A Schinzel theorem on continued fractions in function fields

XCII.4 (2000)

A Schinzel theorem on continued fractions in function fields

by

Weiqun Hu (Nanjing)

1. Introduction and statement of result. It is well known that the expansion of the continued fraction of a positive quadratic irrational num- ber α is periodical. The problem of evaluating and estimating the period p(α) of the continued fraction of α is closely related to many number theo- retic problems such as Pell’s equations and the fundamental unit of the real quadratic field Q(α). In 1961, A. Schinzel [8] proved an interesting result.

Let a, b, c be integers with a ≥ 1, and let f (x) = a

x

+ bx + c be a quadratic polynomial with discriminant d = b

− 4a

c. We define two sets

Z = {n ∈ Z : n ≥ 1 and f (n) is square-free}, e E = {n ∈ e Z : d - (2a

n + b)

}.

Schinzel’s result says that (I) lim

n∈eZ−E

p( p

f (n)) < ∞, (II) lim

n∈E

p( p

f (n)) = ∞.

Later, S. Louboutin [6] and A. Farhane [2] presented more exact and effective lower bounds of p( p

f (n)) for n ∈ E. In this paper we show an analogue of Schinzel’s result in the function field case.

Let F

be the finite field with q elements where q is odd, k = F

(t) be the rational function field over F

, R = F

[t] be its polynomial ring, k

= F

((1/t)) be the completion of k at the infinite place ∞ = (1/t). Each element α ∈ k

− k is a (formal) power series

α = X

n≥l

c

 1 t



, c

∈ F

(n ≥ l), sgn α = c

∈ F

.

Let v

be the normal exponential valuation in k

with v

(1/t) = 1. In this paper we prefer to use the notation deg α = −v

(α) = −l since it is

2000 Mathematics Subject Classification: 11D41, 11R27, 11R58.

Key words and phrases: continued fraction, function field.

[291]

just the ordinary degree if α is a polynomial in R. We define [α] =

X

c

 1 t



∈ R, {α} = X

n≥sup{1,l}

c

 1 t



.

Then we have the continued fraction expansion of α, α = [a

, a

, a

, . . .] = a

+ 1

a

+ 1 a

+ . . . where a

= [α] and α

= 1/{a

}, a

= [α

] for i ≥ 1 so that

α

= [a

, a

, . . .], deg α

= deg a

≥ 1 for i ≥ 1.

Suppose that F (t) is a square-free polynomial in R such that deg F (t) ≥ 1 and p

F (t) ∈ k

. (Note: F (t) is called square-free if there is no A ∈ R with deg A ≥ 1 such that A

| F (t). By Hensel’s lemma, p

F (t) ∈ k

if and only if 2 | deg F (t) and sgn F (t) ∈ F

.) The subfield K = k( p

F (t)) of k

is called a “real” quadratic function field. It is well known that the continued fraction of p

F (t) is periodical. In fact, we need a generalized version of the period in the study of the algebraic structure of the real quadratic function field K.

For each α ∈ k

, α is algebraic over k with degree two if and only if the continued fraction of α = [a

, a

, . . . , a

, . . .] is infinite and quasi-periodical, which means that there exist c ∈ F

and integers n

≥ 0 and l ≥ 1 such that

a

= c

a

for all n ≥ n

.

The smallest positive integer l satisfying this condition is called the quasi- period of the continued fraction α = [a

, a

, . . .] and denoted by p(α). If α = p F (t) and l = p( p

F (t)), then the fundamental unit of the real quadratic function field K = k( p

F (t)) is P

+ Q

p

F (t) where P

/Q

= [a

, a

, . . . , a

].

The continued fraction method of studying the ideal class number and units of real quadratic function fields was initiated by E. Artin [1]. For recent work we refer to D. Hayes [4] and C. D. Gonz´alez [3].

Now we state the result of this paper. Let A, B, C be polynomials in R = F

[t], deg A ≥ 0, f (x) = A

x

+ Bx + C, D = B

− 4A

C, and

R = {N ∈ R : f (N ) is square-free and e p

f (N ) ∈ k

}, E = {N ∈ e R : D - (2A

N + B)

}.

