On a character sum problem of H. Cohn

8 Pages·0.13 MB·English

by  Par Kurlberg

ON A CHARACTER SUM PROBLEM OF H. COHN
1
0 PA¨R KURLBERG
0
2
Abstract. Letf beacomplexvaluedfunctiononafinite fieldF

n
suchthatf(0)=0,f(1)=1,and f(x) =1forx=0. Cohnasked
a | | 6
if itfollowsthatf is anontrivialmultiplicative characterprovided
J
that f(x)f(x+h) = 1 for h = 0. We prove that this is
9 x∈F
− 6
2 the case for finite fields of prime cardinality under the assumption
P
that the nonzero values of f are roots of unity.

]
T
N
.
h
1. Introduction
t
a
m Let p be prime and let F be the finite field with pk elements. Let
pk
[ f : F× C be a nontrivial multiplicative character, and extend f to
pk →
1 a function on F by letting f(0) = 0. It is then easy to see that the
pk
v
following holds:
2
4
2 1 if h = 0
(1.1) f(x)f(x+h) = − 6
1
pk 1 if h = 0
10 xX∈Fpk ( −
0
Cohn asked (see p. 202 in [3]) if the converse is true in the following
/
h sense: if a function f : F C satisfies
pk
t →
a
m (1.2) f(0) = 0, f(1) = 1, and f(x) = 1 for x = 0
| | 6
:
v and equation 1.1, does it follow that f is a multiplicative character?
Xi The problem has recently received some attention. In [2], Choi and
Siu proved that the converse is not true for k > 1. One of the argu-
r
a
ments given is quite pretty, and proceeds as follows: Let λ be a linear
automorphism ofF so thatλ(1) = 1. If f satisfies 1.1 and1.2, so does
pk
f composed with λ. Now, if f is an injective multiplicative character
then the converse being true implies that f composed with λ must
be an injective multiplicative character. On the other hand, a simple
counting argument shows that the number of possible λ’s is greater
than the number of injective characters.
However, the case k = 1 remains unresolved. In [1], Biro proved that
there are only finitely many functions satisfying equation 1.1 and 1.2
Author supported in part by the National Science Foundation (DMS 0071503).
1
2 PA¨R KURLBERG
for each p. Biro also solved the following “characteristic p” version of
the problem ([1], Theorem 2):
Theorem (Biro). Let p be a prime, let F be the finite field with p
p
elements, and F F any field of characteristic p. Assume that there
p
⊃
is given an a F for every i F such that a = 0,a = 1,a = 0 for
i p 0 1 i
∈ ∈ 6
i = 0, and
6 a
i+j
= 1
a −
× i
iX∈Fp
for every j F×. Then a = iA for every i F with some 1 A
∈ p i ∈ p ≤ ≤
p 2.
−
Using this Biro deduces that the converse holds for functions taking
values in 1,0,1 .1 In fact, if m is coprime to p, then the case of
{− }
the nonzero values of f being m-th roots of unity can be deduced in a
similar way: Let O be the ring of integers in Q(e2πi/m), and let P O
⊂
beaprimeideallyingabovep. Theresultthenfollowsfromthetheorem
by letting F = O/P and noting that m-th roots of unity are distinct
modulo p. (Since f(x) = 1 for x = 0 we have f(x) = 1/f(x).)
| | 6
The aim of this paper is to show that the converse is true for the
case k = 1, under the additional assumption that the nonzero values
of f : F C are m-th roots of unity, including the case p m. We
p
→ |
begin by giving a proof that does not depend on Biro’s result for the
case (m,p) = 1, and we then show how to modify the argument for the
general case.
Acknowledgements: I would like to thank Ernest Croot, Andrew
Granville, Robert Rumely, and Mark Watkins for helpful and stimu-
lating discussions. I would also like to thank the referee for several
suggestions on how to improve the exposition, and for pointing out
that the case p m can be deduced independently of Biro’s theorem.
|
2. Preliminaries
In what follows we assume that p is odd since the case p = 2 is
trivial.
We will use the following conventions: if a function f takes values
in C and σ Aut(C/Q), then we let fσ be the function defined by
∈
fσ(x) = σ(f(x)). We regard ψ(x) = e2πix/p as a nontrivial additive
character of F . For an integer t, ψ will denote the character ψ (x) =
p t t
ψ(tx). By ζ we denote the m-th root of unity ζ = e2πi/m.
m m
1
There appearsto be severalindependent proofs ofthis result, see the introduc-
tion in [2].
