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Diophan ine equa ions o e global unc ion ields I:
The Thue equa ion
Is án Gaála,1, Michael Pohs b,∗,2
aUni e si y o Deb ecen, Ma hema ical Ins i u e, H-4010 Deb ecen P . 12, Hunga y
bTechnische Uni e s ä Be lin, Ins i u ü Ma hema ik, S aße des 17. Juni 136, Be lin, Ge many
Recei ed 19 Janua y 2004; e ised 5 July 2005
Communica ed by Da id Goss
Abs ac
We sol e comple ely Thue equa ions in unc ion ields o e a bi a y ini e ields. In he unc ion
ield case such equa ions we e o me ly only sol ed o e algeb aically closed ields (o cha ac e -
is ic ze o and posi i e cha ac e is ic). Ou me hod can be applied o simila ypes o Diophan ine
equa ions, as well.
2005 Published by Else ie Inc.
MSC: 11D59; 11Y50; 11R58
Keywo ds: Thue equa ions; Global unc ion ields
1. In oduc ion
Classical Diophan ine equa ions like Thue equa ion (c . Thue [10]) a e adi ionally
sol ed o e he ings o a ional in ege s o o e he ing o in ege s o a numbe ield
*Co esponding au ho .
E-mail add esses: [email p o ec ed] (I. Gaál), [email p o ec ed] (M. Pohs ).
1Resea ch suppo ed in pa by G an s T 037367 and T 042985 om he Hunga ian Na ional Founda ion o
Scien i ic Resea ch.
2Resea ch suppo ed by he Deu sche Fo schungsgemeinscha .
0022-314X/$ – see on ma e 2005 Published by Else ie Inc.
doi:10.1016/j.jn .2005.10.009
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(c . Bake [1]). Se e al au ho s conside ed he analogous p oblem o e unc ion ields:
mo e exac ly o e he ing o in ege s o a unc ion ield o e an algeb aically closed ield,
see, e.g., [6,8], bo h in case he g ound ield is o ze o o posi i e cha ac e is ic. These
esul s ha e also common gene aliza ions (c . Gy˝
o y [5]).
Ou pu pose is now o in es iga e his p oblem in unc ion ields o e a bi a y ini e
ields. I is well known ha Diophan ine equa ions o e ini e ields play an impo an ole
in c yp og aphy, c . Niede ei e and Xing [7].
Also, om a p ac ical poin o iew his case is much mo e s aigh o wa d han he
case o algeb aically closed g ound ields (which mos ly occu in heo y only).
As i will u n ou , also in his case he uni equa ion plays a c ucial ole and he so-
lu ions can be compu ed easily. Howe e , o cons uc ing he app op ia e unc ion ields,
pe o ming calcula ions in hem, de e mining heigh s, e c. we in ensi ely use he compu e
algeb a package KASH [2].
2. Global unc ion ields
In he ollowing we shall s ongly ely on he a gumen used by Mason [6] o unc ion
ields o e algeb aically closed ields. We show how his ideas can be ans e ed o ou
si ua ion. A gene al desc ip ion o p ope ies o unc ion ields (also o e ini e ields) can
be ound in he book o S ich eno h [9].
We in oduce some no a ions. k=Fqdeno es a ini e ield wi h q=pdelemen s. The
a ional unc ion ield o kis k( ) as usual, and Kis a ini e ex ension o k( ) o deg ee n0
and genus g0. The in eg al closu e o k[ ]in Kis deno ed by oK. We assume ha K
is sepa ably gene a ed o e k( ) by an elemen ybelonging o oKand ha kis he ull
cons an ield o K. Any elemen ∈Khas a unique p esen a ion
=
n0

i=1
hiyi−1,h
i∈k( ).
Conjuga es o elemen s ( ields) a e deno ed by uppe case indices. Le A:=
((y(j))i−1)1⩽i,j⩽n∈Kn×nha e de e minan D. We no e ha Dis he disc iminan o y.
I is nonze o since Kis sepa ably gene a ed. We ob ain he sys em o linea equa ions:
 (1),..., (n)=(h1,...,h
n)A.
Hence, he hia e a ional unc ions in he (j),(y(j))i−1.
The se o all (exponen ial) alua ions o Kis deno ed by V, he subse o in ini e al-
ua ions by V∞. By abuse o no a ion we do no dis inguish be ween places and alua ions.
Fo example, we w i e deg o he deg ee o he di iso belonging o he alua ion ∈V.