Let p( p

f (N )) be the quasi-period of the continued fraction of p

f (N ).

Theorem 1.1.

lim

N ∈ eR−E

p( p

f (N )) < ∞.

Theorem 1.2.

N ∈E

lim p( p

f (N )) = ∞.

Before proving Theorems 1.1 and 1.2, in Section 2 we introduce several basic facts on continued fractions in function fields. Most results are just simple analogues of facts in the theory of ordinary continued fractions in the real number field case (see Hua’s book [5], for instance), so we present the proof of some particular facts for the reader’s convenience and omit the proof of others. In Sections 3 and 4 we prove Theorems 1.1 and 1.2.

2. Continued fractions in function fields. From now on we fix the following notations: k = F

(t) (2 - q), R = F

[t], k

= F((1/t)). For α ∈ k

− k,

α = c

 1 t



+ c

 1 t



+ . . . , c

∈ F

(i ≥ m), c

∈ F

; we set deg α = −m, sgn α = c

. We have the continued fraction (2.1) α = [a

, a

, . . . , a

, . . .].

For n ≥ 0, the nth convergent of the continued fraction (2.1) is P

/Q

= [a

, a

, . . . , a

]

which can be calculated recursively by

(2.2) P

= a

, P

= a

a

+ 1, P

= a

P

+ P

(n ≥ 2), Q

= 1, Q

= a

, Q

= a

Q

+ Q

(n ≥ 2).

We have the following basic facts.

Fact 1. For n ≥ 1,

P

Q

− P

Q

= (−1)

, (2.3)

α = α

P

+ P

α

Q

+ Q

(2.4)

where α

= [a

, a

, . . .].

Let A, B, C ∈ R, deg A ≥ 0, f (x) = A

x + Bx + C, D = B

− 4A

C, and p

f (N ) ∈ k

(N ∈ R).

Fact 2. For α = p

f (N ) = [a

, a

, . . . , a

, . . .], we have

(2.5) α

= U

+ p

f (N ) V

(n ≥ 0)

where U

, V

are polynomials in R and can be calculated recursively by U

= 0, V

= 1, a

= [ p

f (N )], (2.6) U

= a

V

− U

, V

= f (N ) − U

V

,

a

=

 U

+ p f (N ) V



(n ≥ 0).

Moreover , p( p

f (N )) = 2k if and only if k is the smallest integer such that U

/U

∈ F

, while p( p

f (N )) = 2k + 1 if and only if k is the smallest integer such that V

/V

∈ F

.

The next result means that the convergents are the best approximations of α ∈ k

.

Lemma 2.1. For α ∈ k

− k, P, Q ∈ R, Q 6= 0 and (P, Q) = 1, the following statements are equivalent to each other :

(1) P/Q = P

/Q

or P/Q = −P

/Q

for some n ≥ 0;

(2) deg(P

− α

Q

) < deg α;

(3) deg(P − αQ) < − deg Q or deg(P + αQ) < − deg Q.

P r o o f. Let P/Q = [c

, c

, . . . , c

] be the finite continued fraction of P/Q, and P

/Q

= [c

, c

, . . . , c

]. There is β ∈ k

such that

(2.7) α = [c

, . . . , c

, β] = βP + P

βQ + Q

. Namely,

β = −αQ

+ P

αQ − P . From (2.7) and (2.3) we have

α − P

Q = (−1)

Q(βQ + Q

) , deg(P − αQ) = − deg(βQ + Q

).

Therefore

P/Q = P

/Q

⇔ deg β ≥ 1 ⇔ deg(P − αQ) < − deg Q, P/Q = −P

/Q

⇔ deg(P + αQ) < − deg Q.

So we proved the equivalence (1)⇔(3). The equivalence (2)⇔(3) is easy to prove.

The following result is analogous to Theorem H in [8].

Lemma 2.2. Suppose that ξ = [b

, b

, . . .] ∈ k

−k and ξ

= [b

, b

, . . .].

Let p, r, s ∈ R, deg r < deg s and rs = d 6= 0. If ξ

= pξ + r

s , p[b

, b

, . . . , b

] + r

s = [d

, d

, . . . , d

],

then ξ

= [d

, d

, . . . , d

, ξ

] where ξ

= (p

ξ

+ r

)/s

, p

, r

, s

∈ R, p

s

= ±d, deg r

< deg s

.

