1. Introduction. Let r, k, h be positive integers. For any Dirichlet char- acter χ modulo k write

LXXIX.4 (1997)

On the average of character sums for a group of characters

by

D. A. Burgess (Nottingham)

To J. W. S. Cassels

1. Introduction. Let r, k, h be positive integers. For any Dirichlet char- acter χ modulo k write

W

(χ) = X

X

χ(x + m)   

. In [2] it was shown that, if χ is a primitive character, then (1) W

(χ)  kh

+ k

h

,

which implies that

(2) W

(χ)  k

h

for h ≤ k

. In [6] it was shown that

(3) X

primitive χ (mod k)

W

(χ)  k

h

,

so that (2) holds for all h, on average for all primitive characters modulo k.

Thus it is reasonable to conjecture that (2) might hold for some h > k

, on average for the primitive characters of a large subgroup of the characters modulo k. In [8] such a result was obtained, it being shown that, for any prime p,

(4) X

χ mod p³ χ^p2=χ0

W

(χ)  p

h

+ p

h

,

where χ

is the principal character, and thus (2) holds for h ≤ k

, on average for the characters modulo k = p

in the group of order p

.

1991 Mathematics Subject Classification: Primary 11L40.

[313]

In this paper the argument is strengthened to show in the following theorem that, for the non-principal characters of this group, (2) remains true for h ≤ k

, on average.

Theorem 1. Let p be an odd prime number. Let

S = X

χ mod p³ χ^p2=χ₀ χ6=χ₀

p³

X

x=1

X

χ(x + m)   

.

Then in the case r = 2 we have

S  p

h

+ p

h

. From [1] it follows that if k is prime then (5) W

(χ)  kh

+ k

h

for all positive h. By the methods of this paper it is shown that (5) can be improved for h > p

, on average for the non-principal characters modulo k = p

in the group of order p

.

Theorem 2. Let p be an odd prime number. Let

S = X

χ mod p³ χ^p2=χ0

χ6=χ0

p³

X

x=1

X

χ(x + m)   

.

Then in the case r = 3 we have

S  p

h

+ min(h, p)

p

h + min(h, p)

p

.

In Section 8 we shall describe some corollaries of these theorems.

2. Preliminary transformation of the problem. For S as in the statement of both theorems, we have

(6) S = X

χ mod p³ χ^p2=χ₀

p³

X

x=1

X

χ(x+m)   

−

p³

X

x=1

 X

m=1

χ

(x+m)



= S

−S

,

say.

Now

S

= X

m p³

X

x=1

χ

(f

(x))χ

(f

(x)),

where m ∈ Z

satisfies 0 < m

≤ h for 1 ≤ i ≤ 2r, and

f

(x) = (x + m

) . . . (x + m

), f

(x) = (x + m

) . . . (x + m

).

Thus

(7) S

= X

m

p³

X

x6≡−mx=1i(mod p)

1 = X

m

p

#{x : 1 ≤ x ≤ p, p - f

(x)f

(x)}.

We also have S

= X

m p³

X

x=1

X

χ^p2=χ0

χ(f

(x))χ(f

(x)) = X

m

p³

X

x=1 p

-

X

χ^p2=χ0

χ

 f

(x) f

(x)



= p

X

m

#{1 ≤ x ≤ p

, 1 ≤ z < p : p - f

(x)f

(x),

f

(x) − z

f

(x) ≡ 0 (mod p

)}.

Thus, writing

(8) f

(x) = f

(x) − z

f

(x), we have

(9) S

= S

+ S

,

where

(10) S

= p

X

m

X

#{1 ≤ x ≤ p

: x 6≡ −m

(mod p),

f

(x) ≡ 0 (mod p

), f

(x) 6≡ 0 (mod p)}

and

(11) S

= p

X

m

X

#{1 ≤ x ≤ p

: x 6≡ −m

(mod p),

f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p)}.

