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Annals of Mathematics The number of extensions of a number field with fixed degree and bounded discriminant By Jordan S.. The number of extensions of a number field with fixed degree and

Annals of Mathematics

The number of extensions of a number field with fixed degree and bounded discriminant

By Jordan S Ellenberg and Akshay Venkatesh*

The number of extensions of a number field with fixed degree and bounded discriminant

By Jordan S Ellenberg and Akshay Venkatesh*

Abstract

We give an upper bound on the number of extensions of a fixed number field of prescribed degree and discriminant≤ X; these bounds improve on work

of Schmidt We also prove various related results, such as lower bounds for the number of extensions and upper bounds for Galois extensions

1 Introduction

Let K be a number field, and let N K,n (X) be the number of number fields L (always considered up to K-isomorphism) such that [L : K] = n and

NKQD L/K < X Here D L/K is the relative discriminant of L/K, and N KQ is

the norm on ideals of K, valued in positive integers D L=|D L/Q| will refer to

discriminant overQ

A folk conjecture, possibly due to Linnik, asserts that

N K,n (X) ∼ c K,n X (n fixed, X → ∞).

This conjecture is trivial when n = 2; it has been proved for n = 3 by Davenport and Heilbronn [7] in case K =Q, and by Datskovsky and Wright in

general [6]; and for n = 4, 5 and K =Q by Bhargava [3], [2] A weaker version

of the conjecture for n = 5 was also recently established by Kable and Yukie

[11] These beautiful results are proved by methods which seem not to extend

to higher n The best upper bound for general n is due to Schmidt [18], who

showed

N K,n (X)  X (n+2)/4 where the implied constant depends on K and n We refer to [4] for a survey

of results

In many cases, it is easy to show that N K,n (X) is bounded below by a constant multiple of X; for instance, if n is even, simply consider the set of

*The first author was partially supported by NSA Young Investigator Grant MDA905-02-1-0097 The second author was partially supported by NSF Grant DMS-0245606.

quadratic extensions of a fixed L0/K of degree n/2 For the study of lower

bounds it is therefore more interesting to study the number of number fields L

such that [L : K] = n, N KQD L/K < X and the Galois closure of L has Galois

group S n over K Denote this number by N K,n
(X) Malle showed [14, Prop.

6.2] that

N Q,n
(X) > c
n X 1/n

for some constant c
n

The main result of this paper is to improve these bounds, with particular

attention to the “large n limit.” The upper bound lies much deeper than the

lower bound

Throughout this paper we will use  and  where the implicit constant

depends on n; we will not make this n-dependency explicit (but see our

ap-pendix to [1] for results in this direction)

Theorem 1.1 For all n > 2 and all number fields K, we have

N K,n (X)  (XD n

K A [K: n Q])exp(C √ log n)

where A n is a constant depending only on n, and C is an absolute constant Further,

X 1/2+1/n2  K N K,n
(X).

In particular, for all ε > 0

lim sup

X →∞

log N K,n (X) log X  ε n ε , lim inf

X →∞

log N K,n
(X) log X ≥ 1

2 +

1

n2.

(1.1)

Linnik’s conjecture claims that the limit in (1.1) is equal to 1; thus, despite its evident imprecision, the upper bound in Theorem 1.1 seems to offer the first

serious evidence towards this conjecture for large n It is also worth observing

that de Jong and Katz [9] have studied a problem of a related nature where

the number field K is replaced by the function field Fq (T ); even here, where

much stronger geometric techniques are available, they obtain an exponent of

the nature c log(n); a proof of this bound can be found in [8, Lemma 2.4] This suggests that replacing n ε in (1.1) by a constant will be rather difficult

We will also prove various related results on the number of number fields

with certain Galois-theoretic properties For instance, if G ≤S n , let N K,n (X; G)

be the number of number fields L such that [L : K] = n, N KQD L/K < X, and

the action of Gal( ¯K/K) on embeddings K  → C is conjugate to the G-action

on {1, , n} We describe how one can obtain upper bounds on N K,n (X; G) using the invariant theory of G A typical example is:

Proposition 1.2 Let G ≤ S6 be a permutation group whose action is conjugate to the PSL2(F5)-action onP1(F5) Then N Q,6 (X; G)  ε X 8/5+ε

Specializing further, let N K,n (X; Gal) be the number of Galois extensions among those counted by N K,n (X); we prove the following upper bound.