ON A CHARACTER SUM PROBLEM OF H. COHN 3
Let m be even and large enough so that all nonzero values of f
are m-th roots of unity, and write m = npk, where (n,p) = 1. Let
K = Q(ζ ), L = K(ζ ,ζ ), and let G = Gal(L/Q), H = Gal(L/K)
n p pk
denote the Galois groups of L/Q and L/K. By O and O we will
K L
denote the ring of integers in K respectively L.
The “Gauss sum”
p−1
G(f,ψ) = f(x)ψ(x)
x=0
X
is clearly an algebraic integer. As in the case of classical Gauss sums,
the absolute value of G(f,ψ) can easily be determined:
Lemma 1. If f satisfies 1.1, then
√p if t 0 mod p,
G(f,ψ ) = 6≡
t
| | 0 if t 0 mod p.
(
≡
Proof. We have
G(f,ψ ) 2 = f(x)f(y)ψ(t(x y)) = f(x)f(x+h)ψ( th)
t
| | − −
x,Xy∈Fp x,Xh∈Fp
= ψ(0) f(x)f(x)+ ψ( th) f(x)f(x+h)
−
xX∈Fp hX∈Fp× xX∈Fp
p if t 0 mod p,
= p 1 ψ( th) = 6≡
− − − 0 if t 0 mod p,
(
hX∈Fp× ≡
The action of complex conjugation on K is given by an element
in G, and since G is abelian, equation 1.1 is G-invariant. I.e., if f
satisfies 1.2, so does fσ for all σ G. But if σ G then σ(G(f,ψ)) =
∈ ∈
G(fσ,ψ ), where σ(ζ ) = ζt. Since fσ also satisfies 1.1, we find that
t p p
G(fσ,ψ ) = p1/2, and hence the Q-norm of G(f,ψ) is a power of p.
t
| |
The factorization of the principal ideal G(f,ψ)O thus consists only of
L
prime ideals P p.
L
|
It is well known that Q(ζ )/Q is totally ramified over p, and that
pk
Q(ζ )/Q does not ramify at p if (n,p) = 1. Comparing ramification
n
indices gives that if P is a prime ideal in O that divides p, then
K K
P is totally ramified in L. In particular, if P is any prime ideal in
K L
the ring of integers in O that lies above p, then σ(P ) = P for all
L L L
σ H.
∈
Let l = max(1,k). Then H consists of elements σ such that
t
σ (ζ ) = ζt , σ (ζ ) = ζ .
t pl pl t n n
4 PA¨R KURLBERG
Choose t so that σ generates H. Applying σ to the principal ideal
t t
G(f,ψ)O = P η(PL)
L L
PYL|p
we find that
σ (G(f,ψ)O ) = σ ( P η(PL)) = P η(PL) = G(f,ψ)O
t L t L L L
PYL|p PYL|p
and hence σ (G(f,ψ)) = uG(f,ψ) for some unit u.
t
Since the absolute value of any complex embedding ofG(f,ψ) equals
√p, we find that all conjugates of u = σ(G(f,ψ))/G(f,ψ) has absolute
value one. Hence u is in fact a root of unity, and there are integers a,b
such that
(2.1) σ (G(f,ψ)) = ζaζbG(f,ψ).
t pl n
3. The case (m,p) = 1
Since f is fixed by H we find that σ (G(f,ψ)) = G(f,ψ ), and equa-
t t
tion 2.1 can, after the change of variable x t−1x, be written as
→
p−1 p−1
(3.1) f(x)ψ(x) = ζ−aζ−b f(t−1x)ψ(x).
p n
x=1 x=1
X X
Lemma 2. If f takes values in n-th roots of unity for x 0 mod p
6≡
and equation 3.1 holds then a 0 mod p.
≡
Proof. From 3.1 we obtain that
p−1 p−1
(3.2) A ζi = B ζi
i p i p
i=1 i=0
X X
where A = f(i) and B = ζ−bf(t−1(i + a)). (Note that B = 0.)
i i n p−a
Since 1 = p−1ζi we may rewrite 3.2 as
− i=1 p
p−1 p−1
P
(3.3) A ζi = (B B )ζi.
i p i − 0 p
i=1 i=1
X X
Theelements ζ ,ζ2,ζ3,...ζp−1 arelinearlyindependentoverK,hence
{ p p p p }
A = B B . From lemma 1 we have p−1f(x) = 0, which implies
i i − 0 x=0
that p−1A = 0, as well as p−1B = 0. Therefore,
i=1 i i=0 i P
p−1 p−1 p−1
P P
0 = A = (B B ) = B pB = pB .
i i 0 i 0 0
− − −
i=1 i=1 i=0
X X X
ON A CHARACTER SUM PROBLEM OF H. COHN 5
But B = ζ−bf(t−1(0+a)) which is nonzero unless a 0 mod p.
