ON INTEGER VALUES OF KLOOSTERMAN ... - Semantic Scholar
ON INTEGER VALUES OF KLOOSTERMAN SUMS KEIJO KONONEN, MARKO RINTA-AHO, AND KEIJO VÄÄNÄNEN Abstract. In this note we use Carlitz’s lifting formula for Kloosterman sums and some results from the theory of cyclotomic fields to show that a Kloosterman sum over a finite field obtains a rational integer value at a fixed point if and only if the same sum lifted to any extension field remains a rational integer.
1. Introduction Let q be a power of a prime p and n a positive integer. We denote by Fq the finite field of q elements and by Tr the absolute trace function Fq → Fp . Also, let ζp := e2πi/p be the complex primitive pth root of unity. The Kloosterman sum at a point a ∈ Fq is defined by the equation X Kq (a) = χ(x + ax−1 ), x∈F∗q
where χ : Fq → Q(ζp ) is the canonical additive character of the field Tr(x) Fq defined by χ(x) = ζp . Obviously, Kq (a) is an algebraic integer of the pth cyclotomic field Q(ζp ). For every index i ∈ (Z/pZ)∗ we let σi denote the unique Qautomorphism of the field Q(ζp ) mapping ζp to ζpi . Put λp := 1 − ζp and L := hλp i (the prime ideal generated by λp ). It is well known that (1 − ζpi )/λp is a unit of the ring Z[ζp ] for every i ∈ (Z/pZ)∗ and that hpi = L p−1 . One easily sees that σi (Kq (a)) = Kq (i2 a). Consequently the Kloosterman sum values are always real, since the complex conjugate Kq (a) = σ−1 (Kq (a)) = Kq (a). Thus the Kloosterman sum Kq (a) is an algebraic integer of the maximal real subfield Q(ζp + ζp−1 ) of Q(ζp ). Moreover, we have Kq (a) ∈ Z if and only if Kq (i2 a) = Kq (a) for every i ∈ (Z/pZ)∗ . Especially, Kq (a) ∈ Z always in the cases p = 2, 3. Let Fqn be an extension field of Fq . Carlitz has proved (see [Car69] or [LN97, pp. 226-229]) that the Kloosterman sum Kq (a) and the lifted Kloosterman sum Kqn (a) satisfy (1)
Kqn (a) = (−1)n−1 (A1 (a)n + A2 (a)n )
for every a ∈ Fq , where
√ Aj (a) := 21 (Kq (a) + (−1)j D)
Key words and phrases. Kloosterman sums, Cyclotomic fields. 1
2
KEIJO KONONEN, MARKO RINTA-AHO, AND KEIJO VÄÄNÄNEN
are the roots and D := Kq (a)2 − 4q is the discriminant of the equation x2 − Kq (a)x + q = 0. Alternatively Kqn (a) = (−1)n−1 Dn (Kq (a), q), where Dn is the Dickson polynomial (of the first kind) defined by bn c 2 X n−i n Dn (x, q) = (−q)i xn−2i . n − i i i=0 Since Dickson polynomials have got integer coefficients, it follows that Kqn (a) ∈ Z whenever Kq (a) ∈ Z. Here we investigate whether the converse is true. That is, we want to find out if (2)
Kqn (a) ∈ Z =⇒ Kq (a) ∈ Z.