Proposition 1.3 For each n > 4, one has N K,n (X; Gal)  K,n,ε X 3/8+ε

In combination with the lower bound in Theorem 1.1, this shows that if one orders the number fields of fixed degree overQ by discriminant, a random one is not Galois

Although we will use certain ad hoc tools, the central idea will always

be to count fields by counting integral points on certain associated varieties, which are related to the invariant theory of the Galois group These varieties must be well-chosen to obtain good bounds In fact, the varieties we use are

birational to the Hilbert scheme of r points in Pn, suggesting the importance

of a closer study of the distribution of rational points on these Hilbert schemes The results can perhaps be improved using certain techniques from the study of integral points, such as the result of Bombieri-Pila [15] However, the proof of Theorem 1.1 turns out, somewhat surprisingly, to require only elementary arguments from the geometry of numbers and linear algebra

Acknowledgments. The authors are grateful for the hospitality of the American Institute of Mathematics, where the first phase of this work was undertaken We also thank Hendrik Lenstra for useful comments on an earlier draft

2 Proof of upper bound

The main idea of Schmidt’s proof is as follows: by Minkowski’s theorem,

an extension L/K contains an integer α whose archimedean valuations are all

bounded by a function of ∆L= NKQD L/K Since all the archimedean absolute values are bounded in terms of ∆L, so are the symmetric functions of these

absolute values; in other words, α is a root of a monic polynomial in Z[x] whose

coefficients have (real) absolute value bounded in terms of ∆L There are only finitely many such polynomials, and counting them gives the theorem of [18]

The main idea of Theorem 1.1 is to count r-tuples of integers in L instead

of single integers

Let An = Spec(Z[x1, x2, , x n ]) denote affine n-space, which we regard

as being defined over Z We fix an algebraic closure ¯K of K Let ρ1, , ρ n

be the embeddings of L into ¯ K Then the map φ L = ρ1⊕ · · · ⊕ ρ n embedsO L

in ¯K n =An( ¯K), and the direct sum of r copies of this map is an embedding

O r

L → ( ¯ K n)r = (An)r( ¯K) (which map we also, by abuse of notation, call φ L) The affine variety (An)r is naturally coordinatized by functions

{x j,k }1≤j≤n,1≤k≤r The symmetric group S n acts on (An)r by permuting

x 1,k , , x n,k for each k The S n-invariants in the coordinate ring of (An)r

are called multisymmetric functions If f is a multisymmetric function, the composition f ◦ φ L : O r

L → ¯ K takes image in O K It follows that if R ⊂

Z[{x j,k }1≤j≤n,1≤k≤r]S n is a subring of the ring of multisymmetric functions

and A = Spec(R), there is a map of sets

F :

L

O r

L → A(O K)

where the union is over all number fields L with [L : K] = n.

Our overall strategy can now be outlined as follows If x is algebraic over

K, write ||x|| for the maximum of the archimedean absolute values of x For

a positive real number Y , let B(Y ) be the set of algebraic integers x in K with degree n over K and ||x|| < Y Let f1, , f s ∈ Z[{x j,k }1≤j≤n,1≤k≤r]S n

be multisymmetric functions with degrees d1, , d s Put R = Z[f1, , f s],

and set A = Spec(R) Then there is a constant c such that (for any Y ) one

has ||f i (φ L (α1, α2, , α r))|| < cY d i whenever α j ∈ B(Y ) (1 ≤ j ≤ r) and the

α j all belong to some subextension L ⊂ ¯ K, [L : K] = n Let A(O K)Y be the

subset of A( O K ) consisting of points P such that ||f i (P ) || < cY d i Then for

any subset S Y of B(Y ) r, we have a diagram of sets

(2.1)

{(L, α1, α2, , α r ) : [L : K] = n, ∆ L < X, (α1, , α r)∈ (O L)r ∩ S Y } −−−−→ A(O F K)Y





{L : [L : K] = n, ∆ L < X }.

The cardinality of the lower set is precisely N K,n (X) Our goal is to choose

A = Spec(R), Y, and S Y in such a way that that the vertical map in (2.1)

is surjective (by Minkowski’s theorem), while the horizontal mapF has finite

fibers whose cardinality we can bound This will yield the desired bound on

N K,n (X) Since |A(O K)Y |  K (c s Yi d i)[K:Q], it should be our aim to choose

f1, , f s whose total degree is as low as possible

We begin with a series of lemmas about polynomials over an arbitrary

characteristic-0 field F

Let S be any test ring We give An the structure of a ring scheme so that the ring structure on An (S) = S n is the natural one Let Tr be the map An → A1 which, on S-points, induces the map (z1, , z n) ∈ S n