Fo a nonze o elemen ∈Kwe deno e by ( ) he alue o a . Fo in eg al elemen s
his is he highes powe o he di iso belonging o ha di ides he di iso ( ), and his
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concep is ex ended o a ional elemen s in he usual way. Fo he no malized alua ions
N( ) = ( ) ·deg he p oduc o mula holds:

∈V
N( ) =0∀ ∈K {0}.
The heigh o a nonze o elemen o Kis de ined o be
H( ):= 
∈V
max0,
N( ).
Because o he p oduc o mula his is an amoun o
H( )=−
∈V
min0,
N( )
which hen holds o all elemen s o Kincluding 0.
3. Uni equa ions
Le V0be a ini e subse o V. Then he nonze o elemen s γ∈Ksa is ying (γ) =0
o all /∈V0 o m a mul iplica i e g oup in K. These elemen s a e called V0-uni s.Fo
V0=V∞ he V0-uni s a e jus he uni s o he ing oK.
The esolu ion o Thue equa ions (as well as se e al o he ypes o classical Diophan ine
equa ions) is usually educed o equa ions o he o m
γ1+γ2+γ3=0(1)
whe e he γia e V0-uni s o a sui able se V0.
The c ucial inequali y o Mason on he abo e equa ion becomes in ou case:
Lemma 3.1. Le V0be a ini e subse o Vand le γi(1⩽i⩽3)be V0-uni s sa is ying (1).
Then ei he γ1
γ3is in Kpo i s heigh is bounded:
Hγ1
γ3⩽2g−2+
∈V0
deg . (2)
P oo . The p oo is along he lines o he p oo o Lemma 2 in [6] (see [6, p. 14]). Since in
ou case he ield o cons an s kis no algeb aically closed we encoun e a ew addi ional
di icul ies which we will poin ou in wha ollows.
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We may assume ha :=γ1/γ3and he e o e γ2/γ3=− −1 do no belong o k.Le
V1:= ∈V: ( ) < 0,V
2:= ∈V: ( ) > 0,
V3:= ∈V: ( ) =0∧ ( +1)>0.
These a e disjoin subse s o he se V0. Then we ha e
H( )=
∈V1− ( )deg =
∈V2
( )deg =
∈V3
( +1)deg .
This is ue because o 1 =(1+ )+(− ), he p oduc o mula, and he p ope y ( ) < 0
⇔ (1+ )<0. I zdeno es a p ime elemen o he alua ion ∈V hen he di e en ial
1d sa is ies (1d ) = (d /dz) (see [9, Chap e IV]). I is no a p h powe hen he
di iso 1d is canonical, i.e., deg(1d ) =2g−2, since he ield ko cons an s is comple e
(see [9, Chap e I.5]). This yields
2g−2=
∈V
(d )deg =
∈V0
(d )deg =
∈V1∪V2∪V3
(d )deg
⩾
∈V1∪V2 ( ) −1deg +
∈V3 (1+ )−1deg
=−H( )+H( )−
∈V1∪V2
deg +H( )−
∈V3
deg
⩾H( )−
∈V1∪V2∪V3
deg ⩾H( )−
∈V0
deg
whence he asse ion ollows. 2
Taking Φ=−γ1/γ3,Ψ=−γ2/γ3Eq. (1) gi es he uni equa ion in wo a iables
Φ+Ψ=1(3)
whe e Φ,Ψ a e V0-uni s. Because o cha ac e is ic p he numbe o solu ions o such a
uni equa ion can be in ini e.
Fo example, i V0is jus he se o in ini e alua ions and η,1−ηa e bo h uni s o oK
hen also ηκ,(1−η)κis a solu ion o (3) o e e y exponen κ=p. Hence, he e exis
solu ions o a bi a y la ge heigh s in his si ua ion.
The subsequen lemma shows ha o any ini e subse V0o V, he g oup o V0-uni s
o Kcon ains only a ini e numbe , say s,o V0-uni s ηwhich a e no p h powe s and o
which also 1 −ηis a V0-uni . We deno e he se o hese uni s by {η1,...,η
s}.
Lemma 3.2. Le V0be a ini e subse o V. Assume ha a V0-uni Φin Kis a solu ion
o (3).I Φis no a p h powe o ηi(1⩽i⩽s,  ∈Z⩾0) hen Φbelongs o a ini e subse
o Kwhich can be calcula ed.
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Fo he p oo we e e o [6, Lemma 11, p. 98]. The p oo s a s by assuming ha Φis
no a p h powe in Kand he e o e also p o ides he means o calcula e he ηi.
