of X. These points correspond to maximal ideals of , which| Q| are exactly the prime ideals with finite residue field since is a finitely generated O O Z-algebra [EGA IV3, 10.4.11.1(i)]. Let π : X X and π : C C be the normalization morphisms. Then X is the disjoint → i i → i union C . Correspondingly, the integral closure of in K is a finite product of rings . i O O Oi By Lemmae 3.1, K is a globale field and there exists S such that is the ringe of S -integers ` i i Oi i in Ki. Sincee π is an isomorphism above the generice point of each Ci, it is an isomorphisme above X Z for some finite subset Z X . e − ⊂| | Lemma 4.1. The index ( : ) is finite. O O 6 e Proof. Since is a finitely generated Z-algebra, is a finite -module. The finite -module O O O O / is supported on Z, and each p Z has finite residue field. Thus / has a filtration O O ∈ O O with quotients that are finite as sets, so / ise finite as a set. e O O e Our work so far proves the following. e Proposition 4.2. A scheme X is a reduced affine finite-type Z-scheme of pure dimension 1 if and only if X = Spec for some finite-index subring of a finite product of rings of O O Si-integers in global fields Ki.

5. Invariants We retain the notation of Section 4.

5.1. The invariants m, r1, r2 of K. Let m = m(K) be the number of irreducible components of X, so m = #I. Define r1 = r1(K) and r2 = r2(K) so that r1 is the number of ring homomorphisms K R, and 2r is the number of ring homomorphisms K C whose image is not contained → 2 → in R. If K is a number field, these are the usual r1 and r2. If K is a global function field, then r1 = r2 = 0. In the general case K = Ki, we have r1(K) = r1(Ki) and r2(K)= r2(Ki). Q P 5.2. TheP unit group O× and roots of unity. Let × be the unit group of . Later we will prove that × is a finitely generated abelian group.O Let µ( ) be the torsionO subgroup of ×. Let w( O) := #µ( ). O O O O 5.3. The Picard group Pic O and the class number h(O). Let Pic := Pic X = 1 O∗ O H (X, X ) [Har77, Exercise III.4.5]. Later we will prove that Pic is finite. Let h( ) := # Pic . O O O

5.4. The discriminant Disc O. For each global field Ki, we defined Disc Ki in Section 2. Define Disc :=( : )2 Disc K ; this is so that in the case where is an order in the O O O i O ring of integers of a number field, Disc = det (TrK/Q(eiej))1≤i,j≤n for any Z-basis (ei) of Q O . e O

5.5. The logarithmic embedding and the regulator R(O). Let S = Si and Snonarch = (Si)nonarch. For v Si S, let Kv be the completion of Ki at v. Taking the product of ∈ ⊆ ` × log ||v R × λ RS `the homomorphisms Kv yields a homomorphism v∈S Kv . Let φ be the composition −→ −→ Q × K× λ RS. e O −→ v −→ vY∈S × Let φ = φ × . Since ker λ is boundede in K while is a discrete closed subset of |O v∈S v O v∈S Kv, ker φ is finite; on the other hand, the codomain of φ is torsion-free; thus ker φ = µ( ). e Q QO S The group φ( ×) is a direct product of lattices in R i . Later we will prove that × is O 0 O of finite index in ×, so φ( ×) is again a full lattice L( ) in RSi . The covolume of L( ) Q 0 is called the regulatore eO , R( O). O O O 7 Q e 5.6. The zeta function. Since Spec is of finite type over Z, it has a zeta function defined as an Euler product, as in [Ser65, p. 83]:O ζ (s) := 1 q−s −1 . X − p p∈|X| Y The product converges only for s C with sufficiently large real part, but as is well known ∈ and as we will explain, ζO(s) admits a meromorphic continuation to the whole complex plane and has a pole at s = 1 of order m. We have now defined all the quantities appearing in Theorem 1.1.