It turns out that the answer is positive and this is our main result. Theorem 1. Assume that Fq is a subfield of Fqn and that a ∈ Fq . Then Kqn (a) ∈ Z if and only if Kq (a) ∈ Z. 2. Proof of Theorem 1 Recently Marko Moisio [Moi09, Lemma 12] proved the following fact about the minimal polynomials of Kloosterman sums which also turns out to be useful for our purposes. Lemma 2. Let m(x) ∈ Z[x] denote the minimal polynomial of the Kloosterman sum Kq (a) ∈ Z[ζp ] over Q. Then m(x) ≡ (x + 1)t (mod p), where t is the degree of m(x). In particular Kq (a) ≡ −1 (mod p), if Kq (a) ∈ Z. Since every proper subfield of a finite field is contained in at least one maximal subfield, it is enough to consider (2) in the case where Fq is a maximal subfield of Fqn . Then n will obviously be a prime. We already know that the Kloosterman sum is always an integer in the cases p = 2, 3. In the case p > 3 we have the following result. Lemma 3. In the case p > 3, n a prime and n 6= p the statement (2) is true for all a ∈ Fq . Proof. Put b := Kqn (a) ∈ Z. Then Kq (a) will be a zero of the polynomial fn (x) := Dn (x, q) + (−1)n b ∈ Z[x]. Let m(x) ∈ Z[x] denote the minimal polynomial of the algebraic integer Kq (a). Obviously m(x)|fn (x). Let t be the degree of the polynomial m(x). Now fn (x) ≡ xn + (−1)n−1 (mod p), since Kqn (a) ≡ −1 (mod p). It follows from Lemma 2 that (x + 1)t |(xn + (−1)n−1 ) in the ring Zp [x]. If now Kq (a) 6∈ Z, then the polynomial xn + (−1)n−1 has −1 as a multiple zero. But since the formal derivative of this polynomial is nxn−1 and n is a prime, we must have n = p. We shall attack the case n = p by studying the conjugates of the Kloosterman sums Kq (a) and Kqp (a).
ON INTEGER VALUES OF KLOOSTERMAN SUMS
3
Lemma 4. Let a ∈ Fq and assume that Kqp (a) ∈ Z, then every algebraic conjugate of the Kloosterman sum Kq (a) is of the form √ (3) Kq (i2 a) = 21 (ζpk + ζp−k )Kq (a) + 21 (ζpk − ζp−k ) D for a suitable k ∈ Z/pZ. √ Proof. Put F := Q(ζp , D). Then every automorphism σi of the cyclotomic field Q(ζp ) can be extended to a monomorphism σ ˜i : F → C (possibly in two different ways). It follows from (1) that x2 − (−1)p−1 Kqp (a)x + q p = (x − A1 (a)p ) (x − A2 (a)p ) . Since the left-hand side polynomial has got coefficients in Z, σ ˜i permutes the roots of this polynomial. Thus we either have σ ˜i (Aj (a)p ) = Aj (a)p or σ ˜i (Aj (a)p ) = Aj 0 (a)p for j = 1, 2, where j 0 ∈ {1, 2}, j 0 6= j. In the first case σ ˜i (A1 (a)) = ζpl A1 (a), σ ˜i (A2 (a)) = ζpk A2 (a) and in the second σ ˜i (A1 (a)) = ζpk A2 (a), σ ˜i (A2 (a)) = ζpl A1 (a) for suitable k, l ∈ Z/pZ. Moreover, A1 (a)A2 (a) = q, so we must have l = −k. In both of the above cases we get Kq (i2 a) = σi (Kq (a)) = σ ˜i (Kq (a)) = σ ˜i (A1 (a)) + σ ˜i (A2 (a)) = ζp−k A1 (a) + ζpk A2 (a), which together with the definition of the Aj (a) gives (3).
After these preparations we are ready to prove our main result. We already know that Kqn (a) ∈ Z whenever Kq (a) ∈ Z. Lemma 3 and the comments before it also tell us that it is enough to prove the converse direction only in the case n = p. So assume that Kqp (a) ∈ Z. We will show that Kq (a) ∈ Z by proving that Kq (i2 a) = Kq (a) for every i ∈ (Z/pZ)∗ . This is equivalent to showing that √ we can choose k = 0 in the equation (3) for every i. In the case D 6∈ Q(ζp ) this is clear, since all the other √ terms in (3) belong to the cyclotomic field. In the case D ∈ Q(ζp ) our idea is to look at the equation (3) modulo L 2 in the ring Z[ζp ]. Let m(x) ∈ Z[x] be the minimal polynomial of the Kloosterman sum Kq (a). By Lemma 2 m(x) ≡ (x + 1)t (mod p), where t is the degree of the m(x). Thus (Kq (a) + 1)t ≡ 0 (mod L p−1 ) in the ring Z[ζp ]. Because the Kloosterman sum is an element of the maximal real subfield of the cyclotomic field, t ≤ p−1 . Consequently 2 2 Kq (a) ≡ −1 (mod L ). Since 1 − ζpk is either 0 or differs from λp by a unit factor, ζpk + ζp−k = −k ζp [(1 − ζpk )2 + 2ζpk ] ≡ 2 (mod L 2 ). Similarly ζpk − ζp−k = ζp−k [(1 − ζpk )2 + 2(ζpk − 1)] ≡ 2(1 − ζp−k ) (mod L 2 ). Combining these results with (3) gives √ −1 ≡ 21 · 2 · (−1) + 21 · 2(1 − ζp−k ) D (mod L 2 ).
