Short Kloosterman Sums for Polynomials over Finite Fields

Short Kloosterman Sums for Polynomials over Finite Fields William D Banks Department of Mathematics, University of Missouri Columbia, MO 65211 USA bbanks@mathmissouriedu Asma Harcharras Department of Mathematics, University of Missouri Columbia, MO 65211 USA harchars@mathmissouriedu Igor E Shparlinski Department of Computing, Macquarie University Sydney, NSW 2109, Australia igor@icsmqeduau Abstract We extend to the setting of polynomials over a finite field certain estimates for short Kloosterman sums originally due to Karatsuba Our estimates are then used to establish some uniformity of distribution results in the ring F q [x]/m(x) for collections of polynomials either of the form f 1 g 1 or of the form f 1 g 1 + afg, where f and g are polynomials coprime to M and of very small degree relative to M, and a is an arbitrary polynomial We also give estimates for short Mathematics Subject Classification: 11T23, 11T06 1

Kloosterman sums where the summation runs over products of two irreducible polynomials of small degree It is likely that this result can be used to give an improvement of the Brun-Titchmarsh theorem for polynomials over finite fields 1 Introduction Let q be a prime power, F q the finite field with q elements, and R the polynomial ring F q [x] Fix an irreducible polynomial M R of degree deg(m) = m > 0, and let R M denote the field R/(M) Put R m = {f R deg(f) < m}, R m = {f R m f 0}, and observe the natural bijections R m R M, R m R M In particular, for every f R m, there exists a unique element f R m such that ff 1 (mod M) Then f is the inverse of f if both polynomials are viewed as elements of R M For any subset E {0, 1,, m 1} and any two polynomials f, g R m, with f(x) = m 1 j=0 a j x j, g(x) = m 1 j=0 b j x j, write f E g whenever a j = b j for all j E Then E defines an equivalence relation on R m, and we will denote by R m / E the corresponding set of equivalence classes In this paper, we study the distribution in R m / E of polynomials of the form (fg), where f and g are nonzero polynomials of small degree relative to m We show that the polynomials (fg) are uniformly distributed in R m / E provided that the cardinality of E satisfies a certain upper bound Our main result in this direction is Theorem 6 of Section 5 As an application, our Theorem 6 implies the following result: Theorem 1 Let ǫ be a real number such that 0 ǫ < 1/3, and suppose that m ǫ 1 and q m 1 Then for any polynomial F R m and any set 2

E {0, 1,, m 1} of cardinality E m 3ǫ (log m) 3, there exist polynomials f, g R m, with deg(f), deg(g) m 2/3+ǫ log m, such that (fg) E F Moreover, if ǫ is at least 1/12, and m ǫ 1, the result holds for any choice of the prime power q We remark that the conditions of Theorem 1 are independent of the choice of M; the conclusion therefore holds for every irreducible polynomial M of degree m Now for any f R, let {f} be the unique polynomial in R m such that f {f} (mod M) In this paper, we also study the distribution in R m / E of polynomials of the form { (fg) + afg }, where a R m, and f and g are nonzero polynomials of small degree relative to m We show that the polynomials { (fg) + afg } are uniformly distributed in R m / E, assuming again that the cardinality of E satisfies a certain bound Our main result in this direction is Theorem 7 of Section 5, which implies the following: Theorem 2 Let ǫ be a real number such that 0 ǫ < 1/3, and suppose that m ǫ 1 and q m 1 Then for any two polynomials F, a R m and any set E {0, 1,, m 1} of cardinality E m3ǫ (log m) 3, 8 there exist polynomials f, g R m, with deg(f), deg(g) m 2/3+ǫ log m, such that { (fg) + afg } E F Moreover, if ǫ is at least 1/12, and m ǫ 1, the result holds for any choice of the prime power q 3