z1+· · · + z n ∈ S Given an element x = (x j,k)1≤j≤n,1≤k≤r ∈ (S n)r, we denote

by xk ∈ S n the k-th “row” (x 1,k , x 2,k , , x n,k), and by x(j) ∈ S r the j-th

“column” (x j,1 , x j,2 , , x j,r) These correspond to maps x k : (An)r →

An , x (j): (An)r → A r

Let σ = (i1, , i r) be an element of Zr

≥0; we will think of Zr

≥0 as an

additive semigroup, operations being defined pointwise Then σ defines a

S n -equivariant map χ σ : (An)r → A n by the rule

χ σ(x) = xi1

1xi2

2 x i r

r

Here xi1

1 denotes x1 raised to the i1-th power, i.e xi1

1 = x1 × x1· · · · ×x1

(i times), the “multiplication” being taken with respect to ring-scheme

struc-ture on An

In particular, F n=An (F ) has a ring structure, and Tr, χ σinduce maps on

F -points, namely Tr : F n → F, χ σ : (F n)r → F n; we abuse notation and use

the same symbols for these maps The map (x, y)

pairing on F n, with respect to which we can speak of “orthogonal complement” Lemma 2.1 Let x ∈ (F n)r , and let Σ0 be a subset of Zr

≥0 such that the

|Σ0| vectors χ σ(x)σ ∈Σ0 generate a subspace of F n (considered as an F -vector

space) of dimension greater than n/2.

Denote by Σ1 = Σ0 + Σ0 the set of sums of two elements of Σ0 Let

W ⊂ F n be the subspace of F n spanned by χ σ(x)σ ∈Σ1 Then the orthogonal complement of W is contained in a coordinate hyperplane x j = 0 for some j.

Proof Write m for |Σ0| and let v1, , v m be the vectors χ σ (x) as σ ranges

over Σ0 Then W is the space spanned by the products v a v b (the algebra

structure on F n being as noted above) Suppose w is orthogonal to W Then

Tr(v a v b w) = 0

(2.2)

for all a, b; if V is the space spanned by the {v a }, then (2.2) implies that wV

and V are orthogonal This implies in turn that dim wV ≤ n−dim V < dim V ,

so multiplication by w is not an automorphism of F n ; in other words, w lies on

a coordinate hyperplane A subspace of F n contained in a union of coordinate hyperplanes is contained in a single coordinate hyperplane; this completes the proof

For each σ ∈ Z r

≥0 , let f σ : (An)r → A1 be the composition Tr◦ χ σ Then

f σ is a multisymmetric function When Σ is a subset ofZr

≥0 , we denote by RΣ the subring of functions on (An)r generated by {f σ } σ ∈Σ One has a natural

map of affine schemes

FΣ: (An

)r → Spec RΣ.

(2.3)

The goal of the algebro-geometric part of our argument is to show that, by

choosing Σ large enough, we can guarantee that FΣ is generically finite, and even place some restrictions on the locus in (An)r where FΣ has positive-dimensional fibers

Lemma 2.2 Let x be a point of (An)r (F ), and let Σ1 be a subset of Zr

≥0

such that the |Σ1| vectors χ σ(x)σ ∈Σ1 span F n as an F -vector space For each k between 1 and r let e k ∈ Z r

≥0 be the vector with a 1 in the k-th coordinate and

0’s elsewhere Let Σ be a set which contains Σ1+ Σ1, and Σ1+ e k for all k Then the preimage FΣ−1 (FΣ(x))⊂ (A n)r (F ) is finite, of cardinality at most (n!) r

Proof Let y be FΣ(x) Let m = |Σ1| As in the proof of Lemma 2.1,

let v
1, , v m
be the image of x under the {χ σ } σ ∈Σ1 We may suppose by

relabeling that v1
, , v
n form a basis for F n (as an F -vector space) Since

Σ contains Σ1 + Σ1, the determination of y fixes Tr(v a
v
b ) for all a, b; and

since Σ contains Σ1+ e k , we also know the traces Tr(v
axk ) for all a and k It

follows that, for each k, we can represent the action of multiplication by x k

on the F -vector space spanned by v1
, , v
nby a matrix whose coefficients are

determined by y But such a matrix evidently determines xkup to permutation

of coordinates; this proves the desired result

In the proof of Proposition 2.5 below, we will need to show that, by

allowing x to vary over certain subspaces of (F n)r, we can ensure that x can

be chosen in order to verify the hypothesis of Lemma 2.1

Lemma 2.3 Let V be a F -subspace of F n of dimension m, and let Σ0⊂

Zr

≥0 be a subset of size m Let Z ⊂ V r be the subset of points x ∈ V r such that the m vectors χ σ(x)σ ∈Σ0 are not linearly independent (over F ) in F n Then

Z is not the whole of V r If one identifies V r with F mr , Z is contained in

the F -points of a hypersurface, defined over F , whose degree is bounded by a constant depending only on n and Σ0.