P r o o f. For a non-singular matrix A =

over R and α ∈ k

we define the action of A on α by

A(α) = aα + b cα + d .

Let [b

, . . . , b

] = A

/B

and [d

, . . . , d

] = C

/D

. From (2.4) we have ξ

= C

ξ

+ C

D

ξ

+ D

=

 C

C

D

D

 (ξ

), ξ

=

 p r o s

 (ξ) =

 p r o s

  A

A

B

B

 (ξ

).

Therefore ξ

=

t⁰ s⁰

(ξ

) where

 p

r

t

s



=

 C

C

D

D



 p r o s

  A

A

B

B

 (2.8)

= ±

 −D

C

D

−C

  p r o s

  A

A

B

B



= ±

 −pD

−rD

+ sC

pD

rD

− sC

  A

A

B

B



and

t

= ±(pD

A

+ rD

B

− sC

B

)

= ±sD

B

 pA

/B

+ r

s − C

D



= 0, s

= ±(pD

A

+ rD

B

− sC

B

), r

= ∓(pD

A

+ rD

B

− sC

B

), so that deg r

< deg s

. From (2.8) we have

p

s

=     p

r

o s

    = ±

    p r

o s     = ±d.

This completes the proof of Lemma 2.2 (cf. Theorem 3 in Chapter IV of [7]).

3. Proof of Theorem 1.1. Suppose that N ∈ e R − E, which means that f (N ) = A

N

+ BN + C is square-free, p

f (N ) ∈ k

and (3.1) D = B

− 4A

C | (2A

N + B)

. Let

(3.2) ξ

= p

f (N ) = 1 2A

p (2A

N + B)

− D.

From (3.1) we know that there exist U, V, W ∈ R such that (3.3) 2A

N + B = U V W, D = U V

. The formula (3.2) becomes

(3.4) ξ

= V

2A ξ where

ξ = p

U

W

− U = [U W, 2W, 2U W ] = [U W, ξ

], (3.5)

ξ

= [2W, 2U W ] = [2W, 2U W, ξ

].

(3.6)

Now we use Lemma 2.2 repeatedly to get the continued fraction expan- sion of ξ

= p

f (N ). Let (3.7) η

= V U W

2A = 2A

N + B

2A = [a

, a

, . . . , a

].

From (3.5) and Lemma 2.2 we have

(3.8) ξ

= [a

, a

, . . . , a

, ξ

] where

(3.9) ξ

= p

ξ

+ r

s

, p

, r

, s

∈ R, p

s

= ±2AV, deg r

< deg s

. In general, for each i ≥ 1 we have

(3.10) ξ

= p

ξ

+ r

s

, p

, r

, s

∈ R, p

s

= ±2AV, deg r

< deg s

. Let

η

= p

[2W, 2U W ] + r

s

= p

(2W + 1/(2U W )) + r

s

(3.11)

= [a

, a

, . . . , a

].

From (3.6) and Lemma 2.2 we have

(3.12) ξ

= [a

, a

, . . . , a

, ξ

] where

ξ

= p

ξ

+ r

s

, p

, r

, s

∈ R, p

s

= ±2AV, (3.13)

deg r

< deg s

. Since A, B, C and D = B

− 4A

C are fixed polynomials, we choose N such that

(3.14) deg N > max{deg(A

/B), deg(D/A)}.

From (3.3) we know that

deg(U V W ) = deg(A

N ) > deg(AD) = deg(AU V

)

so that deg W > deg(AV ) = deg(p

s

) ≥ deg s

(i ≥ 1). Therefore by (3.11), deg a

= deg[η

] = deg p

W

s

≥ 1 (i ≥ 1)

and by (3.8) and (3.11) we have the continued fraction expansion of ξ

= p f (N ):

(3.15) p f (N )

= [a

, . . . , a

, a

, . . . , a

, . . . , a

, a

, . . . , a

, . . .].

