Distribution of orders in number fields

In this paper we study the distribution of orders of bounded discriminants in number fields. We give an asymptotic formula for the number of orders contained in the ring of integers of a quintic number field.


Introduction
Let K/Q be an extension of degree n with ring of integers O K . An order O is a subring of O K with identity that is a Z-module of rank n. Set In this paper we study the asymptotic growth of N K (B) as B grows.

Results
Let E/Q be the normal closure of K with Galois group G = Gal (E/Q). Then G has a natural embedding in S n as a transitive subgroup. Let V 2 be the vector space whose basis is the set of 2-element subsets of {1, · · · , n}. The group G has a natural action on V 2 . Let r 2 be the dimension of the space of G fixed vectors in V 2 . Then we have the following theorem: Theorem 1. Let K/Q number field of degree n.
1. For n ≤ 5, there is a constant C K > 0 such that 2. For any n > 5, Table 1 lists the transitive subgroups of S n for small n and the corresponding r 2 values. The reference for the list of subgroups up to conjugation is [DM], §2.9. For the computation of r 2 , see §3.5.  In order to study the behavior of N K (B) we form the counting zeta function where O K is the ring of integers of K and O is an order. This series converges absolutely for ℜs large, and in its domain of absolute convergence we have The zeta functionη K has an Euler product of the form and the summation in the last expression is over full rank sublattices of O K ⊗ Z Z p that are subrings with identity. We define the coefficients a i (p) bỹ The number a i (p) is what in §1.3 be be denoted by a 1,< O K (p i ).
A portion of Theorem 1 is a consequence of the following result: Theorem 2. The Euler product f (s) = p unramified (1 + a 1 (p)p −s ) converges absolutely for ℜs large, and it has an analytic continuation to a meromorphic function on an open set containing ℜs ≥ 1 with a unique pole at s = 1 of order r 2 .
Remark 1. It is reasonable to conjecture that for n small the functionη K (s) is holomorphic for ℜs > 1, and has an analytic continuation to a domain containing ℜs ≥ 1 with a unique pole of order r 2 at s = 1. If this is true, then there is a non-zero constant C K such that as B → ∞. The conjecture is true for n ≤ 5 by Theorem 1. The results of Brakenhoff [Br], summarized in §1.2 below, show that for n ≥ 8 there is a pole to the right of ℜs = 1.
Part 1 of Theorem 1 is a consequence of the following theorem combined with Lemma 4.15 of [dSG1]: Theorem 3. Let n ≤ 5. There is a finite set S of primes such that the series p ∈S ∞ i=2 a i (p) p iσ converges for any real σ > 19/20.
We give heuristic reasoning for why this result should hold in the case n = 5. Let b i (p) be the number of subrings with identity of Z 5 p , i.e. orders, whose index is p i . It is reasonable to expect that for all i and p. It is then a consequence of Theorem 13 that the series converges for σ > 19/20. Alas, we have not been able to prove (1)-even though we are confident it is true. Here we employ an alternative method based on p-adic integration.
Part 2 of Theorem 1 is a consequence of the following theorem and Lemma 4.15 of [dSG1]: Theorem 4. Let n > 5. There is a finite set S of primes such that the series p ∈S ∞ i=2 a i (p) p iσ converges for any real σ > n 2 − 7 6 . Remark 2. Note that by Theorem 1.5 of [dSG1], the zeta function η K (s) has an analytic continuation to a domain of the form ℜs > α − ǫ with α > 0 the abscissa of convergence and ǫ > 0.
Remark 3. A byproduct of our methods, stated as Corollary 4 and Corollary 5 in §5, is an improvement of the upper bounds obtained by Brakenhoff [Br], Theorem 5.1 and Theorem 8.1. Given k ∈ N, we define f d (k) to be the number of orders in Z d of index equal to k. Clearly, N d (B) = k≤B f d (k). It is easy to see that the function f d (k) is multiplicative, i.e. if k 1 , k 2 are coprime integers then

More generally if
This is the prototype of the problem that we study in this paper: Despite its innocent appearance this is a very difficult problem, and prior to our work the only cases for which an asymptotic formula is known for N d (B) are d = 2, 3, 4 [L]. Here we obtain an asymptotic formula for N 5 (B), and give non-trivial bounds for N d (B) when d > 5.
Definition. Let d, k ∈ N. We define a < Z d (k) to be number of subrings S of Z d , not necessarily with identity, such that [Z d : S] = k.
A subring S in Z d which is of finite index as an additive group will necessarily be a free Z-module of rank d. Such subrings are called multiplicative sublattices in [L]. An elementary proposition in [L] states that for any d, k ∈ N, d ≥ 2, we have f d (k) = a < Z d−1 (k). As a result, with the notation of §1.3 η Q d (s) = ζ < Z d−1 (s). Determining the asymptotic behavior of N 1 (B) and N 2 (B) is trivial. In this paper we will use the method of p-adic integration as in §1.3 to prove the following theorem: Theorem 5.
1. Let d ≤ 5. There is a positive real number C d such that 2. Suppose d ≥ 6. Then for any ǫ > 0 we have We actually prove a more precise statement and give error estimates; See Theorems 11, 12 and 13. We include the d = 3 case to illustrate our method in a simple case. Our results for d ≥ 5 are new.
Theorem 5 is more than just a prototype of the type of result we can prove. The computations in §4.5 form the backbone of the proof of Theorem 1. In fact, Theorem 7 shows that, essentially, whatever estimate we obtain for the volumes of the sets considered in §4.5 works in general.
We expect the asymptotic formula in Part 1 of Theorem 5 to be valid for d < 8. The formalism of p-adic integration shows that N d (B) has an asymptotic formula of the form CB a (log B) b−1 , for a rational number a and a natural number b, but for d ≥ 8 it is not clear what the numbers a, b should be.
We finish this introduction with the following conjecture: Conjecture 1. Let K/Q be a number field of degree d. There exists a natural number β K such that
Since in this work we are counting sublattices with additional structure we expect slower asymptotic growth. Theorem 1 is trivial for a quadratic field as the counting zeta function is simply the Riemann zeta function ζ(s). For K a cubic or quartic extension of Q, Theorem 1 is due to Datsovsky-Wright [DW] for the cubic case, and Nakagawa [N] for the quartic case.
In the cubic setting, there is a bijection between the set of equivalence classes of integral binary cubic forms and the set of orders of cubic fields. Then it follows from Shintani's theory of zeta functions associated to the prehomogeneous vector space of binary cubic forms combined with a theorem of [DW] thatη In the quartic setting, Nakagawa explicitly computes the local factors of the zeta functionη K using an intricate combinatorial argument involving counting the number of solutions of some very complicated congruences. Due to computational difficulties at the prime 2, Nakagawa's theorem assumes some mild ramification conditions. Under these conditions, he shows that the zeta functionη K (s) has an analytic continuation to ℜs > 2/3. Nakagawa's explicit local computations can also be used to prove Theorem 5 for d = 4. The paper [L] contains a different approach to Theorem 5 using combinatorial arguments. Here, too, the local Euler factors of the counting zeta function are explicitly computed, though the proof follows from elegant recursive formulas, c.f. Propositions 6.2 and 6.3 of [L].
In a series of spectacular papers, Bhargava studies orders in quintic fields. In [B1], he shows that there is a canonical bijection between the set of orbits of GL 4 (Z) × SL 5 (Z) on the space Z 4 ⊗ ∧ 2 Z 5 and the set of isomorphism classes of pairs (R, S) with R a quintic ring and S a sextic resolvent ring of R. An impressive theorem of Bhargava [B2] which is proved using the above bijection says that The thesis [Br] contains an array of interesting results on the distribution of orders in number fields. In keeping with our notation below, for a number field K we let We then let a 1,< (n, m) = max Theorem 5.1 of [Br] is the statement that n−1+d and c 8 (n) given by the following table: One can compute the values of c 7 (n) explicitly as follows: In particular, for n ≥ 8, c 7 (n) > 1.
Theorem 8.1 of [Br], which is used in [B2], is the following result: If K/Q is a quintic field, then for any prime p We improve the upper bounds in these theorems in §5, Corollary 4 and Corollary 5.