We consider the non-singular roots of f

. Since the numbers of non- singular roots modulo p

and p are the same we have from (10) that

S

≤ p

X

m p−1

X

z=1

#{1 ≤ x ≤ p : p - f

(x)f

(x), f

(x) ≡ 0 (mod p)}

= p

X

m

X

-

#{1 ≤ z < p : f

(x) − zf

(x) ≡ 0 (mod p)}

= p

X

m

#{1 ≤ x ≤ p : p - f

(x)f

(x)} = S

from (7). Now from (6) and (9) we have

(12) S ≤ S

.

3. Estimates for solution sets of polynomials. In the proof of our theorems we shall require some lemmas concerning the number of solutions of congruences to a prime power modulus. We shall use the following gen- eralisation of the well known estimate for the number of non-singular roots of a congruence.

Lemma 1. Let F be a polynomial of degree n having integer coefficients.

Let p be a prime, d a positive integer , and α, β and γ be non-negative integers satisfying γ = dα/de. Then

#{1 ≤ x ≤ p

: p

| F (x), p

k F

(x)} ≤ n.

P r o o f. This is Proposition 1 of [7].

We shall require an estimate for the number of solutions of a congruence in many variables to a prime modulus. The following will suffice.

Lemma 2. Let G be a polynomial in x

, . . . , x

which is not identically zero modulo the prime p. Let 0 < h

≤ p for all i. Then the number of x, satisfying 0 < x

≤ h

for all i, for which G(x) ≡ 0 (mod p) is O(( Q

h

)/ min h

).

P r o o f. This is an easy modification of Lemma 5 of [4].

We shall also use the following estimate which, under favourable condi- tions, can provide an optimal estimate for the average number of singular roots of a set of polynomials.

Lemma 3. Let F

(y) (0 < i ≤ n) be polynomials in ν variables y

(0 <

k ≤ ν). Let p be a prime and 2β ≥ α

≥ . . . ≥ α

> β ≥ α

≥ . . . ≥ α

be positive integers. Let H ∈ N

. Write

N = #{y : ∀k ≤ ν 0 < y

≤ H

, ∀i F

(y) ≡ 0 (mod p

)}.

Put λ

= dH

/p

e for all k ≤ ν. Then

N  X

∀k≤ν |BB_k|<λ_k

#



y : ∀k ≤ ν 0 < y

≤ p

, ∀i ≤ m F

(y) ≡ 0 (mod p

),

∀i > m F

(y) ≡ 0 (mod p

), ∀i ≤ m X

B

∂F

∂y

(y) ≡ 0 (mod p

)

 . P r o o f. Clearly

N ≤ #{y : ∀k ≤ ν 0 < y

≤ λ

p

, ∀i F

(y) ≡ 0 (mod p

)}.

For 1 ≤ k ≤ ν write y

= a

+ p

b

, where 0 < a

≤ p

, 0 ≤ b

< λ

. Then, for m < i ≤ n, F

(y) ≡ 0 (mod p

) becomes

F

(a) ≡ 0 (mod p

), while for i ≤ m, F

(y) ≡ 0 (mod p

) becomes

F

(a) + X

∂F

∂y

(a)b

p

≡ 0 (mod p

).

The latter congruences imply, for i ≤ m, F

(a) ≡ 0 (mod p

) so that, say,

∀i ≤ m F

(a) = c

(a)p

. Thus we have

N ≤ X

∀i≤m Fi(a)≡0 (mod pa ^β)

∀i>m F_i(a)≡0 (mod p^αi)

#



b : ∀k ≤ ν 0 ≤ b

< λ

,

∀i ≤ m X

∂F

∂y

(a)b

≡ −c

(a) (mod p

)

 . Now the number of solutions of the inhomogeneous congruences

∀i ≤ m X

∂F

∂y

(a)b

≡ −c

(a) (mod p

)

in the variables b satisfying 0 ≤ b

< λ

for all k ≤ ν is at most the number of solutions of the homogeneous congruences

∀i ≤ m X

∂F

∂y

(a)B

≡ 0 (mod p

)

in the variables B satisfying |B

| < λ

for all k ≤ ν. Thus we have

N ≤ X

∀i≤m F_i(a)≡0 (mod pa ^β)

∀i>m Fi(a)≡0 (mod p^αi)

#



B : ∀k ≤ ν |B

| < λ

,

∀i ≤ m X

∂F

∂y

(a)B

≡ 0 (mod p

)



≤ X

∀k≤ν |BB_k|<λ_k

#



a : ∀k ≤ ν 0 < a

≤ p

, ∀i ≤ m F

(a) ≡ 0 (mod p

),

∀i > m F

(a) ≡ 0 (mod p

), ∀i ≤ m X

B

∂F

∂y

(a) ≡ 0 (mod p

)



.