6. Relating the invariants of and O O Theorem 1.1 for a product of rings of S-integers follows from Theorem 2.6. In particular, e it holds for . To prove it for , we compare the formulas for and term by term. O O O O For maximal ideals p and P , we write P p when π maps the closed point P X to p X. e ⊆ O ⊆ O | e ∈ ∈ e e 6.1. The zeta functions of O and O. Hecke [Hec17], generalizing Riemann’s work, proved that the Dedekind zeta function of a number field has a meromorphic continuation to the entire complex plane and has a simplee pole at s = 1. The analogous result for global function fields was proved by F. K. Schmidt [Sch31]. These imply the analogue for a ring of S-integers in a global field. Taking a product yields the corresponding result of products of m rings of S-integers, except that now the pole has order m; this applies in particular to . Next, by definition, O −s −1 ζ e(s) (1 qP ) e O = P|p − , ζ (s) (1 q−s)−1 O p Q p Y − where, for all but finitely many p, the fraction on the right is 1; cf. [Jen69, Theorem]. Thus ζO(s) too is meromorphic with a pole of order m at 1, and we deduce the following. Proposition 6.1. We have −1 −1 ζ e(s) P p(1 qP ) lim O = | − . s→1 ζ (s) −1 −1 O p Q 1 qp Y − 6.2. The discriminants of O and O. Our definition of Disc immediately implies the following. O e Disc Proposition 6.2. We have O =( : )−2. Disc O O Oe 6.3. Local units. Let c be the annihilatore of the -module / . Then c is also an ⊆ O O O O -ideal, called the conductor of . It is the largest -ideal contained in . O O O O Let p be a maximal ideal of . Let be the localization of the e-algebra at p. Then O Op O O e is a semilocal ring whose maximal ideals corresponde to the finitely many maximal ideals Op P lying above p. Since / is finite,e / / , and / is nontriviale for ⊂ O O O O O ≃ p Op Op Op Op onlye finitely many p. Let cp be the localization of c at p. e e e8 Q e e Lemma 6.3. The natural map × × p/cp p O O× × p → ( /c ) O Oep p is an isomorphism. e Proof. Case 1: c = . Then 1 c , so c = too; thus both sides are trivial. p Op ∈ p p Op Case 2: c = . Then c p P for every maximal ideal P of . If an element p 6 Op p ⊆ Op ⊂ Op a¯ ( /c )× is lifted to an element a , thene a lies outside each P, so a ×. Thus ∈ Op p ∈ Op ∈ Op × ( /c )× is surjective. Similarly, × ( /c )× is surjective. Bothe surjections have Op → Op p Op → Op p the samee kernel 1 + cp, so the result follows.e e Lemmae e 6.4. If c = , then p 6 Op × # /c = # /c 1 q−1 , Op p Op p − P P|p Y #(e /c )× =#( e /c ) 1 q−1 . Op p Op p − p Proof. The maximal ideals of p/cp are the ideals P p/cp for P p. An element of p/cp is a unit if and only if it lies outsideO each maximal ideal.O The probability| that a randomO element −1 of the finite group p/cp lies outsidee P p/cp is 1 qPe , and these events for differente P are independent by theO Chinese remainderO theorem,− so the first equation follows. The second equation is similar (bute easier). e Lemma 6.5. We have × −1 P|p(1 qP ) # Op = #O − . × · 1 q−1 p p p Q p Y Oe Oe Y − Proof. By Lemmas 6.3 and 6.4, × (1 q−1) p p P|p − P #O× = #O −1 ; p p · Q 1 qp Oe Oe − this holds even if cp = p since both sides are 1 in that case. Now take the product of both sides and use the isomorphismO of finite groups

O Op . ≃ p p Oe Y Oe 6.4. Using the Leray spectral sequence to relate units and Picard groups for O O O O× O and . Let X be the structure sheaf of X, and let X be the sheaf of units of X . Deﬁne × O e similarly. X e 1 × Lemma 6.6. The sheaf R π O e on X is 0. ∗ X 1 × Proof. By [Har77, Proposition III.8.1], its stalk (R π O e ) at a closed point p of X is ∗ X p lim Pic π−1U, where U ranges over open neighborhoods of p in X. Since π−1(p) is ﬁnite, U −→every line bundle on π−1U becomes trivial on π−1U ′ for some smaller neighborhood U ′ of p in X. Thus lim Pic π−1U = 0. U −→ 9 1 × Lemma 6.7. We have H (X, π O e ) Pic X. ∗ X ≃ Proof. The Leray spectral sequence e

Hp (X, Rqπ F ) = Hp+q X, F ∗ ⇒ × with F = O e yields an exact sequence e X

1 × 0 1 × 0 H X, π O e Pic X H X, R π O e . −→ ∗ X −→ −→ ∗ X Lemma 6.6 above completes the proof. e Proposition 6.8. The following is an exact sequence of ﬁnite groups:

× × (3) 0 O Op Pic Pic 0. −→ × −→ × −→ O −→ O −→ p p Oe M Oe e × × Proof. View p / p as a skyscraper sheaf on X supported at p. Then we have an exact sequence of sheavesO O on X e × × × p 0 O π O e O 0. −→ X −→ ∗ X −→ × −→ p p M Oe The corresponding long exact sequence in cohomology is