Our method
Given a ring R whose additive group is isomorphic to Z d for some d ∈ N we define a < R (k) := |{S subring of R | [R : S] = k}| . For any k ∈ N, a < R (k) is finite. We define the subring zeta function of R by We view ζ < R (s) not just as a formal series, but as a series converging on some non-trivial subset of the complex numbers. The idea is that the analytic properties of the resulting complex function have consequences for the distribution of subrings of finite index in R. In particular, by various Tauberian theorems e.g. Theorem 8, the location of poles and their orders gives information about the function s < R (B) defined by Similar constructions can be made for subgroups of finitely generated groups and ideals in rings, but in this introduction we only consider subring zeta functions. We have the following theorem which is a summary of results from [GSS,dSG1] Theorem 6. 1. The series ζ < R (s) converges in some right half plane of C. The abscissa of convergence α < R of ζ < R (s) is a rational number. There is a δ > 0 such that ζ < R (s) can be meromorphically continued to the domain {s ∈ C | ℜ(s) > α < R − δ}. Furthermore, the line ℜ(s) = α < R contains at most one pole of ζ < R (s) at the point s = α < R . 2. There is an Euler product decomposition with the local Euler factor given by This local factor is a rational function of p −s ; there are polynomials The polynomials P p , Q p can be chosen to have bounded degree as p varies. The local Euler factors satisfy functional equations.
The functional equation mentioned in the theorem is proved in [V1]; also see Chapter 4 of [dSW]. A corollary of this theorem is that the asymptotic behavior of the function It is a fundamental problem in the subject to relate the numbers α < R , b < R , c < R ∈ R to structure of R.
The paper [GSS] introduced a p-adic formalism to study the local Euler factors ζ < R (s). Fix a Z-basis for R and identify R with Z d . The multiplication in R is given by a bi-additive map Definition. Let M p (β) be the subset of the set of d × d lower triangular matrices M with entries in Z p such that if the rows of M = (x ij ) 1≤i,j≤d are denoted by v 1 , . . . , v d , then for for all i, j satisfying 1 ≤ i, j ≤ d, there are p-adic integers c 1 ij , . . . , c d ij such that Let dM be the normalized additive Haar measure on T d (Z p ). Proposition 3.1 of [GSS] says: Most of the statements of Theorem 6 are proved using this p-adic formulation. The integral appearing in (2) is an example of a cone integral. The beauty of the equation (2) is that it allows us to express the number of subrings of a given index in terms of volumes of certain p-adic domains.
Let D = (f 0 , g 0 , f 1 , g 1 , · · · , f l , g l ) be polynomials in the variables x 1 , . . . , x m with rational coefficients. We call D the cone integral data. For a prime number p we define and we define the cone integral associated to the cone integral data D by with dx is the normalized additive Haar measure. The study of such integrals in special cases was started by Igusa [I1, I2]. Igusa's original method was based on the resolution of singularities. Igusa's approach was generalized by Denef [Dn1] and du Sautoy and Grunewald [dSG1]. Denef [Dn1] also introduced the use of elimination of quantifiers in Q p as an alternative approach. For surveys on cone integrals and their applications to zeta functions of groups and rings, as well as references and examples, see [dSG2,dSW,V2]. In general, calculating cone integrals is difficult and requires explicit desingularizations of highly singular varieties. For a "simple" example, see [dST].
There is a modification of this formalism to treat subrings with identity. Again, let R be a ring with identity whose additive group is isomorphic to Z d and for simplicity assume that the identity of R is sent to (1, 1, . . . , 1) under this isomorphism. For k ∈ N, let a 1,< R (k) := |{S subring with identity of R | [R : S] = k}|. Now define the unitary subring zeta function of R by As before we have an Euler product expansion We let Again suppose after identifying R with Z d , the multiplication on R is given by a bi-additive map β : which extends to a bi-linear map Definition. Let M 1 p (β) be the subset of M p (β) whose rows generate a unitary subring.
Then it is not hard to see that This integral too is a cone integral as we will see in §5. As a result, the asymptotic behavior of s < R (B) is of the form c < R B α < R (log B) b < R −1 as B → ∞. Again, we use the expression (3) to write the number of unitary subrings of a given index in terms of volumes of certain p-adic sets.
In our problems of interest, the ring R is a product of rings of integers of number fields. The two usual methods to study the cone integrals coming from subring zeta functions are resolution of singularities and elimination of quantifiers. Neither of these methods, however, can be applied in any obvious fashion to the problem of counting subrings of such R. This is due to the fact that our cone integrals are too complicated (see Sections 4.4 and 4.5). In general there is no effective algorithm to eliminate quantifiers for a complicated p-adic domain, and resolution of singularities, while in principle computationally tractable, is dreadful for domains of the type considered here. For example, the domain needed to study Z d would involve about d 3 inequalities of the form v p (f (x)) ≤ v p (g(x)) with x a vector of variables of length about d 2 , and f , g ranging over polynomials with integer coefficients of degrees 2 to d.
In this paper, inspired by [STT], we propose a different approach. So far as the determination of the fundamental quantities α < R , b < R is concerned, we do not need explicit computations of the local integrals. Instead, in favorable circumstances such as those under consideration here, we can accomplish this by computing the first two terms of the Euler factors and estimating the rest of the terms. It is precisely for this reason that our method can be applied to more cases that what was treated in the earlier papers [DW, L, N]. Here the difficulty lies in estimating volumes of certain p-adic sets that arise in the split situation of Z d , see §4.4, 4.5, and 4.6. Once this has been accomplished, we will use the results of §2.3 to show that the volume estimates obtained for the Z n setting automatically extend to an arbitrary R of the sort considered here.

Organization of the paper
The rest of the paper is organized as follows. In §2 we recall results by Denef [Dn2], and use them to prove Theorem 7. We prove Theorem 2 in §3, using the outline explained in §3.2. Section 3.1 contains the statements of the Tauberian theorems we use in this work. We discuss the values of r 2 in §3.5. The proof of Theorem 5 is presented in §4. The outline of the proof is sketched in §4.1 and details are postponed to later sections. In Section 4.2 we collect several lemmas used in estimating volumes. Section 4.3 contains the treatment of the simple case of Z 3 . We include this simple case to illustrate the method. In Sections 4.4 and 4.5 we give bounds for the volumes of our domains for n = 4 and n = 5, respectively. These bounds are then used in Sections 4.4.2 and 4.5.2 to establish Theorems 11, 12, and 13 which imply the first part of Theorem 5. The proof of the second part of Theorem 5 is presented in Section 4.6. The paper ends with the proof of Theorem 1 in §5.

Notation
In this paper a ring R is an additive group with a bi-additive multiplication such that the underlying additive group is finitely generated. We write S ≤ R if S is a subring of R. The number [R : S] is defined to be the index of S in R as an additive subgroup. Throughout this paper p is a prime number. When p is used as the index of a sum or product, we will always understand that it ranges through the primes. The symbols Q p and Z p are the field of p-adic numbers and its ring of integers, respectively. We let U p denote the group of units of Z p . We normalize the additive Haar measure on Q p such that vol (Z p ) = 1, and the volume of a subset of Q p is always with respect to this measure. For example, if P (x) is a statement about a p-adic number x, the volume of x ∈ Q p such that P (x) means the normalized Haar measure of the set {x ∈ Q p ; P (x)}. The measure on Q r p for any r > 0 is normalized similarly. The function v p : Q p → Z ∪ {∞} is the p-adic valuation. If f : S → C and g : S → R + are functions defined on a set S to the set of positive real numbers R + and C, respectively, the notation f (x) = O(g(x)) means there is a constant C > 0 such that for all x ∈ S we have |f (x)| ≤ Cg(x); this is also sometimes denoted by f (x) ≪ g(x). If S, T are sets, and f : S → C and g : S × T → R + are functions, the notation f (x) = O y (g(x, y)) means that for every y ∈ T , there is a constant C(y) > 0 such that for every x ∈ S we have |f (x)| ≤ C(y)g(x, y). If For a complex number s, ℜ(s), usually denoted by σ, is the real part of s. We will, without explicit mention, repeatedly use the fact that p prime p a−bs , with a, b real numbers, converges for ℜ(s) > (a+1)/b. The collection of n×n matrices with entries in a ring R is denoted by M n (R). The set of lower triangular matrices in M n (R) is written T n (R). A finite extension K/Q is called a number field, and its absolute discriminant is denoted by disc K . The ring of integers of K is written O K . A subring with identity of O K which is a Z-module of rank equal to the Z-rank of O K is called an order. We write ζ(s) for the Riemann zeta function. If ψ is a property of integers, and f an arithmetic function, p ψ f (p) means the sum of the values of f over all prime numbers p which satisfy ψ; for example if S is a set of integers, p ∈S f (p) means the sum is over all those prime numbers which are not in S.
versations, and especially for her crucial observation Lemma 1, and Eun Hye Lee for numerical computations on UIC's Argo cluster to provide support for equation (1).

Geometry and p-adic integrals
In this section we study a multivariable version of the Igusa zeta integral following the method of [Dn2] and [dSG1]. We start with some geometric preparation.

Resolutions with good reduction
We recall the the material of Section 2 of [Dn2]. In this section K is an arbitrary field of characteristic zero, R a discrete valuation subring of K with field of fractions K, P unique maximal ideal, and residue field K which we assume to be perfect. Let f (X) ∈ K[X], X = (X 1 , · · · , X m ) be a nonzero polynomial. Let A resolution (Y, h) for f over K consists of a closed integral subscheme Y of P k X for some k, and the morphism h : Y → X which is the restriction of the projective morphism P k X → X such that: 1. Y is smooth over Spec K; 2. the restriction h : Y \ h −1 (D) → X \ D is an isomorphism; 3. the reduced scheme (h −1 (D)) red associated to h −1 (D) has only normal crossings.
Let E i , i ∈ T , be the irreducible components of (h −1 (D)) red . For i ∈ T , we define N i to be the multiplicity of E i in the divisor of div h on Y, and let ν i − 1 be the multiplicity of E i in the divisor of h * (dx 1 ∧ · · · ∧ dx m ). We have We think of P k X as an open subscheme of P kX . If Z is a closed subscheme of P k X , we defineZ to be the scheme theoretic closure of Z in P kX . We also set Z =Z × R Spec K, and we call it the reduction of Z mod P .
Leth :Ỹ →X be the restriction toỸ of the projective morphism P kX → X , andh : Y → X be obtained fromh by base extension. We say (Y, h) has good reduction mod P if the following two conditions are satisfied: 1. Y is smooth over Spec K; 2.Ē i is smooth over Spec K, for all i ∈ T , and ∪ iĒi has only normal crossings; and 3. for i = j,Ē i andĒ j have no common irreducible components.
Let K ′ be a field containing K, R ′ a discrete valuation subring of K ′ who fraction field is K ′ , and which contains R, with maximal ideal P ′ containing P , and with perfect residue field. Suppose (Y, h) be a resolution of f over In the arithmetic case, let F be a number field, and O F its ring of integers. Let f (X) ∈ F [X], X = (X 1 , · · · , X m ). Let (Y, h) be a resolution for f . For any maximal ideal p, we consider the discrete valuation ring O F,p with maximal ideal pO F,p . Note that the field of fractions of O F,p is F . Theorem 2.4 of [Dn2] then states that for almost all p, (Y, h) is a resolution with good reduction mod pO F,p . As a corollary, if F p is the p-adic completion of F , and O p its ring of integers, and by abuse of notation p its unique prime ideal, then (Y, h) is a resolution of f over F p with good reduction mod p for almost all p.