4. Proof of Theorem 1. Clearly in proving the theorem we may sup- pose that p > 2. We consider here the case r = 2.

It remains to consider the singular roots. Noting (11) we write

(13) S

= S

+ S

,

where S

= p

X

m

#{1 ≤ x ≤ p

, 1 ≤ z < p : x 6≡ −m

(mod p),

f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p), f

(x) ≡ 0 (mod p)}, and

S

= p

X

m p−1

X

z=2

#{1 ≤ x ≤ p

: x 6≡ −m

(mod p), f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p), f

(x) 6≡ 0 (mod p)}.

Clearly we have S

≤ p

X

m

#{1 ≤ x ≤ p

:

(m

+ m

− m

− m

)x + (m

m

− m

m

) ≡ 0 (mod p

), (m

+ m

− m

− m

) ≡ 0 (mod p)}.

Write

(14) p

= highest common factor(p

, m

+ m

− m

− m

), where 1 ≤ δ ≤ 3. For solubility of the congruence

(m

+ m

− m

− m

)x + (m

m

− m

m

) ≡ 0 (mod p

) we require also

(15) p

| (m

m

− m

m

).

The congruence then has at most p

solutions satisfying 1 ≤ x ≤ p

. (14) and (15) imply

(m

− m

)(m

− m

) ≡ 0 (mod p

).

Suppose that p

| (m

− m

) and p

| (m

− m

). Then the number of such m is

O

 h

 1 + h

p



1 + h p



1 + h p



= O

 h

p

+ h

 . Thus

(16) S

 p

X

δ,ε

p

 h

p

+ h



 ph

+ p

h

.

On the other hand, we have S

 p

X

m

X

#{1 ≤ x ≤ p

: f

(x) ≡ 0 (mod p

),

f

(x) ≡ 0 (mod p), f

(x) 6≡ 0 (mod p)}.

Thus, by Lemma 1,

S

 p

#{m, 1 < z < p : ∃x f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p)}.

Thus

(17) S

 p

#{m, z, x : ∀i 0 < m

≤ h, 0 < x ≤ p, 0 < z < p, f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p)}.

Put

(18) λ =

 h p



, µ =

n p if λ > 1, h if λ = 1.

Now we apply Lemma 3, treating x, z as constants and the m

as our vari- ables, to obtain

(19) S

 p

X

∀k≤4 |BBk|<λ

#



a, x, z : ∀k ≤ 4 0 < a

≤ µ, 0 < x ≤ p,

0 < z < p, f

(x) ≡ 0 (mod p), f

(x) ≡ 0 (mod p),

X

B

∂f

∂m

(x) ≡ 0 (mod p)



if λ > 1, while if λ = 1 this follows immediately from (17).

Given B, x, z, a

, a

we have, from (19),

f

(x) ≡ 2(1 − z)x + (a

+ a

− za

− za

) ≡ 0 (mod p),

from which a

+ a

is determined modulo p. Then also from (19) we have f

(x) ≡ (1 − z)x

+ (a

+ a

− za

− za

)x + (a

a

− za

a

) ≡ 0 (mod p), and so a

a

is also determined modulo p. Thus there are at most two choices for a

, a

. Use these congruences to eliminate a

, a

. We have on writing

π

= a

+ a

, π

= a

a

, %

= a

+ a

, %

= a

a

, the identity

(π

− 4π

)(B

− B

)

− (2B

a

+ 2B

a

− π

(B

+ B

))

= 0.