0 n ≡
Thus
p−1 p−1
(3.4) f(x)ψ(x) = ζ−b f(t−1x)ψ(x)
n
x=1 x=1
X X
and the linear independence of ζ ,ζ2,ζ3,...ζp−1 over K implies that
{ p p p p }
f(t−1x) = f(x)ζb
n
for all x = 0. Thus
6
f(t−k) = f(t−(k−1))ζb = ... = f(1)ζkb = ζkb.
n n n
Taking k = p 1 we find that ζb is a (p 1)-th root of unity, and that
− n −
f is a multiplicative character.
4. The general case
In this case m = npk where (n,p) = 1 and k > 0. We will need the
following:
Lemma 3. If a K and pk−1a ζi K(ζ ) then
i ∈ i=0 i pk ∈ p
pk−1P p−1
(4.1) aiζpik = apk−1jζpj
i=0 j=0
X X
Proof. We may assume that k > 1. The minimal polynomial for ζ
pk
(over K as well as over Q) is given by
xpk 1
− = 1+xpk−1 +x2pk−1 +...+x(p−1)pk−1.
xpk−1 1
−
Hence, by letting˜i [0,pk−1 1] be a representative of i modulo pk−1,
∈ −
we can rewrite the left hand side of equation 4.1 as
(p−1)pk−1−1
(a a )ζi
i − (p−1)pk−1+˜i pk
i=0
X
with no further relations among the ζi ’s, and thus
pk
(p−1)pk−1−1
(a a )ζi K(ζ )
i − (p−1)pk−1+˜i pk ∈ p
i=0
X
if and only if a a = 0 for all i not congruent to zero modulo
i− (p−1)pk−1+˜i
pk−1.
6 PA¨R KURLBERG
Recall from equation 2.1 (note that l = k since k 1) that
≥
σ (G(f,ψ)) = ζa ζbG(f,ψ).
t pk n
Let G˜ = ζs G(f,ψ) where σ (ζs )/ζs = ζ−a. (Such an s exists as
pk t pk pk pk
σ (ζs )/ζs = ζ(t−1)s, and t 1 mod p since σ generates H.) We then
t pk pk pk 6≡ t
have
σ (G˜) = σ (ζs G(f,ψ))
t t pk
= σ (ζs )σ (G(f,ψ)) = σ (ζs )ζa ζbG(f,ψ)) = ζbG˜.
t pk t t pk pk n n
The following lemma shows that G˜ must transform by a nontrivial
n-th root of unity:
Lemma 4. There is no integer s such that ζs G(f,ψ) K.
pk ∈
Proof. We first assume that ζs = 1. Let G(f,ψ)O = P η(PL)
pk L PL|p L
be the factorization of the principal ideal G(f,ψ)O . Since p does not
L
Q
ramify in K, we have pO = P , and hence pO = P e
K PK|p K L PL|p L
where e is the ramification index of P in L.
K
Q Q
Since ψ(x) = ζx is congruent to 1 modulo P for all x, we find that
p L
p−1 p−1
G(f,ψ) = f(x)ψ(x) f(x) mod P
L
≡
x=0 x=1
X X
for all P p. Now, since f(0) = 0, we have p−1f(x) = G(f,ψ )
L| x=1 0
and by lemma 1, G(f,ψ ) = 0. Thus G(f,ψ) P for all P p, i.e.,
0 L L
∈P |
η(P ) > 0 for all P p. But if G(f,ψ) K then e η(P ) for all P p,
L L L L
| ∈ | |
and since complex conjugation permutes the set of primes of O that
L
lies above p, and
p = G(f,ψ)G(f,ψ),
we get that P 2e pO for all P , contradicting that the ramification
L L L
|
index is e.