Multivariable cone integral
For a finite extension F of Q p , we let be O F its ring of integers, q the maximal ideal, |.| F its normalized absolute value, and v F the corresponding discrete valuation. Let q be the size of F , the residue field of F . Let f 1 , · · · , f l and g 1 , · · · , g l be polynomials in the variables X = (X 1 , · · · , X m ) with rational coefficients. We denote by ψ F (X) the first order formula v F (f i (X)) ≤ v F (g i (X)), i = 1, . . . , l.
As before we call the formula ψ F (X) a cone condition, and the polynomials f i , g i , 1 ≤ i ≤ l, the cone data.
We define . . , h k are polynomials in X with rational coefficients, we define the cone integral in k complex variables s = (s 1 , · · · , s k ) ∈ C k with respect to ψ by Our first goal here is to find an explicit formula for Z ψ for p outside a finite set of primes. In this section, following the method of [dSG1] closely we will find an explicit formula for our multivariable cone integral which depends on the numerical invariants of a resolution.
Let r = |T a | and write T a = {i 1 , · · · , i r }. Then in the local ring O mY ,a we write We define vectors w j , 1 ≤ j ≤ r, by Define an integral J a,ψ (s, F ) by the following expression: Here the function θ is defined as follows: Let A j,a = w j and B j,a = N i j (h 0 ) + ν i j for 1 ≤ j ≤ r and A j,a = 0 and B j,a = 1 for j > r. Then The set Λ is the intersection of N m with a rational polyhedral cone C in R m . Write this cone as a disjoint union of simplicial cones C 1 , . . . , C t with Then Λ is the disjoint union of the following sets and put A i,u,I = A i,u,a and B i,u,I = B i,u,a for any a ∈ {x ∈ Y(F ); x ∈ E i if and only if i ∈ I}. Clearly, Putting everything together

The absolute convergence of the integral is guaranteed if
We note that the domain of the absolute convergence depends only on the geometry of our data, and not on the particular choice of the field F .
As in [dSG1], we derive another expression for the integral. Set where t = |T |. This is closed cone. This cone is a disjoint union of open simplicial pieces called R k , 0 ≤ k ≤ w. We assume that the fundamental region for the lattice points of R k has no lattice points in its interior. We will assume that R 0 = (0, . . . , 0) and that R 1 , . . . , R q are all the open one dimensional edges of the cone D T . Write We also set D T = ∆ T . For each I ⊂ T , there is a subset W I ⊂ {0, . . . , w} such that So if we set c F,k = c F,I and I k = I if k ∈ W I , for every non-archimedean local field F where the resolution has good reduction we have In the situation where the resolution is not necessarily of good reduction, following the argument of Proposition 3.3 of [dSG1] one proves that there exists a finite set B F such that for every b ∈ B F there is an associated subset I b ⊂ T and an integer e b such that

Application to some volume computations
Let F be a finite extension of Q p with ring of integers O F and |.| F its normalized absolute value. We fix a uniformizer ̟ F for F . Let q be the size of the residue field of F . For We define vol F and vol F n , to be the normalized Haar measure on F , and on F n , respectively. If k = (k 1 , . . . , k n ) ∈ Z n , and α ∈ F is non-zero, we set α k = (α k 1 , . . . , α kn ); in particular, We will assume that V F (x) is F -round in that it is invariant under the action of units of the local field. With abuse of language, when we say V , we mean the assignment that takes an extension F of Q p and an element x ∈ O n F , and returns the set Definition. Let α = (α 1 , · · · , α n ) ∈ R n , ℓ ∈ N, and P ∈ R[X 1 , . . . , X n ] with positive coefficients. We say V is (ℓ, α, P,

Now here is the theorem:
Theorem 7. Suppose there is α = (α 1 , · · · , α n ) ∈ R n , ℓ ∈ N, P ∈ R[X 1 , . . . , X n ] with positive coefficients, and an infinite set of primes P such that for all p ∈ P the set V is (ℓ, α, P, Q p )-narrow. Then V is (ℓ, α, P, Q p )-narrow for almost all primes p.
In the statement of the theorem "almost all" means all but possibly finitely many.
Proof. Let F = Q p for p ∈ P. In order to prove the theorem, we consider the following integral: On the other hand, we write This is a multivariable cone integral.
Since the set P is infinite, we may assume that p is good in the sense of §2.1. By §2.2 we have with non-negative integer vectors A j and non-negative integers B j . Regrouping terms gives We note that if |I i | > m + n, then c F,i = 0. As a result we may write with P i,k (X) a polynomial with positive integral coefficients which depends only on i and k, and not on the choice of the field F . Further, the number of terms of P i,k depends on k in a polynomial fashion. In particular there are no cancellations between the terms. These observations imply that V is (α, F )-narrow if and only if for each i = 0, . . . , w, we have some N such that This is true if and only if Proposition 4.9 combined with Proposition 4.13 of [dSG1] implies that, after letting p become larger in P, this inequality is true if and only if which is equivalent to Since P is infinite, we can let p → ∞, and as a result an inequality of this nature is valid if and only if it is true for degree reasons. The theorem now follows.
Remark 4. Here is a variation of the above theorem which may be useful in other contexts. There is a finite set S of primes such that every p / ∈ S has the following property: If there is α ∈ R n , ℓ ∈ N, and P ∈ R[X 1 , . . . , X n ] such V is (ℓ, α, P, F )-narrow for every F finite extension of Q p , then for all q / ∈ S, V is (ℓ, α, E)-narrow for every E finite extension of Q q .
3 The proof of Theorem 2

Tauberian theorem
We will use the Tauberian theorem of [CT], Appendix A, in the following form: a n n s be a Dirichlet series with an Euler product Suppose each Euler factor is of the form where a 1 (p) = k, a positive integer independent of p, and a l (p) are nonnegative real numbers. Suppose there is a δ 0 satisfying 1 Then there is a polynomial P of degree k − 1 such that for all ǫ > 0 n≤B a n = BP (log B)

Outline of the proof of Theorem 2
If p is unramified in K, we write where each p i is a prime ideal in O K , and let where O p i is the ring of integers of the completion of K at the prime p i , and the isomorphism class of O K ⊗ Z Z p is determined by the multi-set f p = {f 1 , · · · , f r }, called the type of p. The type of a prime is always a partition of n. We typically write the type of an unramified prime p in the form f p = 1 v 2 w r e 1 1 · · · r e k k , where 1 < 2 < r 1 < · · · < r k are the distinct residue degrees, and u, w, e 1 , · · · , e k are the number of times each of these appears.
The starting point of the proof of the theorem is the following proposition: in particular a 1 (p) depends only on the type f p .
We will present the proof of this proposition in Section 3.3. Given a partition f as above, we let Then we observe that the condition that p has type 1 u 2 w r e 1 1 · · · r e k k is Chebotarev condition in G = Gal (E/Q) in the sense that there are a number of conjugacy classes C i ⊂ G, 1 ≤ i ≤ t, such that p has type 1 u 2 w r e 1 1 · · · r e k k if and only if E/Q p = C i for some i. Here E/Q p is the Frobenius conjugacy class of p in G. Next we use the following fact: Proposition 2. Let L/K be a Galois extension of number fields with Galois group H = Gal (L/K). Let C ⊂ H be a conjugacy class and define Then F C (s) converges absolutely for ℜs > 1. Furthermore, F C (s) |H| has an analytic continuation to a meromorphic function on an open set containing ℜs ≥ 1 with a unique pole of order |C| at s = 1.
We will present the proof of this proposition in Section 3.4. Now suppose a partition f of n is given. On the one hand f can be type of a prime p, and on the other hand p determines a conjugacy class in S n . It is a well-known fact that if p has type f in K/Q, then E/Q p has cycle type f . Given a type f , we define b(f ) be the number of elements of G of cycle type f in S n . Combining everything done so far one concludes that the function f (s) in the statement of Theorem 2 has a pole at s = 1 of order We finally have the following statement: Lemma 1 (B. Srinivasan). We have r = r 2 .
Proof of Lemma. We define a function α on G as follows. If g is of cycle decomposition type f , we set α(g) = a(f ). We note that the expression on the right is equal to α, ψ where ψ is the trivial character of G, and , is the inner product on the space of class functions of G. The function α is character of the permutation representation π of G on the set of 2-element subsets of {1, 2, . . . , n}. In fact, if g is of type f as above, then it is clear that it fixes u 2 + w 2-element sets. Then the expression on the right hand side of (5) is equal to the multiplicity of the trivial representation in π. For every orbit of G on the set of 2-element subsets of {1, 2, . . . , n} we get a copy of the trivial representation in π, and these are the only copies of the trivial representation in π. It is easily seen that if G is transitive the number of such orbits is equal to r 2 .
Theorem 2 now follows from a standard Tauberian argument.