Now from (19) we have X

k=1

B

∂f

(x)

∂m

≡ (B

+ B

− zB

− zB

)x

+ (a

B

+ a

B

− za

B

− za

B

)

≡ 0 (mod p).

Thus eliminating a

, a

we have

(20) (z

%

− 4z%

(1 − z)x − 4x

z(1 − z) − 4z%

)(B

− B

)

−z

(2(B

a

+ B

a

) + 2x(B

+ B

) − (%

+ 2x)(B

+ B

))

≡ 0 (mod p).

Thus from (19), (21) S

 p

X

B

#{x, z, a

, a

:

(z

%

− 4z%

(1 − z)x − 4x

z(1 − z) − 4z%

)(B

− B

)

−z

(2(B

a

+ B

a

) + 2x(B

+ B

) − (%

+ 2x)(B

+ B

))

≡ 0 (mod p)}.

By Lemma 2, for a given choice of B, (20) has at most O(p

µ) solutions in x, z, a

, a

, unless this polynomial is identically zero modulo p. If the coefficient of za

a

is zero we have

−4(B

− B

)

≡ 0 (mod p),

and thus B

≡ B

(mod p). Under this condition if the coefficient of z

a

is zero we have

−(2B

− B

− B

)

≡ 0 (mod p), and if the coefficient of z

a

is zero we have

−(2B

− B

− B

)

≡ 0 (mod p).

Thus if the polynomial is identically zero modulo p we have B

≡ B

≡ B

≡ B

(mod p).

Hence the number of such cases is O(λ(1 + λ/p)

).

Consequently, from (21) we have S

 p



λ

p

µ +

 λ + λ

p

 p

µ



 p

h

+ p

h

. The theorem follows from (12), (13) and (16).

5. Introduction to proof of Theorem 2. We may suppose that p > 2.

We consider here the case r = 3. Noting (11) we write

(22) S

= X

m

#{x, z : 0 < x ≤ p

, 0 < z < p, f

(x)f

(x) 6≡ 0 (mod p), f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p), either f

6≡ 0 (mod p

) or f

(x) 6≡ 0 (mod p)}

and

(23) S

= X

m

#{x, z : 0 < x ≤ p

, 0 < z < p, f

(x)f

(x) 6≡ 0 (mod p), f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p)}

so that

(24) S

= p

S

+ p

S

. We estimate S

and S

in Section 7.

We shall use also the polynomials g

(x) given by

∀i ≤ 3 g

(x) = f

(x)/(x + m

), ∀i ≥ 4 g

(x) = f

(x)/(x + m

).

Thus we have, from (8),

f

(x) = g

(x)(x + m

) − z

g

(x)(x + m

) and

f

(x) = g

(x) + g

(x) + g

(x) − z

g

(x) − z

g

(x) − z

g

(x).

Write

C

(m) = g

(x)(x + m

) − zg

(x)(x + m

),

C

(m) = (2x + m

+ m

)(x + m

) − z(2x + m

+ m

)(x + m

) + (g

(x) − zg

(x)),

C

(m) = 2(x + m

) − 2z(x + m

)

+ ((4x + 2m

+ 2m

) − z(4x + 2m

+ 2m

)),

C

(m) = b

g

(x) + b

g

(x) + b

g

(x) − b

zg

(x) − b

zg

(x) − b

zg

(x)

= (b

(x + m

) + b

(x + m

))(x + m

)

− z(b

(x + m

) + b

(x + m

))(x + m

) + (b

g

(x) − b

zg

(x)) and

C

(m) = b

(2x + m

+ m

) + b

(2x + m

+ m

) + b

(2x + m

+ m

)

− b

z(2x + m

+ m

) − b

z(2x + m

+ m

)

− b

z(2x + m

+ m

)

= (b

+ b

)(x + m

) − z(b

+ b

)(x + m

)

+ (b

(2x + m

+ m

) + b

(x + m

) + b

(x + m

)

− zb

(2x + m

+ m

) − zb

(x + m

) − zb

(x + m

)).