Proof We may assume (by permuting coordinates) that the map

“projec-tion onto the first m coordinates,” which we denote π : F n → F m, induces an

isomorphism V ∼ = F m Suppose there is a nontrivial linear relation



σ ∈Σ

c σ χ σ(x) = 0∈ F n

,

(2.4)

that is, suppose x∈ Z Each σ ∈ Σ0 also defines a map F r → F (derived from

the map χ σ : (An)r → A n with n = 1) so we may speak of χ σ(x(j)) ∈ F for

1≤ j ≤ n By abuse of notation we also use π to denote the projection of (F n)r

onto (F m)r Then the restriction of π to V r is an isomorphism V r ∼ = F mr

Any nontrivial linear relation between the χ σ(x) yields a nontrivial

rela-tion between the m vectors χ σ (π(x)) in F m This in turn implies vanishing of the determinant

D =









χ σ1(x(1)) · · · χ σ1(x(m))

χ σ m(x(1)) · · · χ σ m(x(m))









.

The contribution of each m × m permutation matrix to D is a distinct

monomial in the mr variables, so D is not identically 0 in F [x 1,1 , , x m,r]

Evidently the degree of D is bounded in terms of n and Σ0 Let V (D) be the vanishing locus of D in (F m)r Now the locus in Z is contained in π −1 (V (D)),

which yields the desired result

Finally, we need a straightforward fact about points of low height on the complements of hypersurfaces

Lemma 2.4 Let f be a polynomial of degree d in variables x1, , x n Then there exist integers a1, , a n such that max1≤i≤n |a i | ≤ (1/2)(d + 1) and

f (a1, , a n)= 0.

Proof There are at most d hyperplanes on which f vanishes, which means

that the function g(x2, , x n ) := f (a1, x2, , x n) is not identically 0 for some

a1 with absolute value at most (1/2)(d + 1) Now proceed by induction on n.

Now we are ready for the key point in the proof of Theorem 1.1 The point is to use the lemmas above to construct Σ which is small enough that

Spec RΣ has few rational points of small height, but which is large enough so

that FΣ does not have too many positive-dimensional fibers

Proposition 2.5 Let Σ0 be a subset of Zr

≥0 of size m > n/2; let Σ1 ⊂

Zr

≥0 contain Σ0+ Σ0; and let Σ ⊂ Z r

≥0 contain Σ1+ Σ1 and Σ1+ e k for all k Let L be a finite extension of K with [L : K] = n Then there is an r-tuple

(α1, , α r)∈ O r

L such that

• For every k,

||α k || Σ D 1/d(n −2)

L , where d = [K : Q].

• The set FΣ−1 (FΣ((φ L (α1, , α r)))) ⊂ (A n)r( ¯K) has cardinality at most

(n!) r

• The elements α1, , α r generate the field extension L/K.

Proof First of all, note that if (α1, , α r ), Σ0, Σ1, Σ satisfy the conditions

above, then so do (α1, , α r ), Σ
0, Σ1, Σ for any subset Σ
0 ⊂ Σ0 with |Σ

0| > n/2 So it suffices to prove the theorem in case n/2 < m ≤ (n/2 + 1).

Let 1 = β1, , β nd be a Q-linearly independent set of integers in O L

such that ||β i || is the i-th successive minimum of || · || on O L, in the sense

of Minkowski’s second theorem [20, III, §3] The K-vector space spanned by

β1, , β md has K-dimension at least m, so we may choose γ1, , γ m among

the β i which are linearly independent over K.

Let V ⊂ ¯ K n be the ¯K-vector space spanned by {φ L (γ i)}1≤i≤m Then by

Lemma 2.3 there is a constant C n,Σ0 and a hypersurface Z ⊂ V r of degree

C n,Σ0 such that, for all x not in Z( ¯ K), the m vectors χ σ(x)σ ∈Σ0 are ¯K-linearly

independent in ¯K n

For every field M strictly intermediate between K and L, we let V M ⊂ ¯ K n

be the ¯K-vector subspace φ L (M ) ⊂ ¯ K n Each (V ∩ V M)r is a certain linear

subspace of V r ; note that, since m > n/2, no subspace V M contains V Let Z

be the union of Z( ¯ K) with (V ∩ V M)r , as M ranges over all fields between K and L.