The main results of this paper (Theorems 6 and 7) rely primarily on bounds for character sums of the form χ ( (fg) + afg ), f,g 0 deg(f) d deg(g) e where χ is a nontrivial additive character of R M Such bounds are provided by Theorem 3 for the case a R m, and by Theorem 4 for the case a = 0 (see Section 4) Theorems 3 and 4 are proved without the assumption that M is irreducible, and we remark that Theorems 6 and 7 can be extended (with only minor modifications) to arbitrary polynomials as well For this reason, we do not make explicit use of the isomorphism R M F q m, and we do not formulate Theorems 6 and 7 in terms of finite fields We also consider the interesting special case of sums of the form f,g P d χ ((fg) ), where P d denotes the set of monic irreducible polynomials of degree d that are relatively prime to M For these sums, our techniques provide a much stronger estimate; see Theorem 5 We remark that the analogous estimate for integers has been used to improve the Brun-Titchmarsh theorem Accordingly, we hope that our estimate can be used to improve the function field analogue of the Brun-Titchmarsh theorem as given in [3] Our methods are essentially those of Karatsuba [5] (see also [2, 4]), which we have extended to work over the polynomial ring F q [x] However, several of the underlying results have been unknown for polynomials, and we have had to establish them in the current paper (in fact, our results for polynomials exhibit some new effects that do not occur in the case of integers) Some of these fundamental results may be of independent interest and are likely to find several other applications; for example, see Lemma 2 Finally, we remark that several uniformity of distribution results on the inverses of polynomials from small sets have recently been obtained in [1] by a different method The first author would like to thank Macquarie University for its hospitality Work supported in part by NSF grant DMS-0070628 (W Banks) and by ARC grant A69700294 (I Shparlinski) 4

2 Notation Throughout the paper, k and l denote positive integers, while d and e are nonnegative real numbers Let q be a fixed prime power, and let F q be the finite field with q elements Put R = F q [x], R = F q [x] {0} Given f, g R, we write f g whenever f = ag for some a F q Then the set of equivalence classes in R / can be naturally identified with the set M of monic polynomials in R We denote the greatest common divisor of f 1,, f k R by gcd(f 1,,f k ); by definition, it is the element h M of greatest degree such that h divides f j, j = 1,,k Similarly, the least common multiple will be denoted by lcm[f 1,,f k ]; it is the element h M of least degree such that f j divides h, j = 1,,k For every d 0, let M(d) be the set of monic polynomials f M of degree deg(f) d 3 Preliminary Results For every f R and k 1, let τ k (f) be the number of ordered k-tuples (f 1,,f k ) M k such that f f 1 f k Observe that τ k (f) = τ k (g) whenever f g Lemma 1 For all f, g R and k 1, we have τ k (fg) τ k (f)τ k (g) If gcd(f, g) = 1, then τ k (fg) = τ k (f)τ k (g) Proof : For any f R, let T k (f) M k be the collection of ordered k-tuples defined by T k (f) = {(f 1,,f k ) M k f f 1 f k } By definition, τ k (f) is the cardinality of T k (f) Consider the natural map T k (f) T k (g) T k (fg) given by ( (f1,,f k ), (g 1,,g k ) ) (f 1 g 1,,f k g k ) It can easily be verified that this map is a bijection if gcd(f, g) = 1, hence we obtain the second statement of the lemma 5

If p M is irreducible and α 0 is any integer, one clearly has ( ) α + k 1 τ k (p α ) = k 1 From this it follows that τ k (p α+β ) τ k (p α )τ k (p β ) for all α, β 0 Now for arbitrary f, g R, let p 1,, p r M be the complete set of irreducible polynomials that occur in the factorization of the product fg Then f p α 1 1 pαr r, g pβ 1 1 pβr r, for some uniquely determined integers α j, β j 0, j = 1,, r, so by our previous results, it follows that τ k (fg) = r j=1 τ k ( p α j +β j j ) r j=1 τ k ( p α j j ) ( β τk p j ) j = τk (f)τ k (g) This completes the proof Lemma 2 For all k, l 1 and d 0, we have f M(d) If l = 1, then (1) holds with equality ( d + k τ k (f) l q deg(f) k ) k l 1 (1) Proof : Let l = 1 be fixed for the moment Since τ 1 (f) = 1 for all f R, and d d q deg(f) = q j = 1 = d + 1, f M(d) j=0 f M deg(f)=j we see that (1) holds with equality for all d 0 when k = 1 Proceeding inductively, we now suppose that (1) holds with equality up to k 1, where k 2 Since τ k (f) = τ k 1(f 2 ), f 1,f 2 M f f 1 f 2 j=0 6