Proof of Proposition 1
We first give an overview of the proof of Proposition 1. A result of Grunewald, Segal, and Smith shows that determining a 1 (p) is equivalent to a counting problem about certain lower-triangular matrices. By Lemma 5.18 of [Br] O p := O K ⊗ Z Z p is a Z p -module of rank n. By choosing a special type of basis for O p and then applying elementary row operations the lower-triangular matrices we consider will be of a relatively simple form. We then break up the overall computation of a 1 (p) into a few parts depending on the type of p. The proof of Proposition 1 depends on the following lemmas.
Lemma 2. Let L/Q p be an extension of degree n. If n > 2, the ring of integers O L of L does not have any subrings of index p that are Z p modules of rank n.
This result shows that in order to determine a 1 (p) in general, we need only determine primes of a restricted type.
Lemma 3. Let p be a prime of type f p = 1 v 2 w r e 1 1 · · · r e k k , and let q be a prime of type f q = 1 v 2 w . Then a 1 (p) = a 1 (q).
We will determine a 1 (p) for primes of this type by considering primes of type 1 v and primes of type 2 w separately. The next lemma follows directly from [L] Proposition 1.1.
The proof of Proposition 1 will follow from combining these results in the following way.
Lemma 6. Let p be a prime of type f p = 1 v 2 w . Then a 1 (p) = v 2 + w. We now explain how to interpret a 1 (p) in terms of a counting problem about lower-triangular matrices. The first observation is that a 1 (p) depends only on O p and not on K. We choose any ordered basis of this ring, {v 1 , . . . , v n } and represent a subring L of O p by a matrix M where the ith column corresponds to v i and L is generated by the rows of M. The entries of this matrix are in Z p . By elementary linear algebra, a version of Gauss-Jordan elimination over Z p , we are free to suppose that M is lower triangular. Multiplying a row of M by a unit in Z p does not change the subring generated by M. Therefore, we may suppose that the (i, i) entry of M is equal to p k i for some k i ≥ 0.
Let M(p) denote the set of all lower triangular matrices whose rows generate a subring of O p with respect to this ordered basis. We can now present a slight modification of a proposition of Grunewald, Segal and Smith.
Proposition 3 (Grunewald, Segal, Smith). For every prime p, where |dv| is the additive Haar measure of the p-adic lower triangular matrices.
The index of a subring L ⊆ O p is the determinant of any matrix M ∈ M(p) generating L. By definition, a 1 (p) is equal to the p −s coefficient of the integral in this proposition. We therefore need only consider matrices M ∈ M(p) where exactly one x ii is equal to p and all others are equal to 1.
Suppose the rows of M generate a subring of O p of index p and suppose that x jj = 1 for some j. By adding multiples of the jth row of M to its other rows we can set each of the nondiagonal entries in column j to 0 without changing the subring generated by this matrix. In fact, by applying a version of Gauss-Jordan elimination we can simultaneously accomplish this for each column which has its diagonal entry equal to 1. This gives a matrix that is diagonal except for a single column that may have nonzero entries below the diagonal. We give an example below:  Suppose the rows of M generate a subring of O p of index p, x jj = p for some j, and every other column of M has a single 1 on the diagonal and is 0 otherwise. Let {a 0 , a 1 , a 2 , . . . , a p−1 } be some choice of representatives for O p /pO p with a 0 = 0 and a 1 = 1. By adding multiples of row j to the rows below it, we may suppose that the entries x j+1,j , x j+2,j , . . . , x n,j are all elements of {a 0 , . . . , a p−1 }. These representatives are uniquely defined by the subring, but the elements of a matrix generating this subring can be changed by an arbitrary element of pZ p . We note that the normalized volume of pZ p is p −1 .
This reduction gives a map from subrings of O p of index p given by a matrix M with x jj = p and all other diagonal entries equal to 1 to tuples (x j+1,j , x j+2,j , . . . , x n,j ) where each x i,j ∈ {a 0 , . . . , a p−1 }. Let a 1 (p, j) be the size of the image of this map. In the case j = n, if the matrix M with diagonal entries all equal to 1 except for x n,n = p and all other entries equal to 0 generates a subring of O p of index p, then we define a 1 (p, n) = 1. Otherwise, a 1 (p, n) = 0. This description along with Proposition 3 shows the following.
The particular basis that we choose for O p has a major effect on the componentwise multiplication of rows of the matrix generating a subring. Our next goal is to pick a convenient basis for this module.
Suppose that p is an unramified prime of type f p = 1 v 2 w r e 1 1 · · · r e k k where the r i are distinct and greater than 2. Each residue degree r i that occurs contributes r i basis elements. We choose these basis elements for O p /f O p to be 1, y, y 2 , . . . , y r i −1 , where f (y) is an irreducible polynomial of degree r i over Z p . We get e i such groups of r i basis elements for each r i , including w blocks of two basis elements {1, y} coming from primes of residue degree 2, and v basis elements {1} corresponding to primes of residue degree 1. We choose these basis elements to be orthogonal to each other unless they correspond to the same irreducible polynomial.
The ordering of the basis elements has a large effect on the form of the lower triangular matrices in M(p). We order this basis so that elements corresponding to a single irreducible polynomial are given left to right by increasing powers of y. The e i sets of r i columns corresponding to the primes of residue degree r i are ordered so that they occur in adjacent blocks. We order these groups of e i blocks of r i columns from left to right by decreasing values of r i , except that we switch the positions of the block of v columns corresponding to primes of residue degree 1, and the w pairs of columns corresponding to primes of residue degree 2. We give an example for a lower triangular matrix corresponding to a prime of type 1 2 2 1 3 1 . The first three columns correspond to basis elements corresponding to an irreducible cubic, followed by two columns corresponding to linear polynomials, and finally by a pair of columns from an irreducible quadratic. In the picture below variable names are chosen to emphasize the grouping of columns:      a 1,1 0 0 0 0 0 0 a 2,1 a 2,2 0 0 0 0 0 a 3,1 a 3,2 a 3,3 0 0 0 0 a 4,1 a 4,2 a 4,3 b 4,4 0 0 0 a 5,1 a 5,2 a 5,3 b 5,4 b 5,5 0 0 a 6,1 a 6,2 a 6,3 b 6,4 b 6,5 c 6,6 0 a 7,1 a 7,2 a 7,3 b 7,4 b 7,5 c 7,6 c 7,7 We now give the proof of Lemma 2 on the non-existence of certain kinds of subrings.
Proof of Lemma 2. Let R be a subring of O L of index p. Then clearly pO L ⊂ R, and consequently This means (R/pO L ) ⊂ (O L /pO L ). Now O L /pO L is a field of order p n , and R/pO L will be a subring of O L /pO L and hence a field itself. Since the index is p the number of elements of R/pO L is p n−1 . Thus if F p k is the finite field with p k elements we have F p n−1 ⊂ F p n . This implies either n − 1 = 0 or n − 1 divides n. In the first case we get n = 1 and in the second case we get n = 2. Any larger value of n gives a contradiction.
Corollary 1. Let p be a prime of type f p = r with r ≥ 3. Then M(p) is empty.
These previous two lemmas allow us to compute a 1 (p) by considering a much smaller class of lower triangular matrices.
Proof of Lemma 3. We choose the ordered basis of O p described above. Suppose that column j corresponds to a basis element coming from a prime of residue degree greater than 2. We claim that the diagonal element of this column must be equal to 1.
We argue by contradiction. Suppose that x jj = p. Basis elements that do not correspond to the same irreducible polynomial are orthogonal. Suppose that the columns corresponding to the same irreducible polynomial as the basis element of column j are labeled by c 1 , . . . , c k . We can reorder the basis elements of O p so that these columns are in the lower right corner of this matrix and by elementary row operations can again suppose that the matrix is lower triangular.
We see that projecting onto the last k rows and columns must give a matrix that generates a multiplicatively closed sublattice of index p of a Z pmodule of rank k which is also a Z p -module of rank k. By the argument of Lemma 2, this is not possible.
Therefore every column corresponding to a basis element coming from a prime of residue degree greater than 2 has its diagonal entry equal to 1 and does not contribute to a 1 (p).
Proof of Lemma 5. A subring of O p of index p is generated by a lower triangular matrix M with exactly one diagonal element equal to p and all others equal to 0. We choose the basis of O p so that columns occur in pairs with each pair corresponding to two basis elements {1, y} of O p /f O p where f (y) is an irreducible quadratic polynomial over F p and the column corresponding to 1 occurs first. When p = 2 we can choose f (y) = y 2 − b with b a positive integer which is not a square modulo p. We focus on this case but note that for p = 2 we can take f (y) = y 2 +y +1 and the rest of the argument is similar. Basis elements occurring in distinct pairs are orthogonal to each other. We denote the componentwise product of rows v i and v j by v i • v j .
We will first show that it is not possible that the column with diagonal entry p corresponds to a basis element 1 for some quadratic polynomial. Suppose that it is and let the row which contains this diagonal element be v 1 . Let v 2 be the row which has diagonal element in the column corresponding to the basis element y for the same polynomial. Suppose the entry in row v 2 in the column with diagonal entry p is a ∈ Z p .
We will now give a first example of an argument that will be important throughout the rest of this section. Suppose M spans a sublattice of index p and has diagonal entries equal to 1 except for a single column in which the corresponding entry is p. We note that all vectors in the lattice spanned by M that are zero except in this entry must lie in pZ p since otherwise we could row reduce M and see that the index of this lattice is actually 1. We will use this fact to show that certain columns cannot have the single diagonal entry equal to p.
We see that v 2 •v 2 has two nonzero entries: 2a in the column corresponding to y and a b + a 2 corresponding to 1, since y 2 is b modulo f (y). Since M generates a multiplicatively closed sublattice, and all other entries in the column with diagonal entry in the row v 2 are 0, and so v 2 • v 2 − 2av 2 must be in the row span of v 1 . So there must exist some α 1 ∈ Z p such that This implies that b − a 2 ∈ pZ p , contradicting the fact that b is a nonsquare modulo p. Therefore we may suppose that for each column corresponding to 1 for a quadratic polynomial, the diagonal entry is 1.
There are w columns which correspond to basis elements y for distinct irreducible quadratic polynomials. We will show that if the diagonal element of such a column is equal to p then all other entries of this column are in pZ p . Applying elementary row operations and applying Lemma 7 will complete the proof.
We suppose that row v 1 has its diagonal entry equal to p and that this column corresponds to a basis element y for some irreducible quadratic polynomial. Let v 2 denote the row with diagonal entry corresponding to the basis element 1 for the same quadratic polynomial. Note that v 1 is above v 2 in this matrix and has a single nonzero entry equal to 1. We will show that it is not possible for there to be a row w with an entry that is a unit in the column with diagonal entry p.
Suppose that there is such a row with an entry a ∈ U p in this column and consider w • v 1 . This has a single nonzero entry equal to a in the column corresponding to the diagonal entry p. The argument above shows that such a matrix actually generates O p and not a subring of index p, which is a contradiction. We have shown that there are no units in the column with diagonal entry p, completing the proof.
Proof of Lemma 6. We continue with the notation of the previous proof. Again we consider p = 2 and note that when p = 2 we choose f (y) = y 2 +y+1 for our irreducible quadratic polynomials and the argument is very similar.
We choose the basis elements of O p so that the first v columns correspond to primes of residue degree 1 and the last 2w columns occur in pairs and correspond to primes of residue degree 2. The proof of the previous lemma shows that matrices with diagonal entry equal to p in a column corresponding to a prime of residue degree 2 contribute w to a 1 (p). We now focus only on the entries of the columns of this matrix which correspond to primes of residue degree 1. Suppose x jj = p and that this column corresponds to a prime of residue degree 1. Since L is a subring and not just a multiplicative sublattice, it must contain the identity element of O p and we see that there must be some entry in this column that is a unit. In fact, we will show that there must be a unique entry in this column that is a unit. Each of the v − j rows directly below this diagonal entry can contain any unit in 1 + pZ p , but no other units can occur. Applying Lemma 7 shows that a 1 (p) = w + v j=1 (v −j) = w + v 2 , completing the proof.
We first note that we cannot have two units in rows corresponding to primes of degree 1 in the column with diagonal entry equal to p. If we did, taking v 1 •v 2 for these two rows would give a vector with a single nonzero entry which is a unit in the column with diagonal entry p. This is a contradiction.
Suppose there is a row with diagonal entry corresponding to an irreducible quadratic polynomial which has a unit entry in the column with diagonal entry p. Let v 1 be the row corresponding to the basis element 1 for this polynomial and v 2 be the row corresponding to the basis element y. Suppose the entry in the column with diagonal entry p is a in row v 1 and c in row v 2 . By assumption, at least one of a, c is a unit. We show that this is a contradiction.
We see that v 1 • v 1 − v 1 has an entry of a 2 − a in the column with diagonal entry p and every other entry of this vector is zero. So either a ∈ pZ p or a ∈ 1 + pZ p . We see that v 2 • v 2 − bv 1 has an entry c 2 − ab in the column with diagonal entry p and every other entry is zero. If a ∈ 1 + pZ p then since b is not a square modulo p, we get a contradiction. If a ∈ Z p then we have c 2 ∈ Z p , which is also a contradiction.
Combining Lemma 3 and Lemma 6 completes the proof of Proposition 1.