4. Applica ion o Thue equa ions
4.1. P elimina ies
We wan o apply hese esul s o ( ela i e) Thue equa ions o e K.Le
F(X,Y):=
n

i=0
AiXn−iYi∈oK[X, Y ]
be a bina y homogeneous o m o deg ee a leas 3. Wi hou loss o gene ali y (c . [4,
p. 20]) we can assume ha Fis monic in X, i.e., A0=1. The polynomial F(X,1)∈oK[X]
is equi ed o be sepa able and i educible. Then o a bi a y m∈oK he equa ion
F(x,y) =min x,y ∈oK
is called a Thue equa ion (o e K). Deno e by αa ze o o F(x,1)in K,le L=K(α) and
oL he in eg al closu e o k[ ]in L. Assume ha kis he ull cons an ield o L, oo. I
x,y ∈oKis a solu ion o he Thue equa ion hen
F(x,y) =NL/K (x −αy) =m. (4)
Deno e by γ(j) (j =1,...,n) he conjuga es o any γ∈Lo e K. Assume ha
(x, y) ∈o2
Kis a solu ion o (4). Then β=x−αy is o no m m, ha is βcan be ep e-
sen ed in he o m
β=x−αy =µ·η(5)
whe e ηis a uni in Land µis an elemen o a ini e se So non-associa ed elemen s o L
o no m mo e K. Fo he solu ion o he co esponding no m equa ion we use he usual
me hods om algeb aic numbe heo y, i.e., calcula e sui able S-uni s [3]. Those, oge he
wi h Di ichle ’s uni heo em (c ., e.g., [11]), can be easily ans e ed o he unc ion ield
case, oo.
Deno e by η1,...,η
a se o undamen al uni s in L( ha can be calcula ed by he
compu e algeb a sys em KASH [2]). Se ing k∗=η0, he e a e in ege exponen s
a0,a
1,...,a
such ha
β=x−αy =µ·ηa0
0·ηa1
1···ηa
.(6)

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Fo ixed dis inc i, j, k (wi h 1 ⩽i, j, k ⩽n)se Lij k =L(α(i),α(j),α(k))wi h genus g.
Deno e by V0a ini e se o alua ions o Lij k con aining he in ini e alua ions and such
ha
α(i) −α(j)=0,
α(j) −α(k)=0,
α(k) −α(i)=0i /∈V0
and
µ(i)=0,
µ(j)=0,
µ(k)=0i /∈V0.
Siegel’s iden i y (holding i ially o any solu ion, see [4, Chap e 3]) gi es
α(i) −α(j)β(k) +α(j) −α(k)β(i) +α(k) −α(i)β(j) =0.(7)
By he undamen al Lemma 3.1
τij k =(α(j) −α(k))β(i)
(α(i) −α(j))β(k)
is ei he o bounded heigh o is con ained in Lp
ij k . In he ollowing he heigh unc ion is
applied always in Lij k .
4.2. E ec i e uppe bounds o he solu ions o Thue equa ions
In case Eq. (4) has only ini ely many solu ions we de i e an uppe bound o he
heigh s o he solu ions. I he equa ion has only ini ely many solu ions, hen he e mus
be i, j, k such ha τij k is no a p h powe in Lij k . We keep he abo e no a ion and se
A=max(H(α(i),H(α(j)), H (α(k))).
Theo em 4.1. I τij k is no a p h powe , hen Eq. (4) has only ini ely many solu ions and
o all solu ions (x, y) we ha e
maxH(x),H(y)⩽11A+1
nH(µ)+4g−4+2
∈V0
deg .
P oo . Applying Lemma 3.1 we ge
H(τij k )⩽2g−2+
∈V0
deg =c1.
This implies
Hβ(i)
β(k) =Hx−α(i)y
x−α(k)y⩽H(τij k )+Hα(i) −α(j)
α(j) −α(k) ⩽c1+4A=c2.
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Using an a gumen o Mason [6, Chap e II.1] o y=0weha e
x
y=α(k)β(i)/β(k) −α(i)
β(i)/β(k) −1
whence
Hx
y⩽2A+2c2.
By
yn=µ
n
h=1x
y−α(h)
we de i e
nH(y) ⩽H(µ)+nHx
y+A
whence he asse ion ollows o y. The bound o xcan be ob ained simila ly. 2
4.3. An algo i hm o calcula ing he solu ions o Thue equa ions
We now u n o inding he solu ions o Eq. (4).