we therefore have τ k (f)q deg(f) = f M(d) = = = f M(d) f 1,f 2 M f 1 M(d) f 1 M(d) d j=0 Hence the lemma is proved when l = 1 f f 1 f 2 τ k 1(f 2 )q q deg(f 1) q deg(f 1) deg(f 1 f 2 ) f 2 M(d deg(f 1 )) τ k 1 (f 2 )q deg(f 2) ( ) d deg(f1 ) + k 1 k 1 ( ) ( d j + k 1 d + k = k 1 k Now suppose that the inequality (1) holds up to l 1, l 2, for all k 1 and d 0 Using Lemma 1, it follows that τ k (f) l q deg(f) = τ k (f 1 f k ) l 1 q deg(f 1f k ) f f 1 f k f M(d) f M(d) f 1,,f k M This completes the proof = f 1,,f k M j=1 deg(f 1 f k ) d f 1,,f k M(d) j=1 f M(d) Using Lemma 2, we obtain the estimate f M(d) k τ k (f j ) l 1 q deg(f j) k τ k (f j ) l 1 q deg(f j) τ k (f) l 1 q deg(f) ( ( d ) ) k l 2 k ( + k d + k = k k τ k (f) l q (α 1) deg(f) q α d ( d + k k 7 k ) k l 1 ) k l 1 ), (2)

which is valid for all k, l 1, d 0, and any real number α 0 This will be used to prove the following: Lemma 3 For all k 1 and d 0, let J (k, d) be the number of ordered k-tuples (f 1,,f k ) M k such that and Then the following estimate holds: deg(f 1 f k ) d, f 1 f k 0 (mod lcm[f 2 1,,f2 k ]) J (k, d) q d/2 ( d/2 + k k ) k ( d/3 + k Proof : For any f M, let λ k (f) be the number of ordered k-tuples (f 1,,f k ) M k such that f = f 1 f k and k ) k 2 f 1 f k 0 (mod lcm[f 2 1,,f 2 k]) Clearly, we have J (k, d) = f M(d) If f j, g j M and gcd(f j, g j ) = 1 for j = 1,, k, then λ k (f) (3) lcm[f 2 1,,f 2 k] lcm[g 2 1,,g 2 k] = lcm[(f 1 g 1 ) 2,,(f k g k ) 2 ]; from this it follows that λ k is multiplicative, ie, that λ k (fg) = λ k (f)λ k (g) whenever gcd(f, g) = 1 Thus, if f M and f = p α 1 1 pαr r is a factorization into positive powers of pairwise-distinct monic irreducibles, then λ k (f) = λ k (p α 1 1 ) λ k (p αr r ) Since it is also clear that λ k (p) = 0 for any irreducible p M, every nonzero term in (3) arises from a polynomial f of the form f = p α 1 1 p αr r, α 1,,α r 2, 8

which implies that f = g 2 h 3 for some g, h M Since λ k (f) τ k (f), we have J (k, d) λ k (g 2 h 3 ) τ k (g 2 h 3 ) g,h M deg(g 2 h 3 ) d By Lemma 1, it follows that J (k, d) g M(d/2) τ k (g) 2 g,h M deg(g 2 h 3 ) d τ k (h) 3 h M((d 2deg(g))/3) Applying the estimate (2) with l = 3 and α = 1, we see that h M((d 2 deg(g))/3) ( (d 2 deg(g))/3 + k τ k (h) 3 q (d 2 deg(g))/3 k ( ) k 2 d/3 + k q 2deg(g)/3 q d/3 k Applying (2) again with l = 2 and α = 1/3, we have ( d/2 + k τ k (g) 2 q 2deg(g)/3 q d/2 /3 k ( ) k d/2 + k q d/6 k g M(d/2) ) k ) k 2 The lemma follows 4 Estimation of Character Sums Throughout this section, we assume that M R is a fixed polynomial of degree deg(m) = m > 0 Let R M be the quotient ring R/(M), let R M be the multiplicative group of R M, and let R M = {f R deg(f) < m and gcd(f, M) = 1} We note that the canonical surjection R R M gives rise to a bijection R M For any f R such that gcd(f, M) = 1, we denote by f the R M 9