Now let Y be a hypersurface of V r so that Y ( ¯ K) contains Z
; one may

choose Y so that the degree of Y is bounded in terms of n and Σ0 By

Lemma 2.4, there is a constant H, depending only on n and Σ0, so that, for

any lattice ι : Zmr  → V r (i.e we require ι(Z mr ) spans V r over ¯K) there is a

point p ∈ Z mr , with ι(p) / ∈ Y ( ¯ K), whose coordinates have absolute value at

most H.

It follows that there exists a set of mr integers c 1,1 , , c m,r with |c j,k |

≤ H, such that

x = (φ L (c 1,1 γ1+· · · + c m,1 γ m ), , φ L (c 1,r γ1+· · · + c m,r γ m))

is not in Y ( ¯ K) For each k between 1 and r define α k ∈ O Lvia

α k = c 1,k γ1+· · · + c m,k γ m

Let W ⊂ L be the K-subspace spanned by χ σ (α1, , α r ) as σ ranges over

Σ1 (here we regard χ σ as a map L r → L, cf remarks after (2.4)) Suppose

W is not the whole of L Then there is a nonzero element t ∈ L such that

TrL K tw = 0 for all w ∈ W It follows that φ L (t) ∈ ¯ K lies in the orthogonal

complement (w.r.t the form Tr on ¯K n ) of φ L (W ) ⊂ ¯ K n But the orthogonal complement to the ¯K-span of φ L (W ) is contained in a coordinate hyperplane

by Lemma 2.1 Since ρ j (t) cannot be 0 for any j and any nonzero t, this is a contradiction; we conclude that W = L, and thus that the vectors {χ σ(x)} σ ∈Σ1

span L as a K-vector space.

The bound on the size of the fiber FΣ−1 (FΣ(x)) follows from Lemma 2.2,

and the fact that x / ∈ V r

M for any M implies that α1, , α r generate the

extension L/K.

It remains to bound the archimedean absolute values of the α i The image

of O L in O L ⊗Z R is a lattice of covolume D 1/2

L , so by Minkowski’s second theorem [20, Th 16],

nd



i=1

||β i || ≤ D 1/2

L

The ||β i || form a nondecreasing sequence, so for m < n, we have

||β md || (n −m)d ≤

nd



i=md+1

||β i || ≤ D 1/2

L

Since m ≤ (1/2)n + 1, we get

||β i || < (D L)1/d(n −2)

for all i ≤ m It follows that all archimedean absolute values of γ i for i ≤ m

are bounded by a constant multiple of D 1/d(n −2)

L , the implicit constant being

absolute The result follows, since each α k is an integral linear combination of

the γ i with coefficients bounded by H.

We are now ready to prove the upper bound in Theorem 1.1; what remains

is merely to make a good choice of Σ and apply Proposition 2.5 Let r and c

be positive integers such thatr+c

r

> n/2, and let Σ0 be the set of all r-tuples

of nonnegative integers with sum at most c We shall choose r, c in the end; but r, c, Σ0, Σ will all depend only on n, so that all constants that depend on

them in fact depend only on n.

Now take Σ to be the set of all r-tuples of nonnegative integers with sum

at most 4c, and consider the map

FΣ: (An

)r → Spec RΣ.

By Proposition 2.5, to every field L with [L : K] = n we can associate an

r-tuple (α1, , α r) of integers satisfying the three conditions in the statement

of the proposition Define Q L ∈ (A n)r( ¯K) to be φ L (α1, , α r ), and let P L ∈

Spec RΣ(O K ) be the point FΣ(Q L)

By the second condition on α1, , α r , there are at most (n!) r points in

FΣ−1 (P L ) By the third condition, Q L = Q L
only if L and L
are isomorphic

over K We conclude that at most (n!) r fields L are sent to the same point in Spec RΣ(O K)

We now restrict our attention to those fields L satisfying

NKQD L/K < X.

In this case, for every archimedean valuation | · | of L and every k ≤ r we have

the bound

|α k |  D 1/d(n −2)

K)1/d(n −2)

(2.5)

Now, f σ being as defined prior to (2.3), f σ (Q L) is an element of O K,

which (by choice of Σ) we can express as a polynomial of degree at most 4c (and absolutely bounded coefficients) in the numbers ρ j (α k)∈ ¯ K If | · | is any

archimedean absolute value on K, we can extend |·| to a archimedean absolute

value on L, and by (2.5) we have

|f σ (Q L)|  (XD n

K)4c/d(n −2)

The number of elements ofO K with archimedean absolute values at most

B is ≤ (2B + 1) d (For large enough B, one can save an extra factor of

D 1/2

K ; this is not necessary for our purpose.) In view of the above equation,

the number of possibilities for f σ (Q L) is  (XD n

K A d n)4c/(n −2) where A n is a

constant depending only on n.