× × × p 1 × 0 O Pic X H X, π O e 0. × ∗ X −→ O −→ O −→ p p −→ −→ −→ M Oe e By definition, Pic X = Pic . By Lemma 6.7, the last term is Pic X = Pic . O O The second term in (3) is finite by Lemma 6.5. Finally, is a finite product of rings of O S-integers, each of which has finite Picard group, so Pic is finite.e Thus alle four groups in (3) are finite. O e e Remark 6.9. For a more elementary derivation of (3), at least in the case where is an integral domain, see [Neu99, Proposition I.12.9]. O

Proposition 6.10. We have

× −1 h( ) P|p(1 qP ) #O = O #O − . × h( ) · · 1 q−1 p Q p Oe Oe Oe Y − Proof. Take the alternating product of the orders of the groups in (3) and use Lemma 6.5.

Remark 6.11. Finiteness of ( × : ×) < can also be viewed as a consequence of the O O ∞ finiteness of ( : ), by [BL17, Theorem 1.3]. O O e 10 e 6.5. The regulators of O and O. Let L = L( ) be as in Section 5.5, and define L = L( ) Si O × × O similarly. The group L is a full lattice in R0 . By Proposition 6.8, ( : ) is finite, so S O O L is a full lattice in R i too. e e e 0 Q e e Proposition 6.12. QWe have R( ) × w( ) O #O = O . R( ) · × w( ) O O O Proof. Applying the snake lemma to e e e / / / / 1 / µ( ) / × / L / 0 O _ O _ _

1 / µ( ) / × / L / 0 O O yields an exact sequence e e e µ( ) × L 1 O O 0 −→ µ( ) −→ × −→ L −→ Oe Oe e of finite groups, the last of which has order R( )/R( ). O O 6.6. Conclusion of the proof. To complete the proof of Theorem 1.1, we compare (2) for e to (2) for . The ratio of the left side of (2) for to the left side of (2) for is O O O O ζOe(s) e lim . e s→1 ζO(s) The ratio of the right sides is −1/2 −1 Disc w( ) h( ) R( ) O O O O . Disc w( )! · h( ) · R( ) O O O O e e e e By Propositions 6.1, 6.2, 6.10, and 6.12 and the definition of Disc , both ratios equal O (1 q−1)−1 P|p − P . −1 −1 p Q 1 qp Y − 7. Example 1: a fiber product of rings Let p be an odd prime. Consider the fiber product of rings1

:= Z[1/2] F F [t]= (a, f) Z[1/2] F [t]: a f(0) (mod p) . O × p p { ∈ × p ≡ } Then = Z[1/2] Fp[t] inside K = Q Fp(t). Thus X := Spec is the disjoint union of two “curves”O Spec×Z[1/2] Spec F [t], and× X := Spec is the sameO except that the points ∐ p O (p) Spece Z[1/2] and (t) Spec Fp[t] are attached. Definee e ∈ ∈ p := (a, f) Z[1/2] F [t]: a f(0) 0 (mod p) { ∈ × p ≡ ≡ } P := (a, f) Z[1/2] F [t]: a 0 (mod p) { ∈ × p ≡ } P′ := (a, f) Z[1/2] F [t]: f(0) 0 (mod p) . { ∈ × p ≡ } 1A fiber product of rings does not correspond to a fiber product of schemes. 11 Then p is a prime of (the point of attachment), and P and P′ are the primes of lying O O above p. The conductor of is p viewed as an -ideal. Propositions 7.1 and 7.2O below verify TheoremO 1.1 for by computing the two sidese of O (2) independently. e Proposition 7.1. We have −1 2 1 p lim(s 1) ζX (s)= − . s→1 − 2 log p Proof. We have