4
KEIJO KONONEN, MARKO RINTA-AHO, AND KEIJO VÄÄNÄNEN
√
It follows that L 2 |(1 − ζp−k ) D. If now k 6= 0, 1 − ζp−k = L and √ we must have L | D and thus also L 2 |D. On the other hand D = Kq (a)2 − 4q ≡ (−1)2 − 4 · 0 = 1 (mod L 2 ), which is a contradiction. Thus we must have k = 0, which completes the proof. 3. Further observations Before finishing we are going to make some more remarks about the points and fields where Kq (a) can obtain integer values. Firstly, by using Theorem 1 one can very easily prove that a cannot lie in the prime subfield. Theorem 5. If p > 3 and a ∈ F∗p , then Kq (a) 6∈ Z. Proof. By Theorem 1 it is enough to prove the result in the case q = p. For every i ∈ Fp let Ni denote the number of x ∈ F∗p such that P P −1 x+ax−1 = i. Then Kp (a) = x∈F∗p ζpx+ax = i∈Fp Ni ζpi . Assume now that Kp (a) ∈ Z. Since the minimal polynomial of ζp is xp−1 +· · ·+x+1, we must have N1 = N2 = · · · = Np−1 and Kp (a) = N0 − N1 . Moreover, Pp−1 i=0 Ni = p − 1 implies that either N0 = 0 and all the other Ni = 1 or N0 = p − 1 and all the other Ni = 0. Thus the mapping f : F∗p → Fp , f (x) = x + ax−1 must be either injective or zero everywhere. Notice that f (ax−1 ) = f (x). Also, we can always find such x ∈ F∗p that ax−1 6= x, if p > 3. Thus f cannot be injective when p > 3. Similarly we can always find such a x ∈ F∗p that f (x) 6= 0 whenever p > 3, because f (x) = 0 is equivalent to x2 = −a. In the case q = p rather more than this is known. Hans Salié [Sal32] has proved that the p−1 Kloosterman sums Kp (a) are divided into two classes according to whether a ∈ F∗p is a square or not and that each . A class consists of the roots of an irreducible polynomial of degree p−1 2 more general result applicable to arbitrary finite fields has later been obtained by Daqing Wan. Wan [Wan95, Theorem 1.1] showed that Kq (a) generates the whole maximal real subfield Q(ζp + ζp−1 ) whenever a ∈ Fq is such that Tr(a) 6= 0. One easy property of the Kloosterman sum is that it is invariant under the action of the Frobenius automorphism x 7→ xp i.e. that Kq (ap ) = Kq (a). This property allows us sometimes quite easily to find points a ∈ F∗q where the Kloosterman sum obtains an integer value. Assume that q = pr and that we have found a ∈ Fq such that for every square i2 ∈ (Z/pZ)∗ (4)
s
i2 a = ap for some s ∈ {0, 1, . . . , r − 1}.