unique polynomial in R M such that ff 1 (mod M) In particular, f is the inverse of f if we regard both polynomials as elements of R M For a real number d such that 0 d < m, let R(d) [resp R M (d)] denote the set of polynomials f R [resp f R M ] of degree deg(f) d Lemma 4 Suppose that k 1, d 0, and (2k 1) d < m Let I(k, d) be the number of ordered 2k-tuples (f 1,,f 2k ) R M (d)2k such that Then f 1 + + f k f k+1 + + f 2k (mod M) (4) where J is defined as in Lemma 3 I(k, d) (q 1) 2k J (2k, 2kd), Proof : Suppose that f 1,,f 2k are elements of R M (d) that satisfy (4) Multiplying both sides of (4) by the product f 1 f 2k and using the fact that f j fj 1 (mod M), we obtain g 1 + + g k g k+1 + + g 2k (mod M), where each g j is defined by the relation f j g j = f 1 f 2k Now since we have deg(g j ) (2k 1) d < m for each j = 1,, 2k, this congruence becomes an equality g 1 + + g k = g k+1 + + g 2k By definition, f j divides g l whenever l j, so this equality implies that f j divides g j as well Consequently f 1 f 2k = f j g j 0 (mod f 2 j ), and therefore f 1 f 2k 0 (mod lcm[f1 2,,f2 2k ]) Since deg(f 1 f 2k ) 2kd, the result follows An additive character of R M is a homomorphism χ : R M C For the sake of convenience in what follows, we will also denote by χ the corresponding homomorphism R C which is trivial on the principal ideal 10

(M), obtained by composing χ : R M C with the canonical surjection R R M For any additive character χ of R M, let Ω χ = {α R χ(αβ) = 1 for all β R} Then Ω χ is an ideal in R; since R is a principal ideal domain, it follows that Ω χ is the ideal generated by a (unique) monic polynomial f χ M Since M Ω χ, f χ is a divisor of M If χ is the trivial character, then f χ = 1 On the other hand, if f χ M, then χ is said to be primitive Theorem 3 Suppose that k, l 1, d, e 0, and (2k 1) d < m, (2l 1) e < m Let F and G be arbitrary subsets of R M (d) and R M (e), respectively Then for any primitive character χ of R M and any element a R, the character sum S = χ ( (fg) + afg ) f F g G satisfies the bound S F G, where = ( F 2k G 2l q m+min(d,e)+1 (q 1) 2k+2l J (2k, 2kd)J (2l, 2le) ) 1/2kl, and J is defined as in Lemma 3 Proof : By Hölder s inequality and the fact that (fg) f g (mod M), we have S l F l l 1 χ(f g + afg) f F g G = F l 1 σ l (β, δ)χ(f β + afδ) f F, β R M δ R(e) where σ l (β, δ) denotes the number of ordered l-tuples (g 1,, g l ) in G l such that g 1 + + g l β (mod M), g 1 + + g l δ (mod M) 11

Now for each f F, let arg f denote the argument of the double summation inside the absolute value in the preceding inequality Then S l F l 1 σ l (β, δ) e i arg f χ(f β + afδ) β R M δ R(e) Raising both sides of this inequality to the power k and applying Hölder s inequality once more, we obtain S kl F (l 1)k k 1 σ l (β, δ) β R M δ R(e) β R M δ R(e) f F σ l (β, δ) e i arg f χ(f β + afδ) Applying Cauchy s inequality to the last part of this expression, we therefore see that S kl F (l 1)k (L 1 ) k 1 (L 2 ) 1/2 (L 3 ) 1/2, (5) where L 1 = L 2 = L 3 = β R M δ R(e) β R M δ R(e) f F σ l (β, δ), β R M δ R(e) f F σ l (β, δ) 2, e i arg f χ(f β + afδ) 2k k The first sum L 1 is equal to the total number of ordered l-tuples (g 1,,g l ) G l : L 1 = G l (6) The second sum L 2 is equal to the number of ordered 2l-tuples (g 1,,g 2l ) G 2l such that g 1 + + g l g l+1 + + g 2l (mod M), g 1 + + g l g l+1 + + g 2l (mod M) 12

Since (2l 1) e < m by hypothesis, we can use Lemma 4 to bound L 2, and we obtain L 2 (q 1) 2l J (2l, 2le) (7) For the third sum L 3, we have L 3 = β R M δ R(e) f 1,,f 2k F e i(arg f 1++arg f k arg f k+1 arg f 2k ) χ((f1 + f2k)β + a(f 1 + f 2k )δ) χ((f1 f 1,,f 2k F + f 2k )β + a(f 1+ f 2k )δ) β R M δ R(e) σ k (α, γ) χ(αβ + aγδ) α R M γ R(d) β R M δ R(e) = σ k (α, γ) χ(αβ) χ(aγδ) β R M, α R M γ R(d) δ R(e) where σ k (α, γ) is the number of ordered 2k-tuples (f 1,,f 2k ) F 2k that satisfy f 1 + + f k α + f k+1 + + f 2k (mod M), (8) and f 1 + + f k γ + f k+1 + + f 2k Now since χ is a primitive character, the sum { q m if α = 0, χ(αβ) = 0 otherwise; β R M (mod M) (9) thus L 3 q m γ R(d) σ k (0, γ) δ R(e) χ(aγδ) qm+e+1 γ R(d) σ k (0, γ) since R(e) = q e+1 As the sum γ R(d) σ k (0, γ) 13