lim(s 1)ζZ(s)=1, s→1 − s 1 1 lim(s 1)ζFp[t](s) = lim − = . s→1 − s→1 1 p1−s log p − Taking zeta functions of X p = X P, P′ = (Spec Z (2), (p) ) (Spec F [t] (t) ) −{ } −{ } −{ } ∐ p −{ } yields e −s −s −s −s (1 p ) ζ (s)=(1 2 )(1 p ) ζZ(s) (1 p ) ζF (s) − X − − · − p[t] 2 −s −s (s 1) ζ (s)=(1 2 )(1 p ) ((s 1) ζZ(s)) (s 1) ζF (s) − X − − − − p[t] −1 2 −1 −1 1 1 p lim(s 1) ζX (s)=(1 2 )(1 p ) 1 = − . s→1 − − − · · log p 2 log p Proposition 7.2. We have r1 r2 −1 −1 2 (2π) h( ) R( ) ((1 qv ) / log qv) 1 p O O v∈Snonarch − = − . w( ) Disc 1/2 2 log p OQ| O| Proof. First, r1 = 1 and r2 = 0. The set Snonarch consists of the place 2 of Q and the place of F (t). By definition, ∞ p Disc =( : )2 (Disc Q) (Disc F (t)) = p2 1 p2·0−2 =1. O O O p · · × × × n × Inside = Z[1/2] F = 2 Z F , we have O × p e ±{ }n∈ × p × = (2n, 2n mod p): n Z . e O {± ∈ } In particular, µ( )= 1 , so w( ) = 2. Since × and × agree modulo torsion, O {± } O O O R( )= R( )= R(Z[1/2]) R(F [t]) = (log 2) 1 = log 2. O O p e · By Lemma 6.3, e × ( /P)× ( /P′)× F× F× Op p × p × O × O× × , p ≃ ( /p) ≃ Fp Oe O in which the denominator F× is embeddede diagonallye in F× F×. Substituting into (3) yields p p × p × F× F× p × p 1 O× × Pic Pic 1. −→ −→ Fp −→ O −→ O −→ Oe 12 e F× F× F× × p × p Z F The subgroup 1 p of surjects onto × , and Pic = Pic [1/2] Pic p[t]= 1 , × O Fp O × { } so Pic = 1 . Thus h( ) = 1. e e SubstitutingO { } all these valuesO shows that the expression to be computed equals 21 (2π)0 1 (log 2) ((1 2−1)/ log 2) ((1 p−1)/ log p) 1 p−1 · · · − − = − . 2 11/2 2 log p · 8. Example 2: a non-maximal order in a real quadratic number field Let = Z[√d], where d is an integer such that d 5 and d 5 (mod 8) (other cases O ≥ ≡ could be handled similarly). Then = Z[(1+√d)/2]. Above the prime ideal p := (2, 1+√d) O of with residue field F2 lies the prime ideal (2) of with residue field F4. The scheme X O:= Spec is analogous to a nodale curve for whichO the two slopes at the node p are O conjugate in the quadratic extension F of F . Letǫ ˜ebe the fundamental unit of ×, so 4 2 O ǫ˜ > 1 for the standard real embedding ֒ R. Let n be the order of the image ofǫ ˜ in × × O → ։ ( /(2)) F4 , so n is 1 or 3. Since is the preimage of F2 under F4, the elemente ǫO:=˜ǫn is≃ the smallest power ofǫ ˜ lyingO ine ×. O O ePropositions 8.1 and 8.2 below verify Theorem 1.1 for by computinge the two sides of (2) independently. O Proposition 8.1. We have

2 3 h( ) logǫ ˜ lim(s 1) ζO(s)= O . s→1 − 2√d e Proof. The classical analytic class number formula for , with r = 1, r = 0, R( ) = logǫ ˜, O 1 2 O w( ) = 2, Disc = d, yields O O e e h( ) logǫ ˜ e e lim(s 1)ζ e(s)= O . s→1 − O √d e On the other hand, Proposition 6.1 with qp = 2 and qP = 4 yields

ζ e(s) (1 1/4)−1 2 lim O = − = . s→1 ζ (s) (1 1/2)−1 3 O − Dividing the ﬁrst equation by the second gives the result. Proposition 8.2. We have

r1 r2 −1 2 (2π) h( ) R( ) ((1 q ) / log qv) 3 h( ) logǫ ˜ O O v∈Snonarch − v 1/2 = O . w( Q) Disc 2√d O | O| e Proof. First, r = 1, r = 0, and w( ) = 2. By deﬁnition, S = . By Proposition 6.2, 1 2 O nonarch ∅ Disc =( : )2 Disc =4d, and R( ) = log ǫ = n logǫ ˜. The exact sequence (3) is O O O O O × F× e e 4 1 O× × Pic Pic 1, −→ −→ F2 −→ O −→ O −→ Oe 13 e so h( )=(3/n) h( ). Thus the expression to be computed equals O O 21 (2π)0 (3/n) h( ) n log ˜ǫ 3 h( ) logǫ ˜ e · O · = O . 2 (4d)1/2 2√d · e e Acknowledgments It is a pleasure to thank Tony Scholl for helpful discussions. We thank Carlos J. Moreno for bringing the article [St¨o98] to our attention, and John Voight for suggesting the reference [Wei82]. Finally, we thank the referee for several excellent suggestions.

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Department of Mathematics, Baruch College, The City University of New York, One Bernard Baruch Way, New York, NY 10010-5526, USA E-mail address: [email protected] Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139-4307, USA E-mail address: [email protected] URL: http://math.mit.edu/~poonen/