Then all the algebraic conjugates σi (Kq (a)) will be equal and consequently Kq (a) ∈ Z. We illustrate by calculations in F52 . Let α be a primitive element and a = αj . Condition (4) is now equivalent to −a = a5 . This holds if and
ON INTEGER VALUES OF KLOOSTERMAN SUMS
5
2
only if 4j ≡ 5 2−1 (mod 52 − 1). Thus we find four points giving integer values: α3 , α9 , α15 and α21 . Computer calculation confirms that these are the only points a in F∗52 for which Kq (a) ∈ Z. On the other hand in the case of the field F53 we don’t find any points by using criteria (4) although direct calculations show that there are some points giving integer values for the Kloosterman sum. Remarkably enough there exist some fields Fq for which the Kloosterman values are distinct except for the action of the Galois group Gal(Fq /Fp ) generated by the Frobenius automorphism. More precisely s Kq (a) = Kq (b) if and only if b = ap for some s. For example Fisher [Fis92] has proved that this holds for all fields satisfying the bound p > (2 · 4r + 1)2 . The referee of Fisher’s paper conjectured that much stricter bound p ≥ 2r should actually hold. As far as we are aware this conjecture appears to be still largely open. For these kind of fields, for which the kloosterman values are distinct modulo the action by Gal(Fq /Fp ), Kq (a) ∈ Z if and only if (4) holds. Note that (4) can never be satisfied if the number p−1 of squares i2 ∈ Fp exceeds the 2 number r of Frobenius conjugates that is if p > 2r + 1. Thus, if the above-mentioned referee’s conjecture is true, the fields Fpr satisfying p > 2r + 1 don’t have any points a 6= 0 for which Kpr (a) ∈ Z. References [Car69] Leonard Carlitz. Kloosterman sums and finite field extensions. Acta Arithmetica, 16:179–193, 1969. [Fis92] Benji Fisher. Distinctness of Kloosterman Sums. In Contemporary Mathematics: p-Adic Methods in Number Theory and Algebraic Geometry, pages 81–102. American Mathematical Society, 1992. [LN97] Rudolf Lidl and Harald Niederreiter. Finite fields, volume 20 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, second edition, 1997. [Moi09] Marko Moisio. On certain values of Kloosterman sums. IEEE Transactions on Information Theory, 55(8):3563–3564, 2009. [Sal32] Hans Salié. Über die Kloostermanschen Summen S(u, v; q). Mathematische Zeitschrift, 34:91–109, 1932. [Wan95] Daqing Wan. Minimal Polynomials and Distinctness of Kloosterman Sums. Finite Fields and Their Applications, 1(2):189–203, 1995. E-mail address: [email protected] E-mail address: [email protected] E-mail address: [email protected] Department of Mathematical Sciences, University Of Oulu, P.O. BOX 3000, FIN-90014 Oulun yliopisto, Finland
1. Introduction Let q be a power of a prime p and n a positive integer. We denote by Fq the finite field of q elements and by Tr the absolute trace function Fq → Fp . Also, let ζp := e2πi/p be the complex primitive pth root of unity. The Kloosterman sum at a point a ∈ Fq is defined by the equation X Kq (a) = χ(x + ax−1 ), x∈F∗q
where χ : Fq → Q(ζp ) is the canonical additive character of the field Tr(x) Fq defined by χ(x) = ζp . Obviously, Kq (a) is an algebraic integer of the pth cyclotomic field Q(ζp ). For every index i ∈ (Z/pZ)∗ we let σi denote the unique Qautomorphism of the field Q(ζp ) mapping ζp to ζpi . Put λp := 1 − ζp and L := hλp i (the prime ideal generated by λp ). It is well known that (1 − ζpi )/λp is a unit of the ring Z[ζp ] for every i ∈ (Z/pZ)∗ and that hpi = L p−1 . One easily sees that σi (Kq (a)) = Kq (i2 a). Consequently the Kloosterman sum values are always real, since the complex conjugate Kq (a) = σ−1 (Kq (a)) = Kq (a). Thus the Kloosterman sum Kq (a) is an algebraic integer of the maximal real subfield Q(ζp + ζp−1 ) of Q(ζp ). Moreover, we have Kq (a) ∈ Z if and only if Kq (i2 a) = Kq (a) for every i ∈ (Z/pZ)∗ . Especially, Kq (a) ∈ Z always in the cases p = 2, 3. Let Fqn be an extension field of Fq . Carlitz has proved (see [Car69] or [LN97, pp. 226-229]) that the Kloosterman sum Kq (a) and the lifted Kloosterman sum Kqn (a) satisfy (1)
Kqn (a) = (−1)n−1 (A1 (a)n + A2 (a)n )
for every a ∈ Fq , where
√ Aj (a) := 21 (Kq (a) + (−1)j D)
Key words and phrases. Kloosterman sums, Cyclotomic fields. 1
2
KEIJO KONONEN, MARKO RINTA-AHO, AND KEIJO VÄÄNÄNEN
are the roots and D := Kq (a)2 − 4q is the discriminant of the equation x2 − Kq (a)x + q = 0. Alternatively Kqn (a) = (−1)n−1 Dn (Kq (a), q), where Dn is the Dickson polynomial (of the first kind) defined by bn c 2 X n−i n Dn (x, q) = (−q)i xn−2i . n − i i i=0 Since Dickson polynomials have got integer coefficients, it follows that Kqn (a) ∈ Z whenever Kq (a) ∈ Z. Here we investigate whether the converse is true. That is, we want to find out if (2)
Kqn (a) ∈ Z =⇒ Kq (a) ∈ Z.