We define λ and µ by (18).

6. Minor lemmas Lemma 4. Write

(25) D

=

  g

(x) −zg

(x) 2x + m

+ m

−z(2x + m

+ m

)

    . Then X

∀i |bb_i|<λ

#{m, x, z : 0 < m

≤ µ, 0 < x ≤ p, 0 < z < p,

f

(x)f

(x) 6≡ 0 (mod p), C

(m) ≡ C

(m) ≡ 0 (mod p), D

≡ 0 (mod p)}

 pµ

λ

. P r o o f. From C

(m) ≡ C

(m) ≡ D

≡ 0 (mod p) it follows that

g

(x) ≡ zg

(x) (mod p),

from which z is uniquely determined. Then from C

(m) ≡ 0 (mod p) it follows that m

= m

. Finally,

D

= z((m

+ m

− m

− m

)x

+ 2(m

m

− m

m

)x + m

m

(m

+ m

) − m

m

(m

+ m

)),

so that, by Lemma 2, D

/z ≡ 0 (mod p) has O(pµ

) solutions as a function of m

, m

, m

, m

, x. Thus the required estimate follows trivially.

Lemma 5. Write D

=

m

m

−zm

m

0

m

+ m

−z(m

+ m

) m

m

− zm

m

b

m

+ b

m

−zb

m

− zb

m

b

m

m

− b

zm

m

      .

Then X

∀i |bbi|<λ D₂identically 0 (mod p)

#{m

, m

, m

, m

, x, z :

0 < m

≤ µ, 0 < x ≤ p, 0 < z < p}

 p

µ

λ

 λ p + 1



. P r o o f. We have

D

= (b

− b

)zm

m

m

+ (b

− b

)zm

m

m

+ (b

− b

)zm

m

m

m

+ (b

− b

)zm

m

m

m

+ (b

− b

)z

m

m

m

m

+ (b

− b

)z

m

m

m

m

+ (b

− b

)z

m

m

m

+ (b

− b

)z

m

m

m

.

This is identically 0 (mod p) only if

b

≡ b

≡ b

≡ b

≡ b

≡ b

(mod p).

The required estimate follows trivially.

Lemma 6. We have X

∀i |bbi|<λ

#{m, x, z : 0 < m

≤ µ, 0 < x ≤ p,

0 < z < p, f

(x)f

(x) 6≡ 0 (mod p), C

(m) ≡ C

(m) ≡ C

(m) ≡ 0 (mod p), D

≡ 0 (mod p)}

 pµ

λ

, where D

is defined by (25).

P r o o f. From C

(m) ≡ C

(m) ≡ D

≡ 0 (mod p) it follows that g

(x) ≡ zg

(x) (mod p),

from which z is uniquely determined. Then from C

(m) ≡ 0 (mod p) it follows that m

= m

and, since f

(x)f

(x) 6≡ 0 (mod p), from C

(m) ≡ 0 (mod p) that

(2x + m

+ m

) − z(2x + m

+ m

) ≡ 0 (mod p).

Now substituting into C

(m) ≡ 0 (mod p) we obtain z = 1 and so also m

+ m

≡ m

+ m

(mod p). But also we have g

(x) ≡ g

(x) (mod p) and so m

m

≡ m

m

(mod p). Thus m

, m

is a permutation of m

, m

. The required estimate follows trivially.

Lemma 7. Write D

=

g

(x) g

(x) 0

2x + m

+ m

2x + m

+ m

g

(x)

1 1 2x + m

+ m

      and

D

=      

g

(x) g

(x) 0

2x + m

+ m

2x + m

+ m

g

(x)

1 1 2x + m

+ m

      . Then X

∀i |bbi|<λ

#{m

, m

, m

, m

, x :

0 < m

≤ µ, 0 < x ≤ p, D

≡ D

≡ 0 (mod p)}

 pµ

λ

. P r o o f. The conditions D

≡ D

≡ 0 (mod p) expand to give

(g

(x) − g

(x))((2x + m

+ m

)(2x + m

+ m

) − g

(x))

−g

(x)(m

+ m

− m

− m

)(2x + m

+ m

)

≡ (g

(x) − g

(x))((2x + m

+ m

)

− g

(x))

− g

(x)(m

+ m

− m

− m

)(2x + m

+ m

) ≡ 0 (mod p).