Proof of Proposition 2
To fix notation we give a quick review of basic class field theory [Ne]. Let K be a number field, and let J K be the free group generated by the finite primes of K. There is a natural map ι : K × → J K . A modulus, called a cycle in [Ne], is a finite formal product of primes of K with non-negative exponents p p np . If m = p p np is a modulus, and x ∈ K, we write x ≡ 1 mod m to mean: • For each finite p|m, x ≡ 1 mod p np ; • for each real prime ν|m, we have x v > 0.
If S is a finite set of primes, we let J S K be the subgroup of J generated by the primes not in S. For a modulus m we let J m K be J S K where S is the set of finite primes that divide m. Set This class group is finite. A congruence subgroup modulo m is a subgroup H m of J m K which contains P m K . We recall the following two main theorems of class field theory: Theorem 9 (Artin Reciprocity Law). For L/K an Abelian extension of number fields, there is a modulus m divisible by all the ramified primes of L/K such that the sequence Theorem 10. For any congruence subgroup H m , there is a unique Abelian extension L/K such that L is the class field of K of the congruence class group J m K /H m . We have the following lemma: Lemma 8. Let K be a number field, m a modulus, and H m a congruence subgroup. If C is a coset of J m K /H m , we set Then f C (s) is holomorphic for ℜs > 1. Furthermore, Then g C (s) = f C (s) r , r = |J m K /H m |, has an analytic continuation to an open set containing ℜs = 1 with a unique pole at s = 1. Assuming GRH, s = 1 is the only pole for ℜs > 1/2. We do not need the additional convergence provided but assuming GRH to prove Proposition 2, but include this statement to give a better idea of the analytic behavior of this function.
This is holomorphic for ℜs > 1/2. We then write with H(s) a function that is holomorphic for ℜs > 1/2. Hence The lemma now follows from results on zero free regions of L-functions, e.g. Ch. 2 of [Mu-Mu].
Next we can prove Proposition 2: Proof of Proposition 2. If L/K is Abelian, this follows from the above lemma and class field theory. In general, let σ ∈ C, and let H = σ . Let M = L H . Note that L/M is an Abelian Galois extension. Let where S is the set of primes of L H satisfying We will also consider where S ′ is the set of primes p of M such that (L/M/p) = σ. We know from what we proved before that F ′ H (s) |H| has a simple pole at s = 1. By the computations of Ch. V, §6 of [Ne] we know that F ′ H (s)/F H (s) is holomorphic for ℜs > 1/2. Thus F H (s) |H| has a simple pole at s = 1 and otherwise holomorphic in an open set containing ℜs ≥ 1.
Next, it follows from the reduction step of the proof of the Chebotarev density theorem, Theorem 6.4 of [Ne], that The proposition is now immediate.

Some remarks on r 2
Suppose we have a finite group G acting on a finite set A. Let O 1 , . . . , O r be the distinct orbits of the action of G. Then G has an induced representation on the vector space V = ⊕ a∈A C.

Proof.
For the first part we show that A n acts transitively on the two element subsets of {1, . . . , n}. For this we notice that for three distinct elements a, b, c, the even permutation (a c)(b a) maps the set {a, b} to the set {b, c}.
For C n and D n , write n = 2k or n = 2k + 1, depending on the parity of n. Suppose C n = (1 2 . . . n) . It is easy to see that for each 1 ≤ i ≤ k, the set is an orbit of the action of C n on the set of two element subsets of {1, . . . , n}. Furthermore, these are all the possible orbits. To see the result for D n , we consider the generators (1 2 . . . n), σ, with σ = (1 n)(2 n − 1) . . . (k k + 1).

We observe that each orbit O i is invariant under the action of σ.
For the case where n is a prime number, we have the following proposition: Proposition 5. Let G be a transitive subgroup of S p , p prime. Then one of the following two possibilities occurs: 1. G is doubly transitive and r 2 (G) = 1; 2. G is solvable in which case p | |G| and r 2 (G) = gcd |G| p , p−1 2 . Proof. A theorem of Burnside [Bu, Mu] says that a transitive subgroup of S p is either doubly transitive or solvable. If the action of G is doubly transitive, then r 2 (G) = 1. If G is solvable, a classical theorem of Galois ( [Hu], p. 163) 1 asserts that G contains a unique normal subgroup C of order p, and is contained in the normalizer of C. Furthermore, G/C is a cyclic group of order dividing p − 1. Up to conjugation we may assume that C = (1 2 . . . p) .
The normalizer of C is the split extension of the group C by the cyclic group Z of order p − 1 consisting of the elements σ k , 1 ≤ k ≤ p − 1 identified by for x ∈ {1, . . . , n}; that the group Z is cyclic is the theorem of the primitive root in elementary number theory. Let σ g be a generator of Z. Since G is transitive, G is equal to C ⋉ σ j g for some j|p − 1. By the description of orbits of C on the two element subsets of {1, . . . , p}, we just need to know the number of orbits of σ j g , σ p−1 2 g on (Z/pZ) * . The latter is equal to 4 The proof of Theorem 5 4.1 Outline of the proof of Theorem 5 Let d ∈ N, and let R = Z d equipped with componentwise addition and multiplication. Namely To emphasize the dependence of M p (β) from Definition 1.3 on d, we write it as M d (p). For d = 2, 3, 4, we will give an explicit description of M d (p) in Sections 4.3, 4.4 and 4.5.
Definition. If k = (k 1 , . . . , k d ) is a d-tuple of non-negative integers, we set It is easy to see that Intuitively what this means is that we have multiplied the rows by units to make the diagonal entries a p-power. We note that this does not change the lattice generated by the rows.
Warning. The volume of M d (p; k) are used to count subrings of finite index in Z d , and orders of finite index in Z d+1 . The reader should be careful about the distinction between subrings and orders.
We have the following lemma which is equivalent to Lemma 4 given during the proof of Proposition 1.
For a proof see [L] Proposition 1.1. The quantity a < Z d (p) is equal to f d+1 (p) of that reference. By Theorem 8, Theorem 5 is proved if we can show the following statement: there is an ǫ > 0 such that for Since by Equation (6) a < Z d (p k ) = k=(k 1 ,...,k d ) in order to prove the lemma we need to estimate µ p (k). The relevant computations are performed in Sections 4.3, 4.4, and 4.5.
The results are stated in Theorems 11, 12, and 13. These theorems form part 1 of Theorem 5.
The proof of part 2 of Theorem 5 appears in §4.6.