It turns out that the answer is positive and this is our main result. Theorem 1. Assume that Fq is a subfield of Fqn and that a ∈ Fq . Then Kqn (a) ∈ Z if and only if Kq (a) ∈ Z. 2. Proof of Theorem 1 Recently Marko Moisio [Moi09, Lemma 12] proved the following fact about the minimal polynomials of Kloosterman sums which also turns out to be useful for our purposes. Lemma 2. Let m(x) ∈ Z[x] denote the minimal polynomial of the Kloosterman sum Kq (a) ∈ Z[ζp ] over Q. Then m(x) ≡ (x + 1)t (mod p), where t is the degree of m(x). In particular Kq (a) ≡ −1 (mod p), if Kq (a) ∈ Z. Since every proper subfield of a finite field is contained in at least one maximal subfield, it is enough to consider (2) in the case where Fq is a maximal subfield of Fqn . Then n will obviously be a prime. We already know that the Kloosterman sum is always an integer in the cases p = 2, 3. In the case p > 3 we have the following result. Lemma 3. In the case p > 3, n a prime and n 6= p the statement (2) is true for all a ∈ Fq . Proof. Put b := Kqn (a) ∈ Z. Then Kq (a) will be a zero of the polynomial fn (x) := Dn (x, q) + (−1)n b ∈ Z[x]. Let m(x) ∈ Z[x] denote the minimal polynomial of the algebraic integer Kq (a). Obviously m(x)|fn (x). Let t be the degree of the polynomial m(x). Now fn (x) ≡ xn + (−1)n−1 (mod p), since Kqn (a) ≡ −1 (mod p). It follows from Lemma 2 that (x + 1)t |(xn + (−1)n−1 ) in the ring Zp [x]. If now Kq (a) 6∈ Z, then the polynomial xn + (−1)n−1 has −1 as a multiple zero. But since the formal derivative of this polynomial is nxn−1 and n is a prime, we must have n = p. We shall attack the case n = p by studying the conjugates of the Kloosterman sums Kq (a) and Kqp (a).
ON INTEGER VALUES OF KLOOSTERMAN SUMS
3
Lemma 4. Let a ∈ Fq and assume that Kqp (a) ∈ Z, then every algebraic conjugate of the Kloosterman sum Kq (a) is of the form √ (3) Kq (i2 a) = 21 (ζpk + ζp−k )Kq (a) + 21 (ζpk − ζp−k ) D for a suitable k ∈ Z/pZ. √ Proof. Put F := Q(ζp , D). Then every automorphism σi of the cyclotomic field Q(ζp ) can be extended to a monomorphism σ ˜i : F → C (possibly in two different ways). It follows from (1) that x2 − (−1)p−1 Kqp (a)x + q p = (x − A1 (a)p ) (x − A2 (a)p ) . Since the left-hand side polynomial has got coefficients in Z, σ ˜i permutes the roots of this polynomial. Thus we either have σ ˜i (Aj (a)p ) = Aj (a)p or σ ˜i (Aj (a)p ) = Aj 0 (a)p for j = 1, 2, where j 0 ∈ {1, 2}, j 0 6= j. In the first case σ ˜i (A1 (a)) = ζpl A1 (a), σ ˜i (A2 (a)) = ζpk A2 (a) and in the second σ ˜i (A1 (a)) = ζpk A2 (a), σ ˜i (A2 (a)) = ζpl A1 (a) for suitable k, l ∈ Z/pZ. Moreover, A1 (a)A2 (a) = q, so we must have l = −k. In both of the above cases we get Kq (i2 a) = σi (Kq (a)) = σ ˜i (Kq (a)) = σ ˜i (A1 (a)) + σ ˜i (A2 (a)) = ζp−k A1 (a) + ζpk A2 (a), which together with the definition of the Aj (a) gives (3).