Case I. Conside i s he case when τij k is o bounded heigh . Simila ly as in he p oo o
Theo em 4.1 we ob ain
Hx−α(i)y
x−α(k)y⩽H(τij k )+Hα(i) −α(j)
α(j) −α(k) ⩽c1+Hα(i) −α(j)
α(j) −α(k) =c
2.(8)
By (6) we ha e
x−α(i)y
x−α(k)y=µ(i)
µ(k) η(i)
1
η(k)
1a1
···η(i)
η(k)
a
whence using (8) we ob ain
Hη(i)
1
η(k)
1a1
···η(i)
η(k)
a ⩽c
2+Hµ(k)
µ(i) =c3.
This means o any in ini e alua ion o Lij k we ha e
a1· η(i)
1
η(k)
1+···+a · η(i)
η(k)
⩽c3.
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No e ha by in e changing iand kwe ge he same exp ession on he le -hand side wi h
opposi e sign: o his eason he inequali ies a e also alid wi h absolu e alues:

a1· η(i)
1
η(k)
1+···+a · η(i)
η(k)

⩽c3.(9)
No e ha he uni s in he abo e o mula ha e ze o alues a ini e alua ions. The in-
equali ies o ype (9) (ob ained o di e en choices o i, k) can be used o de e mine all
possible alues o he exponen s a1,...,a
.
Fo any possible exponen ec o a1,...,a
we can de e mine η=ηa1
1···ηa
in (5).
Then he sys em o equa ions
x−α(1)y=µ(1)·η(1),x−α(2)y=µ(2)·η(2)
can be used o de e mine he co esponding x,y.
Case II. I in (7) we ha e
τij k =(α(j) −α(k))β(i)
(α(i) −α(j))β(k) ∈Lp
ij k ,
hen using (5) we ob ain
(α(j) −α(k))µ(i)
(α(i) −α(j))µ(k) ·η(i)
η(k) ∈Lp
ij k .
He e he las e m is a uni in Lij k hence o any ini e alua ion o Lij k
(α(j) −α(k))µ(i)
(α(i) −α(j))µ(k) 
mus be di isible by p. This usually does no hold and he e is no Case II solu ion. O he -
wise, τij k is a p h powe , say τij k =ψp
ij k , we eplace τij k by ψp
ij k and epea he a gumen .
Rema k. In he abo e calcula ions se e al elemen s (e.g., α(i) −α(j)) a e con ained in
sub ields o ype Lij =K(α(i),α(j))o Lij k . Since o elemen s in Lij he alues a any
alua ion o Lij k can be easily calcula ed om he alues o he co esponding alua ions
o Lij , hence in ac almos all calcula ions can be pe o med in he sub ields Lij which
a e much easie o deal wi h, especially o la ge deg ees n.
5. Examples
Example 1. In he i s example we do no need o apply he undamen al lemma.
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Le k=F5,le K=k( ),le αbe a oo o
y3+ 7+ =0
and le L=K(α). Conside he Thue equa ion
NL/k( )(x −αy) =1inx,y ∈k[ ].(10)
Deno e by α(1),α(2),α(3) he conjuga es o α. Using symme ic polynomials we ha e
α(3)=−α(1)−α(2)
and subs i u ing i in o α(1)α(3)+α(2)α(3)+α(1)α(2)=0 we ob ain
α(2)2+α(1)α(2)+α(1)2=0
whence
α(2)=4α(1)±α(1)√−3
2=34α(1)±α(1)√2.(11)
Obse e ha √2 is con ained in F25, a quad a ic ex ension o K, hence in his case
M=Lα(1),α(2),α(3)=F25( )(α).
Deno e by wa gene a ing elemen o he mul iplica i e g oup F∗
25 o F25 wi h
2=w6,3=w18,4=w12.
By (11) we ha e
α(2)=w16α(1),α
(3)=w8α(1).
Siegel’s iden i y ge s he o m
w8x−α(1)y+x−α(2)y+w16x−α(3)y=0.(12)
In ou case Mhas one in ini e alua ion. In he abo e equa ion all e ms a e uni s ha -
ing ze o alues a all ini e alua ions. By he p oduc o mula hei alue a he in ini e
alua ion is also 0, hence hey a e con ained in he cons an ield F25.