counts the total number of solutions to (8) with α = 0, and (2k 1) d < m by hypothesis, we have by Lemma 4: L 3 q m+e+1 (q 1) 2k J (2k, 2kd) (10) Substituting the estimates (6), (7) and (10) into in (5), we obtain the bound stated in the theorem except that we now have q m+e+1 instead of the term q m+min(d,e)+1 The correct bound follows by symmetry When M divides a, we can improve the bound stated in Theorem 3 Theorem 4 Using the notation of Theorem 3, the character sum S = f F g G satisfies the bound S F G, where χ ( (fg) ) = ( F 2k G 2l q m (q 1) 2k+2l J (2k, 2kd)J (2l, 2le) ) 1/2kl Proof : By Hölder s inequality, we have S l F l 1 l χ(f g ) = F l 1 σ l (β)χ(f β), f F g G f F β R M where σ l (β) denotes the number of ordered l-tuples (g 1,,g l ) in G l such that g1 + + g l β (mod M) For each f F, let arg f denote the argument of the summation inside the absolute value in the preceding inequality Then S l F l 1 σ l (β) e iarg f χ(f β) β R M f F Raising both sides of this inequality to the power k and applying Hölder s inequality once more, we obtain ( ) k 1 k S kl F (l 1)k σ l (β) σ l (β) e i arg f χ(f β) β R M β R M f F 14

Applying Cauchy s inequality, we see that S kl F (l 1)k (L 1 ) k 1 (L 2 ) 1/2 (L 3 ) 1/2, where L 1 = β R M σ l (β), L 2 = β R M σ l (β) 2, L 3 = β R M e i arg f χ(f β) f F 2k The sums L 1 and L 2 can be estimated as in Theorem 3 For the third sum, we have L 3 = e i(arg f 1+ arg f 2k ) χ((f1 + f 2k )β) β R M f 1,,f 2k F χ((f1 + f 2k )β) f 1,,f 2k F β R M σ k (α) χ(αβ), α R M β R M where σ k (α) is the number of ordered 2k-tuples (f 1,, f 2k ) F 2k that satisfy f1 + + f k α + f k+1 + + f 2k (mod M) Using (9) and Lemma 4, we have L 3 q m σ k (0) q m (q 1) 2k J (2k, 2kd) The result follows Theorem 5 Suppose that (2k 1) d < m Then for any primitive character χ of R M, the character sum S = f,g P d χ ((fg) ) 15

satisfies the bound S (k!) 1/k2 P d 2 1/k q m/2k2 Proof : From the Hölder inequality, we obtain S k P d k 1 χ ((fg) ) f P d g P d = P d k 1 ϑ f f P d k g 1,,g k P d χ (f (g 1 + + g k)), where ϑ f is such that ϑ f = 1 Denoting by T k (ψ) the number of solutions of the congruence g 1 + + g k ψ (mod M), g 1,,g k P d, we derive that S k P d k 1 T k (ψ) ϑ f χ (ψf ) ψ R M f P d Applying the Hölder inequality again, we have ( ) 2k 2 S 2k2 P d 2k2 2k T k (ψ) T k (ψ) 2 ψ R M ψ R M ψ R M ϑ f χ (ψf ) f P d Let W(k, d) denote the number of solutions of the congruence f1 + + f k f k+1 + + f 2k (mod M), f 1,, f 2k P d (11) 2k Now, we have ψ R M T k (ψ) = P d k and ψ R M T k (ψ) 2 = W(k, d) 16