These will have only O(1) solutions x unless both polynomials are identically 0 (mod p). But in the first of these the coefficient of x

is m

+m

−m

−m

, and if this is 0 (mod p) then the coefficient of x

is 3(m

m

− m

m

). Thus if both polynomials in x are identically 0 (mod p) then the pair m

, m

is a permutation of m

, m

. This contributes

(26)  λ

pµ

to our estimate.

Now consider the other case in which at least one of D

and D

is not identically 0 (mod p). Then there are only O(1) values for x. The two polynomial congruences D

≡ D

≡ 0 (mod p) are cubics. The difference between these polynomials is

g

(x) g

(x) 0

2x + m

+ m

2x + m

+ m

g

(x) − g

(x)

1 1 m

+ m

− m

− m

      . By row and column operations this simplifies to

m

m

m

m

0

m

+ m

m

+ m

m

m

− m

m

1 1 m

+ m

− m

− m

      ,

which is a polynomial in m

, m

, m

, m

. This polynomial is −1 when m

= m

= 1, m

= m

= 0. Thus it is not identically 0 (mod p) and so has O(µ

) solutions in m

, m

, m

, m

. Thus this contributes

(27)  λ

µ

to our estimate. The lemma follows from (26) and (27).

Lemma 8. Write D

=

m

m

−m

m

0

m

+ m

−(m

+ m

) m

m

b

m

+ b

m

−b

m

− b

m

b

m

m

×      

m

m

− m

m

m

m

0 m

+ m

− m

− m

m

+ m

m

m

0 1 m

+ m

−      

m

m

−m

m

0

m

+ m

−(m

+ m

) m

m

b

m

+ b

m

−b

m

− b

m

b

m

m

×      

m

m

− m

m

m

m

0 m

+ m

− m

− m

m

+ m

m

m

0 1 m

+ m

  .

Then

X

∀i |bb_i|<λ D5identically 0 (mod p)

#{m

, m

, m

, m

: 0 < m

≤ µ}  µ

λ

 λ p + 1



.

P r o o f. Substitute m

= 0 in D

to obtain (b

− b

)m

m

m

. Thus if D

is identically 0 (mod p) then b

≡ b

(mod p). Similar arguments give

b

≡ b

≡ b

, b

≡ b

≡ b

(mod p).

Substituting this in D

, and putting m

= m

= m

= 1 we obtain

−(b

− b

)(1 − m

)

m

also. The required estimate follows trivially.

7. Proof of Theorem 2 Lemma 9. We have

S

 h

+ p

hµ

, where µ is defined by (18).

P r o o f. From Lemma 1 applied to (22) it follows that S

 p X

m

#{z : 0 < z < p, ∃x f

(x)f

(x) 6≡ 0 (mod p),

f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p)}.

We can rewrite this as (28) S

 p

X

p−1

X

z=1

#{m : f

(x)f

(x) 6≡ 0 (mod p),

f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p)}, say. Write

(29) N = #{m : f

(x)f

(x) 6≡ 0 (mod p), f

(x) ≡ 0 (mod p

), f

(x) ≡ 0 (mod p)}.

Thus by Lemma 3,

(30) N  X

∀i |bb_i|<λ

#{m : ∀i 0 < m

≤ µ, f

(x)f

(x) 6≡ 0 (mod p), 0 ≡ C

(m) ≡ C

(m) ≡ C

(m) (mod p)}

if λ > 1, and follows immediately from (29) if λ = 1. Thus from (28) we have

S

 p X

∀i |bb_i|<λ

#{m, x, z : 0 < m

≤ µ, 0 < x ≤ p, 0 < z < p,

f

(x)f

(x) 6≡ 0 (mod p), 0 ≡ C

(m) ≡ C

(m) ≡ C

(m) (mod p)}.