General facts about volumes
We begin with some lemmas that allow us to bound the volumes of certain sets that arise in our volume computations. Let U p denote the set of units of Z p and v p (·) be the p-adic valuation. Recall that for α, Proposition 6. For fixed y, z ∈ Z p , k ≥ 0, the volume of x ∈ Z p such that v p (xy − z) ≥ k is at most p −(k−vp(y)) .
Proof. We first note that for y = 1, the volume of x such that v p (x − z) ≥ k is p −k , since we are just fixing the first k digits in the p-adic expansion of x to coincide with those of z. Similarly, for any unit u ∈ U p the volume of . This holds on a set of volume at most p −(k−vp(y)) if k ≥ v p (y) and on a set of volume 1 if v p (y) ≥ k.
Now if v p (z) < k and v p (y) ≤ v p (z) then we can write y = p vp(y) u for some unique unit u ∈ U p , and z = p vp(y) z ′ for some unique , which holds on a set of volume at most p −(k−vp(y)) .
Proposition 7. For fixed z ∈ Z p , the combined volume of x, y ∈ Z 2 p such that v p (xy − z) ≥ k is at most (k + 1)p −k .
Proof. If v p (y) ≥ k, then there are two cases. Either v p (z) ≥ k in which case any x will work, or v p (z) < k in which case no x works. So assume 0 ≤ v p (y) < k. Then given y with l = v p (y) we need x such that x ∈ p −l (p k Z p +z). So the total volume is Proposition 8. For any fixed z ∈ Z p , the combined volume of x, y ∈ Z 2 p such that v p (x(y − z)) ≥ k is at most (k + 1)p −k .
Proof. This proposition is very similar to the previous one. We have v p (x) ≥ k on a set of volume p −k . Suppose that this does not hold and set v p (x) = m. We see that for any fixed z the volume of y such that v p (y − z) ≥ k − m is p −(k−m) . Summing over the k possible values of m gives the result.
Proposition 9. Suppose z ∈ Z p , k, l ≥ 0 are given. Then the volume of Proof. If there is no such x then the volume is zero and there is nothing to prove. Assume that the volume is non-zero. For simplicity of notation, let y = p l . If v p (t) ≥ k and v p (x(x − y) − z) ≥ k, then x + t also satisfies the same inequality. Given y and z modulo p k , we must determine the number of x modulo p k such that x(x − y) − z ≡ 0 mod p k . If this number is N, the volume of our domain is N · p −k . Suppose X, X + u are both solutions of the congruence This implies that u satisfies the congruence We count the number of nonzero solutions u of this congruence equation.
There are at most p k−⌈k/2⌉ choices for u. If not, then we write 2X − y ≡ p s q mod p k with s < k and (q, p) = 1. We write u = p r m mod p k . By assumption, (m, p) = 1 and r < k. Since we have u + (2X − y) ≡ 0 mod p k−r . If 2r ≥ k, then r ≥ ⌈ k 2 ⌉, and as above there are at most p k−⌈k/2⌉ choices for u.
If 2r < k, then s = r and Equation 7 implies that u and 2X − y match up in the first k − r ≥ ⌈ k 2 ⌉ digits of their p-adic expansions. This gives at most p k−⌈ k 2 ⌉ ≤ p ⌈ k 2 ⌉ choices for u. Multiplication by p −k gives the result.
We point out that in the most general possible case it is not possible to improve this result by more than a factor of 2. Suppose l ≥ ⌈k/2⌉. Then v p (x) + v p (x − p l ) ≥ k if and only if v p (x) ≥ ⌈k/2⌉, which holds on a set of volume at most p −⌈k/2⌉ . However, in some cases we can say something stronger.
Proposition 10. Suppose z ∈ Z p , k, l ≥ 0 are given. Then there is a constant C, which for odd p may be taken to be 6, such that the volume of is bounded by Cp −(k−l) except when p = 2 and v 2 (z) = 2l − 2 < k. In this exceptional situation: 1. If v 2 (z + 2 2l−2 ) ≥ k, the volume is bounded by 2 −⌈k/2⌉ , and this is the best bound possible.
Proof. The proposition will have no content unless l < k. First we consider the case where p is odd. We recognize two basic cases: In this case the result follows from Proposition 10.
2. If v p (z) < k, then our inequality can be valid only when v p (x(x−p l )) = v p (z). Since v p (z) < k, we write z = ζp u with u < k. We are looking for • If v p (x) > l, then we must have v p (x) + l = u, and as a result u − l > l which means u > 2l. Write x = ǫp u−l . Then we need This is a quadratic equation in ǫ with at most two solutions modulo p. Hensel's lemma says that the volume of ǫ satisfying this last inequality is at most 2p −(k−u) . The volume for x is then at most 2p −(u−l) · p −(k−u) = 2p −(k−l) .
An application of Hensel's lemma then says that the volume of ǫ satisfying this inequality is at most 2p −(k−2l) . Since x = p l ǫ, the volume of x is at most 2p −(k−l) . Now we examine the situation for p = 2. Except for the step marked (*) every other step of the proof works verbatim. The argument (*) can be adjusted as follows. We let r = l − u 2 and s = k − u. Then r ≥ 1 and we are trying to determine the volume of ǫ ∈ U p such that v 2 (ǫ(ǫ − 2 r ) − ζ) ≥ s. for a given unit ζ. Rewrite this inequality as v 2 ((ǫ − 2 r−1 ) 2 − (ζ + 2 2r−2 )) ≥ s.
As above, the volume of such ω is bounded by 4·2 −s+2t . The volume of ǫ then is bounded by 4 · 2 −s+t . The volume of x is then bounded by 4 · 2 −k+l · 2 t .

Volume estimates
First we give a description of M 2 (p).
Lemma 11. The set M 2 (p) is the collection of matrices Proof. Let v 1 and v 2 be the first and the second rows of M respectively. Then since entries are in Z p it is clear that We note that the sublattice corresponding to a matrix M as above has finite index if and only if det M = 0.

Subrings
We now prove the following theorem: Theorem 11. There is a polynomial P 3 of degree 2 such that for all ǫ > 0 Proof. By Theorem 8 and Lemma 10, it suffices to prove the following statement: If σ > 1 2 the series p k+l≥2 converges. Here µ p (k, l) is as in Definition 4.1.
Case II. k ≥ 2, l = 0. Then by Proposition 9 and as a result our subseries is majorized by p k≥2 p −kσ which converges for σ > 1 2 .
For the second assertion in the statement of the theorem we observe that

Volume estimates
Lemma 12. The domain M 3 (p) is the collection of 3 × 3 lower triangular matrices   x 11 x 21 x 22 x 31 x 32 x 33   with entries in Z p such that the following inequalities hold: Proof. We want to determine the conditions on matrices such that x 11 , x 21 , x 22 , x 31 , x 32 , x 33 ∈ Z p and for 1 The condition that v 2 • v 2 = α 1 v 1 + α 2 v 2 gives the same condition that we had for the case n = 3. That is, v Clearly α 3 = 0. We have α 2 x 22 = x 32 x 22 , so α 2 = x 32 . So we have α 1 x 11 + x 32 x 21 = x 21 x 31 . This implies We must have α 3 = x 33 . So α 2 x 22 + x 33 x 32 = x 2 32 . This implies Therefore v p (x 22 ) ≤ v p (x 2 32 − x 32 x 33 ). We also have α 1 x 11 + x −1 22 (x 2 32 − x 32 x 33 )x 21 + x 33 x 31 = x 2 31 . This implies Suppose that v p (x 11 ) = k, v p (x 22 ) = l and v p (x 33 ) = r. By multiplying by appropriate units, we can suppose that x 11 = p k , x 22 = p l and x 33 = p r . Note that this does not change the lattice generated by the rows. Then we can define µ p (k; l; r) as in Definition 4.1.
Proof. We divide the proof into three steps. We give two different bounds on µ p (k; l; r) and then take an average.
Remark 5. This is not the best possible bound one can prove. In fact using a more complicated argument similar to the proof of Step I of Theorem 15 we can prove a bound of Cp −9k/8 p −l/2 in Step I of the above theorem. This leads to the bound µ p ≤ Cp −5k/4 p −l/2 after averaging. This however will not improve the bound in Theorem 12 unless one has an analogue of Theorem 17 for r = 1. Such a theorem is easy to prove, but the resulting estimate would still not be as good as the one obtained in [L]. For this reason we decided to include only the simplest non-trivial estimate.
Proposition 12. Let p be odd. If r = 0 and k, l ≥ 1, then Proof. Proposition 9 implies that inequality [4 − 1] holds on a set of x 21 of volume at most 2p −⌈k/2⌉ . Proposition 10 implies that inequality [4 − 3] holds on a set of x 32 of volume at most 2p −l . For fixed x 21 , x 32 , Proposition 10 implies that inequality [4 − 4] holds on a set of x 31 of volume at most 6p −k .
We see that our total volume is bounded by 24p −k−l−⌈k/2⌉ .
Proposition 13. Let p be odd. Then Proof. If k = r = 0, then inequality [4 − 3] and Proposition 9 give the result. Now suppose l = r = 0. Then we have v p (x 21 ) + v p (x 21 − 1) ≥ k which determines two possibilities for x 21 : The volume of such x 31 is 2p −k . As a result the whole volume is at most 2p −2k .
and the two dimensional volume of (x 31 , x 32 ) satisfying this inequality is at most p −k . This gives a bound on the entire volume of p −2k .
Adding up gives the result.