Consequently S 2k2 P d 4k2 4k W(k, d) ψ R M f 1,,f 2k P d χ ( ψ ( f 1 + + f k f k+1 f 2k P d 4k2 4k W(k, d) Applying (9), we see that f 1,,f 2k P d )) k ν=1 ϑ fν χ ( ψ ( f1 + + f k f k+1 f )) 2k ψ R M S 2k2 P d 4k2 4k q n W(k, d) 2 2k ν=k+1 To estimate W(k, d), we remark that (11) is equivalent to the congruence k ν=1 2k i=1 i ν f i 2k 2k f i ν=k+1 i=1 i ν (mod M) Since the degrees of the polynomials on the both sides of this congruence are at most (2k 1)d < n, this congruence yields an equality over F q [X]: Hence, k ν=1 2k i=1 i ν f i = 2k 2k ν=k+1 i=1 i ν f i f 1 + + f k = f k+1 + + f 2k Recalling that the polynomials f 1,,f 2k are irreducible and comparing the denominators of the expressions on both sides of this equation, we see that equality is possible if and only if Therefore and the result follows {f 1,,f k } = {f k+1,,f 2k } W(k, d) k! P d k, 17 ϑ fν

5 Results on Uniform Distribution Throughout this section, let M R be a fixed irreducible polynomial of degree deg(m) = m > 0 Put R m = {f R deg(f) < m}, R m = {f R m f 0}, and for any real number d with 0 d < m, let R (d) = {f R deg(f) d} Note that R m = R M and R (d) = R M (d) in our previous notation, since gcd(f, M) = 1 for all f R m As in the previous section, for each f R m, let f be the unique polynomial in R m such that ff 1 (mod M) Then f is an inverse for f in the multiplicative group R M Since M is irreducible, R M = R/(M) is a field; consequently, an additive character χ of R M is primitive if and only if it is nontrivial Lemma 5 Let k and d be positive integers such that m d =, 2k δ where 0 < δ < 1 Then for every nontrivial character χ of R M, the character sum S = χ ( (fg) ) f,g R (d) satisfies the bound S R (d) 2 exp( ), where δm log q = 2k 2 (2k δ) + log q k + 12k log m Proof : Set e = d, l = k, and F = G = R (d) Since (2k 1)d (2k 1) (2k δ) m < m, we see that all of the conditions of Theorem 4 hold; thus S R (d) 2, 18

where ( ) 2k2 = R (d) 4k q m (q 1) 4k J (2k, 2kd) 2 Since R (d) = q d+1 1, we have by Lemma 3: First, we estimate ( ) 4k q 1 ( ) 2k2 = q m J (2k, 2kd) 2 q d+1 1 q m 4kd J (2k, 2kd) 2 ( ) 4k ( kd + 2k 2kd/3 + 2k q m 2kd 2k 2k m 2kd < m 2k ) 8k 2 ( ) m 2k δ 1 = 2k δm 2k δ Next, since k 1, we have kd (2k 1)d < m, hence kd + 1 m Consequently, ( ) kd + 2k (kd + 1) 2k m 2k 2k Similarly, and the result follows ( ) 2kd/3 + 2k m 2k, 2k Recall that for a set E of nonnegative integers and two polynomials f(x) = j 0 a j x j, g(x) = j 0 b j x j, we write f E g to indicate that a j = b j for all j E Then E defines an equivalence relation on R Theorem 6 Let k and d be positive integers such that m d =, 2k δ 19

where 0 < δ < 1 Fix an arbitrary subset E {0, 1,, m 1} of cardinality E = n and a polynomial F R, and let N be the number of ordered pairs (f, g) in R (d) 2 such that (fg) E F Then N R (d) 2 < R (d) 2 exp( ), where is defined as in Lemma 5 In particular, if then n q n δm 2k 2 (2k δ) 1 k 12k log m, log q 0 < N < 2 R (d) 2 q n Proof : Without loss of generality, we can assume that deg(f) < m Let X E be the set of polynomials in R m whose coefficients vanish on E; that is, X E = {f R m f(x) = j E a j x j } Note that X E is an additive subgroup of R: X E + X E = X E Let Q be the number of representations of the form F = (fg) + φ ψ, where f, g R (d) and φ, ψ X E Since (fg) E F if and only if F (fg) lies in X E, and X E = q m n, we have Now Q = 1 f,g R (d) φ,ψ X E Q = q m n N χ ( (fg) F φ + ψ ), q m χ = 1 χ(f) χ(ψ φ) χ ( (fg) ) q m χ φ,ψ X E f,g R (d) = 1 2 χ(f) χ(φ) χ ( (fg) ) q m χ φ X E f,g R (d) = R (d) 2 q m 2n + 1 2 χ(f) χ(φ) q m φ X E χ 1 20 f,g R (d) χ ( (fg) )