The total number of tuples (p, r, s) satisfying ps = ±2AN and deg r <

deg s is at most M = q

, thus from (3.10) we know that there exist l and j, 1 ≤ l < j ≤ M, such that ξ

= ξ

. From the expansion (3.15) we know that

(3.16) p( p

f (N )) ≤

j−1

X

i=l

(µ

− µ

).

We use Lemma 2.2 again to estimate µ

. Let 2W p

+ r

s

= [c

, c

, . . . , c

].

By Lemma 2.2 and (3.11) we have η

=



c

, c

, . . . , c

, p

(2U W ) + r

s



where p

s

= ±2AV and deg r

< deg s

, so that deg( p

p

(2U W ) + r

/s

) ≥ 1. Let

p

(2U W ) + r

s

= [c

, c

, . . . , c

].

Then η

= [c

, c

, . . . , c

, c

, . . . , c

] and µ

− µ

= t + λ by (3.11).

From the recursive formula for Q

in (2.2) and deg c

≥ 1 (0 ≤ i ≤ t + λ − 1) we know that

t ≤ deg s

≤ deg(AV ), λ ≤ deg s

≤ deg(AV ).

Therefore µ

− µ

= t + λ ≤ 2 deg(AV ) and by (3.16), p( p

f (N )) ≤ (j − l)2 deg(AV ) ≤ 2M deg(AV ) ≤ 2 deg q

(AV )q

provided the formula (3.14) is satisfied. Since there are only finitely many N such that deg N ≤ max{deg(A

/B), deg(D/A)}, this completes the proof of Theorem 1.1.

4. Proof of Theorem 1.2. Now we assume that f (N ) = A

N

+BN +C and D = B

− 4A

C - (2A

N + B)

. Let l = p( p

f (N )) and

p f (N ) = [a

, a

, . . . , a

, . . .],

α

= [a

, a

, . . .], P

/Q

= [a

, a

, . . . , a

], ϕ

= P

+ Q

p

f (N ), ϕ

= P

− Q

p f (N ).

Then ε = ϕ

is the fundamental unit of the quadratic function field K = k( p

f (N )). Let

G = (A, B), β = DG

, U = 2AG

, V = BG

. We choose

A

= AU N + V = (2A

N + B)G

, B

= U = 2AG

and

X = A

+ B

p f (N ).

It is easy to see that

Norm(X) = XX = A

− B

f (N ) = β where Norm denotes the norm for F

(t, p

f (N ))/F

(t). For each k ≥ 1, let X

= A

+ B

p

f (N ), A

, B

∈ R.

The polynomials A

and B

can be calculated recursively by (4.1) A

= A

A

+ B

B

f (N ), B

= A

B

+ B

A

. Let

(4.2) D

= (A

, B

), A

= A

/D

, B

= B

/D

, N

= A

− B

f (N ).

Then (A

, B

) = 1. Finally we choose

(4.3) M =

 deg f (N ) 2 deg β



− 1.

Lemma 4.1. For 1 ≤ k ≤ M , there exists i

such that A

/B

= P

/Q

. P r o o f. We have

deg(A

− B

f (N )) = deg(A

− B

f (N )) − 2 deg D

≤ deg(A

− B

f (N )) = deg(Norm(X

)) = k deg β

≤ M deg β <

deg f (N ).

Then the conclusion follows from Lemma 2.1 and deg(A

+ B

p

f (N )) ≥ 0.

Lemma 4.2. i

− i

≥ 2 for 1 ≤ k ≤ M − 1.

P r o o f. From the condition D - (2A

N + B)

we know that there is an irreducible polynomial P = P (t) in R such that v

(D) > 2v

(2A

N + B), which means that v

(β) > 2v

(A

) where v

is the normal P -adic expo- nential valuation. From β = Norm(X) = A

− B

f (N ) we have v

(A

) = v

(B

f (N )). Then (A

, B

) = 1 and f (N ) ∈ e R imply that

(4.4) v

(B

) = v

(f (N )) = v

(A

) = 0.

Now we prove that v

(A

) = v

(B

) = 0 for all k ≥ 1 by induction.