Forthegeneralcase, thepreviousargumentcarriesthroughbynoting
that ζs is a unit (and thus multiplication of G(f,ψ) by ζs does not
pk pk
change the ideal factorization) and that G(f,ψ) P if and only if
L
∈
ζ G(f,ψ) P .
pk L
∈
Since σ has order pk−1(p 1) and (n,p) = 1 we find that ζb must
t − n
be a nontrivial (p 1)-th root of unity. Hence there exists a non-
−
trivial multiplicative character χ of F× such that χ(t−1) = ζb. But
p n
σ (G(χ,ψ)) = G(χ,ψ ) = χ(t−1)G(χ,ψ) and thus
t t
G˜
δ =
G(χ,ψ)
ON A CHARACTER SUM PROBLEM OF H. COHN 7
is σ -invariant and hence an element of K. Moreover, δ = 1 (for all
t
| |
complex embeddings) since G˜ = G(χ,ψ) = p1/2.
| | | |
Write f(x) = f (x)f (x) where f (x) takes values in pk-th roots of
1 2 1
unity and f (x) takes values in n-th roots of unity. We will show that
2
f (x) must be constant.
1
Lemma 5. Let
a = f (x)
i 2
x:ζpskf1(Xx)ψ(x)=ζpik
If
p−1 p−1
(4.2) ζs f(x)ψ(x) = δ χ(x)ψ(x),
pk
x=1 x=1
X X
then a = 0 unless i = pk−1j for j = 1,2,... ,p 1, in which case
i
| | −
a = 1. In particular, ζs f (x)ψ(x) ranges over all nontrivial p-th
| i| pk 1
roots of unity.
Proof. Collecting terms in 4.2 according to the values of ζs f (x)ψ(x),
pk 1
we obtain
pk−1 p−1
(4.3) a ζi = δ χ(i)ζi K(ζ ).
i pk p ∈ p
i=0 i=1
X X
Clearly a K and a = 0 for at most p 1 values of i. Letting
i i
∈ 6 −
A = a we may, by lemma 3, write equation 4.3 as
i pk−1i
p−1 p−1
A ζi = δ χ(i)ζi.
i p p
i=0 i=1
X X
Since 1 = p−1ζi we get that
− i=1 p
p−1 p−1 p−1
P
(A A )ζi = A ζi = δ χ(i)ζi
i − 0 p i p p
i=1 i=0 i=1
X X X
and hence A A = δχ(i) for all i.
i 0
−
Since a = 0 for at most p 1 values of i, A = 0 implies that A = 0
i 0 j
6 − 6
for some j = 0, and thus A = δχ(j) A = 1. Since
0 j
6 | | | − |
p−1 p−1 p−1
0 = δ χ(i) = (A A ) = A pA ,
i 0 i 0
− −
i=1 i=1 i=0
X X X
we find that p−1A = p A = p. On the other hand, p−1A
| i=0 i| | 0| | i=0 i| ≤
p−1 f (x) = p 1. Thus A = 0, and it follows that A = δχ(i)
x=1| 2 | P − 0 Pi
for i = 0. In other words, apk−1j = Aj = δχ(j) for j = 1,2,... ,p 1,
P 6 −
8 PA¨R KURLBERG
and since there are at most p 1 nonzero values among the a ’s, the
i
−
remaining ones must all be equal to zero.
Now, the lemma gives that ζs f (1)ψ(1) = ζs ζ is a p-th root of
pk 1 pk p
unity, hencepk−1 mustdivides,andthenonzerovaluesoff (x)ψ(x)are
1
thus distinct p-th roots of unity. Replacing ψ by ψ , for r 0 mod p,
r
6≡
inthepreviousargumentgivesthatf (x)ψ(rx)alsorangesoverdistinct
1
p-th roots of unity. On the other hand, if f (x) is not constant, then
1
there exists r 0 mod p such that the set f (x)ψ (x) p−1 contains
6≡ { 1 r }x=1
strictly less than p 1 elements. (If f (x ) = f (x ), write f (x ) =
1 1 1 2 1 1
− 6
ζy1,f (x ) = ζy2 and take r (y y )(x x )−1 mod p.) Hence
p 1 2 p ≡ − 2 − 1 2 − 1
f (x) must be constant, and since f (1) = 1, we find that the nonzero
1 1
values of f(x) are in fact n-th roots of unity. The result has thus been
reduced to the case (m,p) = 1.
References
1. A.Biro,Notes on a problem of H. Cohn, J.ofNumber Theory77 (1999),no.2,
200–208.
2. K.K.SChoiandM.K.SiuCounter-ExamplestoaProblemofCohnonClassifying
Characters, to appear in J. of Number Theory.
3. H.Montgomery, Ten lectures on the interface between analytic number theory
and harmonic analysis, American Mathematical Society, Providence, RI, 1994.
Department of Mathematics, University of Georgia, Athens GA
30602 ([email protected])