Subrings
In this section we prove the following theorem: Theorem 12. There is a polynomial P 4 of degree 5 such that for all ǫ > 0 as B → ∞.
We now consider the second piece of the sum. We consider three cases.
Case I. r ≥ 2. By Theorem 11 the relevant sum is bounded by This sum is equal to This sum is converges for σ > 5 6 .

Volume estimates
We will begin with the set of inequalities defining our region of integration.
Lemma 13. M 4 (p) is the collection of matrices with entries in Z p     x 11 x 21 x 22 x 31 x 32 x 33 x 41 x 42 x 43 x 44     whose entries satisfy: The proof of this lemma is very similar to the proof of Lemma 12. By multiplying by appropriate units, we can suppose that x 11 = p k , x 22 = p l , x 33 = p r and x 44 = p t . We define µ p (k; l; r; t) as in Definition 4.1.
We start with a lemma: Lemma 14. Let p be a prime. Then there is a polynomial with positive coefficients R ∈ R[x] such that µ p (k; l; r; t) ≤ R(k)p −2k−l .
Proof. In this proof we will suppress the dependence of R(k) on k, and will simply write R. The value of the polynomial R does not affect the convergence of the sum we consider, so we do not compute it. The key to our argument will be that once our other variables are fixed, there are several different bounds available to us for the volume of x 31 such that inequalities [5 − 4] and [5 − 10] hold.
More specifically, we use Proposition 9 to give a bound on the volume of the possible set of x 32 , then give a bound on the set of possible x 43 . Once these two values are fixed we again use Proposition 9 to give a bound on the set of x 42 , which then bounds the set of possible x 21 . Finally, we combine a few different possible bounds for the set of x 31 so that these inequalities simultaneously hold.
Proposition 9 implies that inequality [5−3] holds on a set of x 32 of volume at most 2p −l/2 .
Suppose that v p (x 43 (x 43 − x 44 )) = r + z. Inequality [5 − 8] implies that z ≥ 0. This inequality holds on a set of x 43 of volume at most 2p −r/2−z/2 . Fix such an x 43 . Now for fixed x 32 , x 43 , Proposition 9 implies that inequality [5 − 9] holds on a set of x 42 of volume at most 2p −l/2 .
We now consider inequality [5 − 5]. For fixed x 42 , Proposition 8 implies that the total volume of x 21 , x 41 such that this inequality holds is at most (k + 1)p −k .
Finally, we consider x 31 . We begin with inequality [5 − 10]. For fixed values of x 21 , x 32 , x 41 , x 42 , x 43 , we can write this as where y, τ ∈ Z p with v p (y) = r + z. We see that this holds on a set of x 31 of volume at most p −(k−z) .
Consider inequality [5 − 4]. By Proposition 9, this holds on a set of x 31 of volume at most 2p −k/2 . Using 2p −(k−z) as our bound for the volume of x 31 gives a bound on our total volume of R 1 p −2k−l−(r−z)/2 , for some polynomial R 1 . This is enough for our result if r ≥ z. Suppose that this is not the case. By the proof of Proposition 10, we see that the total volume of x 31 such that v p (x 31 (x 31 − x 33 ) − z) ≥ k, is at most 6p −(k−r) unless p = 2, v p (x 31 ) = r − 1 and v p (z) = 2r − 2 < k. If we are not in this exceptional situation the total volume is at most R 2 p −2k−l−(z/2−r/2) . Since r < z, this is at most Rp −2k−l , completing the proof. Suppose that we are in the situation where p = 2, v p (x 31 ) = r − 1 and v p (z) = 2r − 2 < k.
First suppose that v p (x 31 ) = v p (x 32 ). Then v p (x 31 −x 32 ) ≤ v p (x 31 ) = r−1. Inequality [5 − 2] now holds on a set of x 21 of volume at most p −(k−r) . Using this bound for the volume of x 21 , 2p −l/2 for the volume of x 32 and 2p −k/2 for the volume of x 31 , gives the total bound Then v p (x 32 (x 32 − x 33 )) = 2r − 2, and we must have v p (x 21 ) = l. Now consider inequality [5 − 7]. We write x 21 = αp l , x 31 = βp r−1 , and x 32 = γp r−1 for units α, β, γ. Factoring out p l+r−1 , the inequality is now For fixed values of x 21 , x 31 , x 32 , x 42 , x 43 , this holds on a set of x 41 of volume at most p −(k−r) . Using 2p −k/2 as our bound for x 21 and x 31 , this gives total bound R 4 p −2k−l−(z−r)/2 , which is at most Rp −2k−l , completing the proof.
Proposition 14. Let p be any prime. Suppose that k, l, r, t ≥ 0. Then for a polynomial A ∈ R[x] with positive coefficients we have Proof. The value of the polynomial A does not affect the convergence of the sum we will consider so we do not compute it. For example in the collection of equations (11), (12), and (13) the polynomials A will not be the same.
We have two steps: Step I. Here we show that the following three inequalities hold: We proceed as follows. Inequality [5 − 1] holds on a set x 21 of volume at most the minimum of 2p −k/2 and 2p −(k−l) . Inequality [5 − 3] holds on a set x 32 of volume at most the minimum of 2p −l/2 and 2p −(l−r) . Inequality [5 − 8] holds on a set of x 43 of volume at most 2p −(r−t) .
When p = 2, we can use Proposition 10 for the remaining three variables (See the proof of Theorem 15 for details). For p = 2, some care is required. By Proposition 9 we always have the following. For any fixed x 21 and x 32 inequality [5 − 4] holds on a set of x 31 of volume at most 2p −k/2 . For any fixed x 32 , x 43 inequality [5 − 9] holds on a set of x 42 of volume at most 2p −l/2 . For any fixed x 21 , x 31 , x 32 , x 42 , x 43 inequality [5 − 10] holds on a set of x 41 of volume at most 2p −k/2 . Inequality (11) follows from taking 2p −k/2 for the volume of x 21 , x 31 , x 41 , taking 2p −(l−r) for the volume of x 32 , taking 2p −l/2 for the volume of x 42 , and taking 2p −(r−t) for the volume of x 43 .
For inequality (12) we take 2p −k/2 as our bound for the volume of x 21 and x 31 , 2p −l/2 as the bound for x 32 and x 42 , and 2p −(r−t) as the bound for the volume of x 43 . We must now show that when all other variables are fixed, the total volume of x 41 satisfying our inequalities is at most Ap −(k−2t) .
Suppose we are not in the special case in which we cannot apply Proposition 10. We have that the volume of x 41 satisfying inequality [5 − 10] is at most 6p −(k−t) , completing this case.
Inequality (13) will be proved in a few steps. First we suppose that we are in the case where we can apply Proposition 10 to inequality [5 − 4] and conclude that the volume of x 31 satisfying this inequality is at most 6p −(k−r) . As above, we see that either one of x 21 , x 31 has valuation at most 2t, giving a bound of p −(k−2t) , or both have valuation at least 2t, in which case we can apply Proposition 10 and conclude that the total volume of x 41 is at most 6p −(k−t) . Using 2p −k/2 as our bound for x 21 , 2p −l/2 as our bound for x 32 and x 42 , and 2p −(r−t) as our bound for x 43 , we get total volume Ap −5k/2−l+3t , completing this case. Now suppose that we are in the case where we cannot apply Proposition 10 to inequality [5 −4]. Then v p (x 31 ) = r −1. We now consider two subcases. First suppose that v p (x 31 ) = v p (x 32 ). Then inequality [5 − 2] implies that v p (x 21 ) ≥ k − v p (x 31 ) > k − r, which holds on a set of x 21 of volume at most p −(k−r) . We use 2p −k/2 as the bound for the volume of x 31 satisfying inequality [5 − 4]. Now using the same argument given above, the volume of x 41 satisfying these inequalities is at most 6p −(k−2t) . Combining these estimates gives total volume bounded by Ap −5k/2−l+3t , completing this case.
Finally, suppose that v p (x 31 ) = v p (x 32 ) = r − 1. Now for fixed x 32 , x 43 , the total volume of x 42 satisfying inequality [5 − 6] is at most p −(l−r) . We use 2p −(k−l) as the bound on the volume of x 21 satisfying inequality [5 − 1], 2p −k/2 as the bound on the volume of x 31 , 2p −(r−l) as our bound on the volume of x 32 , and 2p −(r−t) as the bound on the volume of x 43 . Using the same argument given above, we can use 6p −(k−2t) as our bound on the volume of x 41 . This gives total bound Ap −5k/2−l+r+3t , completing Step I.
Setting n = 11 gives the result.
We now state several results for odd primes p.
Proposition 15. Let p be odd. Suppose that k, l, r, t ≥ 0. Then there is a polynomial B ∈ R[x] with positive coefficients such that Proof. We have two steps: and µ p (k; l; r; t) ≤ Bp −5k/2−3l/2+3t .
We will use (14) in the proof of Theorem 17. We proceed as follows. Inequality [5 − 1] holds on a set x 21 of volume at most the minimum of 2p −k/2 and 2p −(k−l) . Inequality [5 − 3] holds on a set x 32 of volume at most the minimum of 2p −l/2 and 2p −(l−r) . Inequality [5 − 8] holds on a set of x 43 of volume at most 2p −(r−t) . For any fixed x 21 and x 32 inequality [5 − 4] holds on a set of x 31 of volume at most the minimum of 2p −k/2 and 6p −(k−r) . For any fixed x 32 , x 43 inequality [5 − 9] holds on a set of x 42 of volume at most 6p −(l−t) . For any fixed x 21 , x 31 , x 32 , x 42 , x 43 inequality [5 − 10] holds on a set of x 41 of volume at most 6p −(k−t) . Hence the total volume is bounded by
Simplification gives the result.
Step II. Here we consider an appropriate average of the previous inequalities to prove the theorem. As constants play no role we ignore them. By Lemma 14 and Step I we have This means for all n ≥ 1 Setting n = 17 gives the result. set x 32 of volume at most the minimum of 2p −⌈l/2⌉ and 2p −(l−r) . We also have that inequality [5 − 8] holds on a set of x 43 of volume at most the minimum of 2p −⌈r/2⌉ and 2p −(r−t) = 2p −(r−1) . For any fixed values of x 21 , x 32 , we see that inequality [5 − 4] holds on a set of x 31 of volume at most the maximum of 2p −⌈k/2⌉ and 6p −(k−r) . For any fixed values of x 32 , x 43 , we see that inequality [5 − 9] holds on a set of x 42 of volume at most the maximum of 2p −⌈l/2⌉ and 6p −(l−1) . For any fixed values of x 21 , x 31 , x 32 , x 42 , x 43 , we can write inequality [5 − 10] as k ≤ v p (x 41 (x 41 − x 44 ) − z), for some z ∈ Z p . This holds on a set of x 41 of volume at most the maximum of 2p −⌈k/2⌉ and 6p −(k−1) .
We now combine these inequalities to get bounds on the total volume satisfying inequalities [5 − 1] through [5 − 10]. Note that if k − l ≥ ⌈k/2⌉ and l − r ≥ ⌈l/2⌉, then k − r ≥ ⌈k/2⌉. By using 2p −⌈k/2⌉ as the bound for the volume of x 21 and x 31 , and 2p −(l−r) as the bound for x 32 , we see that our total volume is bounded by Therefore, we are done if l ≥ 4, or if l ≥ 3 and k is odd. Suppose that this is not the case.
Suppose that l ≤ 3. Using 2p −⌈l/2⌉ instead of 2p −(l−r) as our bound for the volume of x 32 , our total bound is now Therefore, we are done if ⌈l/2⌉ + r ≥ 4, or ⌈l/2⌉ + r ≥ 3 and k is odd. Suppose that these conditions do not hold.
First suppose that l = 3. Then r ≤ 1. We can use 2p −⌈r/2⌉ as a bound for the total volume of x 43 satisfying inequality [5 − 8] instead of 2p −(r−1) . We use 2p −(l−r) as our bound for the volume of x 32 satisfying inequality [5 − 3]. We see that our total volume is bounded by Since r ≤ 1, this is at most Dp −2k−l−1 , completing this case. Now suppose that l ≤ 2. For fixed x 32 , x 43 , Proposition 9 implies that the total volume of x 42 satisfying inequality [5 − 9] is at most 2p −⌈l/2⌉ . We use this bound instead of 6p −(l−1) . Our total volume is now bounded by and we are done unless r ≤ 2. In this case ⌈r/2⌉ ≥ r − 1, so we use 2p −⌈r/2⌉ as our bound for the volume of x 43 satisfying inequality [5 − 8]. Now our bound is First suppose r = 2. Then if l is odd or k is odd, we are done. If l = 0, then we can use 2p −k as our bound for the volume of x 21 satisfying inequality [5 − 1], giving Dp −2k−⌈k/2⌉ , as our bound. Therefore, we are done unless k = 0. In this case, k = l = 0, we have that the total volume is at most the total volume of x 43 satisfying inequality [5 − 9], which is at most 2p −1 , which completes this case. Now suppose r = l = 2. This is the most difficult case to consider. If k is odd then 2⌈k/2⌉ = k + 1, and we are done. If k ≥ 6, then we can use 2p −(k−l) as our bound for x 21 , which is enough to complete this case. If k = 0, then we use 1 as our bound for x 41 instead of 6p −(k−1) , and our total bound is Dp −l−1 , completing this case. We now must consider k = 2 and k = 4.
Next suppose l ≤ 2 and r = 1. We have the bound If l = 1, we are done. Suppose l = 2. Then we can use 2p −l as our bound for the volume of x 32 satisfying inequality [5 − 3], and we are done. If l = 0, then we can use 2p −k as the bound for x 21 satisfying inequality [5 − 1], and our bound is which completes this case unless k = 0. If k = l = 0 and r = t = 1, then our total volume is at most the volume of x 43 satisfying inequality [5 − 8], which is 2p −1 , and we are done. Finally, suppose r = 0 and l ≤ 2. We can use 2p −l as our bound for the volume of x 32 satisfying inequality [5 − 3], and for fixed x 21 , x 32 , we use 6p −k as our bound for the volume of x 31 satisfying inequality [5 − 4]. We also use 2p −⌈k/2⌉ as our bound for the volume of x 41 satisfying inequality [5 −10]. Our total volume is now bounded by Dp −2k−⌈k/2⌉−l−⌈l/2⌉ .
Since k + l + r ≥ 1, we are done.
We set n = 7 to get the result.
Remark 7. The case by case analysis of the small values of parameters in the proofs of Theorems 16 and 17 can be avoided if instead one uses the results of [L] for f n (p k ) for small k. In [L] these values are worked out for k up to 5. This is not sufficient for our purposes, but computing the missing data is not difficult using the results of Liu. Here we chose instead to present the above elementary treatment to make the argument self-contained.
Remark 8. The choices of the parameter n in the proofs of Theorems 14, 15, 16, and 17 are made to optimize the error estimate in Theorem 13.