This is true for k = 1 from (4.4). Assume that v

(B

) = v

(A

) = 0 for some i ≥ 1. Since v

(f (N )) = 0, v

(β) > 2v

(A

) = 0 and β

= A

− B

f (N ), we know that A

≡ B

f (N ) (mod P ), which means that

f (N ) P

= 1 and P is split in the quadratic field K = F

(t)( p

f (N )) where

P

means the Legendre symbol. Let P = pe p be the decomposition in K.

Then X = A

+ B

p

f (N ) has to be divisible by p or e p. We can assume that

A

+ B

p f (N ) ≡ 0 (mod p).

From v

(B

) = 0 we know that v

(B

) = 0 so that A

B

≡ − p

f (N ) (mod p).

Moreover, A

+ B

p

f (N ) = (A

+ B

p

f (N ))

≡ 0 (mod p). From the assumption v

(B

) = 0 we have

A

/B

≡ − p

f (N ) (mod p).

Then from (4.1) we see that B

B

B

= A

B

+ A

B

≡ −2 p

f (N ) 6≡ 0 (mod p),

which means that v

(B

) = 0 so that v

(B

) = 0. Finally we deduce that v

(A

) = 0 from β

= A

−B

f (N ). Thus we have proved that v

(A

) = v

(B

) = 0 for all k ≥ 1. Then v

(D

) = 0 since D

= (A

, B

).

From (2.3) and (2.4) we have p f (N ) − P

Q

= (−1)

Q

(α

Q

+ Q

) . Therefore

deg ϕ

= deg(P

− Q

p f (N )) = − deg(α

Q

+ Q

)

= − deg(a

Q

+ Q

) = − deg Q

. From deg(A

− B

p f (N )) < 0 and D

| D

(by definition) we know that

− deg Q

= deg ϕ

= deg(P

− Q

p f (N ))

= deg(A

− B

p f (N ))

= − deg D

+ deg(A

− B

p f (N ))

= − deg D

+ k deg(A

− B

p f (N ))

> − deg D

+ (k + 1) deg(A

− B

p f (N ))

= deg ϕ

= − deg Q

.

Therefore deg Q

> deg Q

which means that i

> i

. If i

= i

+ 1, then

P

Q

− P

Q

= P

Q

− P

Q

= (−1)

, so that v

(P

Q

− P

Q

) = 0. On the other hand,

v

(P

Q

− P

Q

) = v

(A

B

− A

B

)

= v

(A

B

− A

B

)

(since v

(D

) = v

(D

) = 0)

= v

((A

A

+ B

B

f (N ))B

− A

(A

B

+ A

B

))

= v

(B

(A

− B

f (N ))) = v

(β

) (since v

(B

) = 0)

= kv

(β) > 0 (for k ≥ 1).

This contradiction shows that i

− i

≥ 2, which completes the proof of Lemma 4.2.

Lemma 4.3. Let M

=

 deg f (N ) − 2 deg U 2 deg β



− 1 (≤ M ).

Then deg ϕ

< deg ε for all 1 ≤ k ≤ M

where ε = ϕ

is the fundamental unit in K = F

(t, p

f (N )).

P r o o f. If deg ϕ

≥ deg ε for some s (1 ≤ s ≤ M

), let s be the smallest one satisfying this condition. Then

ϕ

ε = (P

+ Q

p

f (N ))(P

− Q

p

f (N )) = A

+ B

p f (N ) where A

, B

∈ R, and

deg(A

− B

f (N )) = deg(Norm(ϕ

)) = deg(Norm(ϕ

ε)) <

deg f (N ).

From Lemma 2.1 and deg(ϕ

ε) = deg ϕ

− deg ε ≥ 0 we know that there exists j such that ϕ

ε = αϕ

for some α ∈ F

. Therefore D

X

= ϕ

= α

εϕ

(α

∈ F

). If j ≥ i

, then

s deg X ≥ deg(D

X

) = deg(εϕ

) ≥ deg(εϕ

) = deg(εX) > deg X > 0.

Thus s ≥ 2 and

deg ϕ

≥ deg(D

D

ϕ

) = deg(D

X

) ≥ deg ε, which contradicts the definition of s. Therefore j < i

. Since

deg(ϕ

ϕ

) = deg(ϕ

ϕ

) = deg(β

D

), ϕ

= (−1)

(α

Q

+Q

)

and deg α

= deg a

, we know that deg ϕ

= − deg Q

= − deg Q

=

− deg U and

deg(β

U ) ≥ deg(β

D

U ) ≥ deg(β

D

ϕ

) = deg ϕ

≥

deg f (N ).