Equa ion (12) leads o he uni equa ion
w4x−α(1)y
x−α(3)y+w20 x−α(2)y
x−α(3)y=1
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γ1=α(2)−α(3)x−α(1)y,
γ2=α(3)−α(1)x−α(2)y,
γ3=α(1)−α(2)x−α(3)y,
hen we ha e
γ1+γ2+γ3=0.(26)
The ield Mhas genus 13. I has eigh in ini e alua ions, all o deg ees 1. The x−α(i)y
a e uni s, ha ing nonze o alues only a he in ini e alua ions. The quo ien s
α(1)−α(2)
α(1)−α(3),α(2)−α(3)
α(2)−α(1),α(3)−α(1)
α(3)−α(2)
ha e nonze o alues all oge he a ou ini e alua ions, wo o hem being o deg ee 4,
he o he wo o deg ee 6. Deno e by V0 he se o he eigh in ini e and hese ou ini e
alua ions. Then by he undamen al Lemma 3.1 γi/γjis ei he o bounded heigh , o is
con ained in M5.
Case I. Assume
Hγ1
γ3⩽2·13 −2+(8+12 +8)=52.
This implies
Hx−α(1)y
x−α(3)y⩽52 +Hα(1)−α(2)
α(2)−α(3)=65.(27)
As we men ioned abo e, Khas uni ank 3. We deno e by ε1,ε2,ε3 he undamen al uni s.
Conside ing he alues o
ε(i)
h
ε(k)
h
(a in ini e alua ions) o h=1,2,3, by (27) and (9) we become
x−αy =µ·εa1
1·εa2
2·εa3
3
wi h a oo o uni y µin kwhe e among o he s he exponen s sa is y
|50a1+6a2+54a3|⩽65,
|51a1+5a2+54a3|⩽65,
|49a1+3a2+54a3|⩽65.

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The e a e abou 8000 solu ions (a1,a
2,a
3)o he abo e sys em o linea inequali ies.
Tes ing all possible exponen ec o s we ound ha Eq. (24) has only he i ial solu-
ions (x, y) =(1,0), (2,0), (3,0), (4,0)( hese yield in ac he mul iplies o x−αy o
x=1,y=0 wi h oo s o uni y in k).
Case II. To exclude γ1
γ3∈M5we conside
γ1
γ3=α(2)−α(3)
α(1)−α(2)·x−α(1)y
x−α(3)y.
The second e m on he igh -hand side is a uni , hence
α(2)−α(3)
α(1)−α(2)
should be di isible by 5 a all ini e alua ions . This is no sa is ied, howe e . (Simila ly
o γ1/γ2and γ2/γ3.)
Compu a ional expe iences. All compu a ions used in he examples we e pe o med by
using he compu e algeb a sys em KASH [2], unning on 1 GHz PC-s. The calcula ions
ook jus some seconds wi h he excep ion o he es o abou 8000 possible exponen
ec o s in Example 4 which ook abou 90 minu es.
Acknowledgmen
The au ho s a e hank ul o he e e ee o his/he aluable ema ks ha lead o an im-
p o emen o he pape .
Re e ences
[1] A. Bake , T anscenden al Numbe Theo y, Camb idge Uni . P ess, Camb idge, 1990.
[2] M. Dabe kow, C. Fieke , J. Klüne s, M. Pohs , K. Roegne , K. Wildange , KANT V4, J. Symbolic Com-
pu . 24 (1997) 267–283.
[3] C. Fieke , Übe ela i e No mgleichungen in algeb aischen Zahlkö pe n, PhD hesis, Be lin, 1997.
[4] I. Gaál, Diophan ine Equa ions and Powe In eg al Bases, Bi khäuse Bos on, Bos on, 2002.
[5] K. Gy˝
o y, Bounds o he solu ions o no m o m, disc iminan o m and index o m equa ions in ini ely
gene a ed in eg al domains, Ac a Ma h. Hunga . 42 (1983) 45–80.
[6] R.C. Mason, Diophan ine Equa ions O e Func ion Fields, Camb idge Uni . P ess, Camb idge, 1984.
[7] H. Niede ei e , C. Xing, Ra ional poin s on cu es o e ini e ields, in: London Ma h. Soc. Lec u e No e
Se ., ol. 285, Camb idge Uni . P ess, Camb idge, 2001.
[8] W.M. Schmid , Thue’s equa ion o e unc ion ields, J. Aus al. Ma h. Soc. Se . A 25 (1978) 385–422.
[9] H. S ich eno h, Algeb aic Func ion Fields and Codes, Sp inge , Be lin, 1993.
[10] A. Thue, Übe Annähe ungswe e algeb aische Zahlen, J. Reine Angew. Ma h. 135 (1909) 284–305.
[11] E. Weiss, Algeb aic Numbe Theo y, New Yo k, 1963.