By Lemma 5, we have Q R (d) 2 q m 2n 1 χ(φ) χ q m 2 ( (fg) ) χ 1 φ X E f,g R (d) R (d) 2 2 exp( ) χ(φ) q m φ X E Using the estimate χ(φ) χ 1 φ X E we have The result follows 2 = q 2m 2n + χ χ 1 φ,ψ X E χ(ψ φ) = q 2m n q 2m 2n, Q R (d) 2 q m 2n < R (d) 2 q m n exp( ) Using Theorem 6, we can now give a proof of Theorem 1 as stated in the introduction Proof : Put λ = (1/2) 1/7 < 1, and consider the collection D of integers d in the interval λm 2/3+ǫ log m d m 2/3+ǫ log m For every d D, we have m 1/3 ǫ log m m d m1/3 ǫ λ log m If m ǫ 1, the closed interval [ m 1/3 ǫ / log m, m 1/3 ǫ /(λ log m) ] has length (λ 1 1) m1/3 ǫ log m > 2 + (1 λ) On the other hand, if d and d + 1 both lie in D, then m d m d + 1 < m d 1 < (1 λ) 2 λ 2 m 1/3+2ǫ (log m) 2 provided that m ǫ 1 Consequently, for some d D, there exists an integer k such that m/d lies in the open interval ( 2k 1, 2k 1 + (1 λ) ) Let k 21

and d be fixed with these properties, and set δ = 2k m/d Then we have λ < δ < 1, and k = m 2d + δ 2 > 0, hence all of the conditions of Theorem 6 are satisfied Applying the theorem, we see that N > 0 provided that E Now for all m ǫ 1, we have thus that is, δm 2k 2 (2k δ) > Since 1/k 1, and δm 2k 2 (2k δ) 1 k k < m 2d + 1 2 m1/3 ǫ 2λ log m + 1 2 < m1/3 ǫ 2λ 2 log m, λm 2k 2 (m/d) = λd > 2k 2 δm 2k 2 (2k δ) > 2λ6 m 3ǫ (log m) 3 12k log m log q 12k log m (12) log q > 6m1/3 ǫ λ 2 log q, it follows that the right side of (12) is bounded below by and this is bounded below by provided that The theorem follows 2λ 6 m 3ǫ (log m) 3 6m1/3 ǫ λ 2 log q 1, 2λ 7 m 3ǫ (log m) 3 = m 3ǫ (log m) 3 log q > 6m 1/3 ǫ (λ λ 2 )m 3ǫ (log m) 3 λ 2 λ 2 m 2/3+ǫ log m 2 (m 1/3 ǫ /(2λ 2 log m)) 2; For the rest of this section, we study the distribution in R M of polynomials of the form (fg) + afg, where a is a fixed element of R, and f and g run through the sets R (d) and R (e), respectively 22

Lemma 6 Let k, l, d and e be positive integers such that m m d =, e =, 2k δ 2l γ where 0 < δ, γ < 1 Suppose that d e Then for every nontrivial character χ of R M and any polynomial a R, the character sum S = χ ( (fg) + afg ) f R (d) g R (e) satisfies the bound S R (d) R (e) exp( ), where ( = δm 4k 2δ γm ) log q 4l 2γ + k + l + d + 1 2kl + (6k3 + 6l 3 ) log m kl Proof : Set F = R (d) and G = R (e) Since (2k 1)d (2k 1) (2l 1) m < m, (2l 1)e (2k δ) (2l γ) m < m, all of the conditions of Theorem 3 hold; thus where S R (d) R (e), ( ) 2kl = R (d) 2k R (e) 2l q m+d+1 (q 1) 2k+2l J (2k, 2kd)J (2l, 2le) The lemma now follows as in the proof of Lemma 5 For any f R, we denote by {f} the unique element of R m such that f {f} (mod M) Theorem 7 Let k, l, d and e be positive integers such that m m d =, e =, 2k δ 2l γ where 0 < δ, γ < 1 Suppose that d e Fix a subset E {0, 1,, m 1} of cardinality E = n and two polynomials F, a R, and let N be the 23