After these preparations we are ready to prove our main result. We already know that Kqn (a) ∈ Z whenever Kq (a) ∈ Z. Lemma 3 and the comments before it also tell us that it is enough to prove the converse direction only in the case n = p. So assume that Kqp (a) ∈ Z. We will show that Kq (a) ∈ Z by proving that Kq (i2 a) = Kq (a) for every i ∈ (Z/pZ)∗ . This is equivalent to showing that √ we can choose k = 0 in the equation (3) for every i. In the case D 6∈ Q(ζp ) this is clear, since all the other √ terms in (3) belong to the cyclotomic field. In the case D ∈ Q(ζp ) our idea is to look at the equation (3) modulo L 2 in the ring Z[ζp ]. Let m(x) ∈ Z[x] be the minimal polynomial of the Kloosterman sum Kq (a). By Lemma 2 m(x) ≡ (x + 1)t (mod p), where t is the degree of the m(x). Thus (Kq (a) + 1)t ≡ 0 (mod L p−1 ) in the ring Z[ζp ]. Because the Kloosterman sum is an element of the maximal real subfield of the cyclotomic field, t ≤ p−1 . Consequently 2 2 Kq (a) ≡ −1 (mod L ). Since 1 − ζpk is either 0 or differs from λp by a unit factor, ζpk + ζp−k = −k ζp [(1 − ζpk )2 + 2ζpk ] ≡ 2 (mod L 2 ). Similarly ζpk − ζp−k = ζp−k [(1 − ζpk )2 + 2(ζpk − 1)] ≡ 2(1 − ζp−k ) (mod L 2 ). Combining these results with (3) gives √ −1 ≡ 21 · 2 · (−1) + 21 · 2(1 − ζp−k ) D (mod L 2 ).
4
KEIJO KONONEN, MARKO RINTA-AHO, AND KEIJO VÄÄNÄNEN
√
It follows that L 2 |(1 − ζp−k ) D. If now k 6= 0, 1 − ζp−k = L and √ we must have L | D and thus also L 2 |D. On the other hand D = Kq (a)2 − 4q ≡ (−1)2 − 4 · 0 = 1 (mod L 2 ), which is a contradiction. Thus we must have k = 0, which completes the proof. 3. Further observations Before finishing we are going to make some more remarks about the points and fields where Kq (a) can obtain integer values. Firstly, by using Theorem 1 one can very easily prove that a cannot lie in the prime subfield. Theorem 5. If p > 3 and a ∈ F∗p , then Kq (a) 6∈ Z. Proof. By Theorem 1 it is enough to prove the result in the case q = p. For every i ∈ Fp let Ni denote the number of x ∈ F∗p such that P P −1 x+ax−1 = i. Then Kp (a) = x∈F∗p ζpx+ax = i∈Fp Ni ζpi . Assume now that Kp (a) ∈ Z. Since the minimal polynomial of ζp is xp−1 +· · ·+x+1, we must have N1 = N2 = · · · = Np−1 and Kp (a) = N0 − N1 . Moreover, Pp−1 i=0 Ni = p − 1 implies that either N0 = 0 and all the other Ni = 1 or N0 = p − 1 and all the other Ni = 0. Thus the mapping f : F∗p → Fp , f (x) = x + ax−1 must be either injective or zero everywhere. Notice that f (ax−1 ) = f (x). Also, we can always find such x ∈ F∗p that ax−1 6= x, if p > 3. Thus f cannot be injective when p > 3. Similarly we can always find such a x ∈ F∗p that f (x) 6= 0 whenever p > 3, because f (x) = 0 is equivalent to x2 = −a. In the case q = p rather more than this is known. Hans Salié [Sal32] has proved that the p−1 Kloosterman sums Kp (a) are divided into two classes according to whether a ∈ F∗p is a square or not and that each . A class consists of the roots of an irreducible polynomial of degree p−1 2 more general result applicable to arbitrary finite fields has later been obtained by Daqing Wan. Wan [Wan95, Theorem 1.1] showed that Kq (a) generates the whole maximal real subfield Q(ζp + ζp−1 ) whenever a ∈ Fq is such that Tr(a) 6= 0. One easy property of the Kloosterman sum is that it is invariant under the action of the Frobenius automorphism x 7→ xp i.e. that Kq (ap ) = Kq (a). This property allows us sometimes quite easily to find points a ∈ F∗q where the Kloosterman sum obtains an integer value. Assume that q = pr and that we have found a ∈ Fq such that for every square i2 ∈ (Z/pZ)∗ (4)