Subrings
In this section we prove the following theorem: Theorem 13. There is a polynomial P 5 of degree 9 such that for all ǫ > 0 as B → ∞.
Proof. By Theorem 8, it suffices to prove the following statement: for σ > 33 34 the expression p m≥2 a < Z 4 (p m ) p mσ converges. In our analysis we will ignore all constants as they will have no bearing on convergence. We write p m≥2 a < Z 4 (p m ) p mσ = m≥2 a < Z 4 (2 m ) 2 mσ + p odd m≥2 a < Z 4 (p m ) p mσ If we use Proposition 14 we see very easily that the first piece converges for σ > 33 34 . So we concentrate on the sum corresponding to the odd primes. We will show that for m ≥ 2 and p odd we have for a polynomial A(m).
It is clear that this will be sufficient for the proof of the theorem. In order to prove (15)  Remark 9. Using Proposition 14 for odd primes instead of Proposition 15 in the proof of Theorem 13 would have produced a weaker error term.

Subrings of Z d for d > 5
In this section we prove part 2 of Theorem 5. The idea is to find non-trivial volume bounds for M 5 (p), and then use an inductive argument to obtain bounds for M d (p) for d > 5.
We begin by defining M 5 (p).
Suppose v 1 , . . . , v d−1 are the rows of M ′ . We now bound the volume of the set of vectors v d with x dd = p k d such that v d • v d = c 1 v 1 + · · · + c d v d with c i ∈ Z p . It is clear that c d = x dd . We then see that for 1 ≤ j ≤ d − 1 If c k , x kj are given for j + 1 ≤ k ≤ d, then the existence of such a a c j is equivalent to c k x kj ≥ k j .
Proposition 9 implies that the volume of x dj is bounded by 2p −⌈k j /2⌉ . Induction will give the result.
for some polynomial B(m). Clearly this implies that p ∈T m≥2 a 1,< O K (p m ) p mσ converges for σ > 19/20. This shows that σ 0 (5) = 19/20 works. The proof of the statement that σ 0 (n) = n/2 − 7/6 works for n ≥ 6 follows the same reasoning, except that we use Corollary 3. This finishes the proof of the theorem.
The following corollary is immediate from equation (22). This is an improvement of Theorem 8.1 of [Br]. As in the introduction we set a 1,< (n, m) = max K/Q extension of degree n a 1,< O K (m).