number of ordered pairs (f, g), with f R (d) and g R (e), such that { (fg) + afg } E F Then N R (d) R (e) < R (d) R (e) exp( ), where is defined as in Lemma 6 In particular, if n δm 4kl(2k δ) + q n γm 4kl(2l γ) k + l + d + 1 2kl (6k3 + 6l 3 ) log m, kl log q then 0 < N < 2 R (d) R (e) q n Proof : Using Theorem 6, the proof is very similar to the proof of Theorem 6; details are left to the reader Using Theorem 7, we can now give a proof of Theorem 2 Proof : Put λ = (1/2) 1/9 < 1, and consider the collection D of pairs of integers (d, e) such that For all such pairs, we have λm 2/3+ǫ log m 2d λ e m2/3+ǫ log m 2m 1/3 ǫ λ log m m d 2m1/3 ǫ λ 2 log m, m 1/3 ǫ log m m e m1/3 ǫ λ log m If m ǫ 1, the closed intervals [ 2m 1/3 ǫ /(λ log m), 2m 1/3 ǫ /(λ 2 log m) ] and [ m 1/3 ǫ / log m, m 1/3 ǫ /(λ log m) ] have lengths greater than 2 + (1 λ) On the other hand, if (d, e) and (d + 1, e + 1) lie in D, then m d m d + 1 < m d 4 < (1 λ), 2 λ 4 m 1/3+2ǫ (log m) 2 m e m e + 1 < m e 1 < (1 λ), 2 λ 2 m 1/3+2ǫ (log m) 2 provided that m ǫ 1 Consequently, for some (d, e) D, there exist integers k and l such that m/d lies in the open interval ( 2k 1, 2k 1 + (1 λ) ), 24

and m/e lies in the open interval ( 2l 1, 2l 1 + (1 λ) ) Let k, l, d and e be fixed with these properties, and set δ = 2k m/d, γ = 2l m/e Then λ < δ, γ < 1, and k = m 2d + δ 2 > 0, l > m 2e + γ 2 > 0, thus all of the conditions of Theorem 7 are satisfied Applying the theorem, we see that N > 0 if E is less than or equal to Since δm 4kl(2k δ) + γm 4kl(2l γ) k + l + d + 1 2kl it follows that N > 0 provided that E Now for m ǫ 1, we have γm = γe > λe 2d, (2l γ) δm 4kl(2k δ) k + l + 1 2kl k < m 2d + 1 2 m1/3 ǫ λ 2 log m + 1 2 < m1/3 ǫ λ 3 log m, (6k3 + 6l 3 ) log m kl log q (6k3 + 6l 3 ) log m (13) kl log q and Consequently, l < m 2e + 1 2 m1/3 ǫ 2λ log m + 1 2 < m1/3 ǫ 2λ 2 log m δm 4kl(2k δ) = δd 4kl > (λ 3 m 2/3+ǫ log m)/2 4 ( m 1/3 ǫ /(λ 3 log m) )( m 1/3 ǫ /(2λ 2 log m) ); that is, We also have δm 2k 2 (2k δ) > λ8 m 3ǫ (log m) 3 4 k + l + 1 2kl > 3 2 25

Finally, since it follows that kl > m 2d m 2e m 2 (λm 2/3+ǫ log m)(2m 2/3+ǫ log m) = (6k3 + 6l 3 ) log m kl log q Thus the right side of (13) is bounded below by and this is bounded below by provided that The theorem follows m2/3 2ǫ 2λ(log m) 2, ( 3 > λ + 3 ) m 1/3 ǫ 8m1/3 ǫ 10 8λ 7 log q > log q λ 8 m 3ǫ (log m) 3 4 λ 9 m 3ǫ (log m) 3 log q > 4 8m1/3 ǫ log q 3 2, = m3ǫ (log m) 3 8 32m 1/3 ǫ (λ 8 λ 9 )m 3ǫ (log m) 3 6 References [1] W Banks and I E Shparlinski, Distribution of inverses in polynomial rings, Indag Math, 12 (2001), 303 315 [2] J Friedlander and H Iwaniec, The Brun Titchmarsh theorem, Analytic Number Theory, Lond Math Soc Lecture Note Series 247, 1997, 363 372 [3] C-N Hsu, The Brun-Titchmarsh theorem in function fields, J Number Theory, 79 (1999), 67 82 [4] A A Karatsuba, Fractional parts of functions of a special form, Izv Ross Akad Nauk Ser Mat (Transl as Russian Acad Sci Izv Math), 55(4) (1995), 61 80 (in Russian) [5] A A Karatsuba, Analogues of Kloosterman sums, (Russian) Izv Ross Akad Nauk Ser Mat 59 (1995), no 5, 93 102 26