Therefore s ≥ (deg f (N ) − 2 deg U )/(2 deg β), which contradicts s ≤ M

. This completes the proof of Lemma 4.3.

Now Theorem 1.2 is a direct consequence since Lemma 4.2 says that i

− i

≥ 2 for 1 ≤ k ≤ M

− 1 and Lemma 4.3 says that 1 ≤ i

≤ p( p

f (N )) for all 1 ≤ k ≤ M

, so that p( p

f (N )) ≥ 2M

−1 = 2

 deg f (N ) − 2 deg U 2 deg β



−1 → ∞ as deg N → ∞.

This completes the proof of Theorem 1.2.

Example. Let f (x) = x

+ C where C = 2a + 1, a ∈ R, deg a ≥ 1. We have A = 1, B = 0, β = −4C = D and U = 1. For each k ≥ 2, we choose N = C

+ a so that f (N ) = (C

+ a)

+ C. Since D = −4C - (2A

N + B)

= 4(C

+ a)

, N belongs to the set E for all k ≥ 2. We use Fact 2 to determine the quasi-period p( p

f (N )) as follows:

U

= 0, V

= 1, a

= [ p

f (N )] = C

+ a, U

= C

+a, V

= C, a

=

 2C

+ 2a C



=

 2C

+ C − 1 C



= 2C

+1, U

= C

+ a + 1, V

= −2C

, a

=

 2C

+ C

−2C



= −C, U

= C

− a − 1, V

= C + (2C

− 1)C

−2C

= −C

, a

=

 2C

− 1

−C



= −2C

,

U

= C

+ a + 1, V

= 2C

, a

= C

, U

= C

− a − 1, V

= C

, a

= 2C

, . . . .

In general, V

= αC

, V

= α

C

, where α, α

∈ F

. If k = 2n, then V

/V

∈ F

and p( p

f (N )) = 4n + 1 = 2k − 1. If k = 2n, then V

/V

∈ F

and p( p

f (N )) = 4n − 1 = 2k − 1. Therefore p( p

f (N )) = 2k − 1 for all k ≥ 2. On the other hand,

2M

− 1 = 2

 deg f (N ) − 2 deg U 2 deg β



− 1 = 2 deg(C

+ a)

deg C − 1 = 2k − 1.

Therefore p( p

f (N )) = 2M

− 1, which shows that our general estimate p( p

f (N )) ≥ 2M

− 1 is the best one.

Acknowledgments. I would like to thank an anonymous referee for several suggestions that improved the exposition of this paper.

References

[1] E. A r t i n, Quadratische K¨orper im Gebiet der h¨oheren Kongruenzen I , II , Math. Z.

19 (1924), 154–246.

[2] A. F a r h a n e, Minoration de la p´eriode du d´eveloppement de √

a

n

+ bn + c en frac- tion continue, Acta Arith. 67 (1994), 63–67.

[3] C. D. G o n z ´a l e z, Class number of quadratic function fields and continued fractions, J. Number Theory 40 (1992), 38–59.

[4] D. H a y e s, Real quadratic function fields, in: CMS Conf. Proc. 7, Amer. Math. Soc., 1987, 203–236.

[5] L. K. H u a, Introduction to Number Theory, Springer, 1982.

[6] S. L o u b o u t i n, Une version effective d’un th´eor`eme de A. Schinzel sur longueurs des p´eriodes de certains d´eveloppements en fractions continues, C. R. Acad. Sci. Paris S´er. I 308 (1989), 511–513.

[7] B. d e M a t h a n, Approximations diophantiennes dans un corps local, Bull. Soc.

Math. France M´em. 21 (1970).

[8] A. S c h i n z e l, On some problems of the arithmetical theory of continued fractions, Acta Arith. 6 (1961), 393–413.

The Fundamental Science Department Nanjing Agriculture College

Nanjing 210038, China

E-mail: binhuang@public1.ptt.js.cn

Received on 22.5.1997

and in revised form on 12.4.1999 (3194)