s
i2 a = ap for some s ∈ {0, 1, . . . , r − 1}.
Then all the algebraic conjugates σi (Kq (a)) will be equal and consequently Kq (a) ∈ Z. We illustrate by calculations in F52 . Let α be a primitive element and a = αj . Condition (4) is now equivalent to −a = a5 . This holds if and
ON INTEGER VALUES OF KLOOSTERMAN SUMS
5
2
only if 4j ≡ 5 2−1 (mod 52 − 1). Thus we find four points giving integer values: α3 , α9 , α15 and α21 . Computer calculation confirms that these are the only points a in F∗52 for which Kq (a) ∈ Z. On the other hand in the case of the field F53 we don’t find any points by using criteria (4) although direct calculations show that there are some points giving integer values for the Kloosterman sum. Remarkably enough there exist some fields Fq for which the Kloosterman values are distinct except for the action of the Galois group Gal(Fq /Fp ) generated by the Frobenius automorphism. More precisely s Kq (a) = Kq (b) if and only if b = ap for some s. For example Fisher [Fis92] has proved that this holds for all fields satisfying the bound p > (2 · 4r + 1)2 . The referee of Fisher’s paper conjectured that much stricter bound p ≥ 2r should actually hold. As far as we are aware this conjecture appears to be still largely open. For these kind of fields, for which the kloosterman values are distinct modulo the action by Gal(Fq /Fp ), Kq (a) ∈ Z if and only if (4) holds. Note that (4) can never be satisfied if the number p−1 of squares i2 ∈ Fp exceeds the 2 number r of Frobenius conjugates that is if p > 2r + 1. Thus, if the above-mentioned referee’s conjecture is true, the fields Fpr satisfying p > 2r + 1 don’t have any points a 6= 0 for which Kpr (a) ∈ Z. References [Car69] Leonard Carlitz. Kloosterman sums and finite field extensions. Acta Arithmetica, 16:179–193, 1969. [Fis92] Benji Fisher. Distinctness of Kloosterman Sums. In Contemporary Mathematics: p-Adic Methods in Number Theory and Algebraic Geometry, pages 81–102. American Mathematical Society, 1992. [LN97] Rudolf Lidl and Harald Niederreiter. Finite fields, volume 20 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, second edition, 1997. [Moi09] Marko Moisio. On certain values of Kloosterman sums. IEEE Transactions on Information Theory, 55(8):3563–3564, 2009. [Sal32] Hans Salié. Über die Kloostermanschen Summen S(u, v; q). Mathematische Zeitschrift, 34:91–109, 1932. [Wan95] Daqing Wan. Minimal Polynomials and Distinctness of Kloosterman Sums. Finite Fields and Their Applications, 1(2):189–203, 1995. E-mail address: [email protected] E-mail address: [email protected] E-mail address: [email protected] Department of Mathematical Sciences, University Of Oulu, P.O. BOX 3000, FIN-90014 Oulun yliopisto, Finland
