C OMPOSITIO M ATHEMATICA
D INESH S. T HAKUR
On characteristic p zeta functions
Compositio Mathematica, tome 99, n
o3 (1995), p. 231-247
<http://www.numdam.org/item?id=CM_1995__99_3_231_0>
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On Characteristic p Zeta Functions
Dedicated to Bhaikaka on his 75th
Birthday
DINESH S. THAKUR*
Department of Mathematics, University of Arizona, Tucson, AZ 85721
e-mail: thakur@math.arizona.edu
Received 18 May 1994; accepted in final form 30 August 1994
© 1995 Kluwer Academic Publishers. Printed in the Netherlands.
Abstract. We show that in contrast to the various known analogies, the orders of vanishing of the
characteristic p zeta functions introduced by Goss sometimes follow interesting pattems involving
base q-digits, providing a challenge to understand them in a general framework as in the classical
case. We answer some questions raised by Goss about the connection of the zeta values and periods.
We also generalize and simplify the proofs of some results.
1. Introduction
It is known in various contexts, that the order of
vanishing
and theleading
term ofzeta function have a lot
of interesting
arithmetic information encoded in them. The characteristic p zeta functions studiedby
Carlitz and Goss are noexception.
Theirvalues relate to the
periods,
the divisibilities of Bernoulli numbersanalogues arising
from them
give
information on class numbers ofcyclotomic
extensions in a wayanalogous
to the classical case andthey
are related to the classical L-functions via congruence formulae. Animportant ingredient missing
is that there is no functionalequation
insight,
eventhough
there areanalogues
of gamma functions and Gausssums. In the classical case, the gamma factors control the orders of
vanishing
ofthe zeta functions at the
negative integers.
Using
the congruence formula with the classicalL-functions,
Goss hasprovided [G4]
lower bounds for orders ofvanishing
of these zeta functions at thenegative integers
and has raised aquestion
whetherthey
are exact. We show(Section 3)
thatthere can be extra
vanishing,
even when these lower bounds are naiveanalogues
of the exact orders of
vanishing
in the classical case. Thepatterns involving
theq-digits
which control these extra zeros are aninteresting phenomenon giving
us achallenge
to put these zeros in ageneral framework,
as is doneclassically.
In the other
sections,
we answer somequestions
about the connection of the zeta values at thepositive integers
with theperiods
and refine some results of Goss.We also
give
altematesimple proofs
for some of the results in the literature.* Supported in part by NSF grants DMS 9207706 and DMS 9314059
Let us
briefly
recall the classical case. If .F is a numberfield,
the Dedekindzeta function
03B6F(s)
e C isdefined,
for s ~ C with realpart greater
than1,
as
03A31/Norm(I)s
=03A0(1 - 1/Norm (P)s)-1,
where the sum is over all non-zero ideals I of the
ring
ofintegers
of F and theproduct
is over theprimes
P.(For
F =Q,
this is the Riemann Zetafunction).
This function hasmeromorphic
continuation to the whole
complex plane
withonly pole
at s = 1 and has asimple
functional
equation connecting 03B6F(s)
to03B6F(1 - s). Clearly,
for s >1,
there areno zeros and hence
analyzing
the factors in the functionalequation (essentially
gamma
functions),
we see that atnegative integer
s, the zeta function vanishesto order r 1 + r2
(r2 respectively)
if s is even(odd respectively).
Inaddition,
ifF is
totally real,
then values atnegative integers
are rational orequivalently, by
functional
equation,
for s apositive
eveninteger, ((F( s)/ (27rÍ)rl)2
eQ.
For a function field over a finite field
Fq,
acomplex
valued zeta function wasintroduced and studied
by
Artin andWeil, by thinking
of the norm as the numberof residue classes and
using
similar definitions. It has functionalequation
andinteresting theory
ofspecial values,
but it tums out to be a rational function ofq-S
and we do not have connection to
203C0i,
forexample.
If we want to consider the norm in the usual relative sense
instead,
as was doneby
Goss,(and
in thesimplest
caseby Carlitz),
we have to deal with the fact that there are neithernaturally given
infiniteplaces
orrings
ofintegers
nor is therea
naturally given
base field(like Q).
Even the rational functionfield,
which canbe considered as a
subfield,
sits in the function field in various different ways.Even when you choose the base and the infinite
places,
the norm of an ideal is anideal in the base and it may not be
principal,
if the base has class number greater than 1 and even if it isprincipal,
there may be several units toconsider,
toget
awell-defined
generator
out of it.Finally,
once we get anelement,
we can raise it toan
integral
poweronly,
not to anyarbitrary complex
power. Thissuggests
eitherrestricting
to a class number 1 base withonly
oneplace
atinfinity (as
is the case forQ
or animaginary quadratic field)
so that the unit group is a finite group or moregenerally restricting
to the extensions for which the norms areprincipal
ideals ordefining
idealexponentiation.
This boils down tochoosing representatives
of theideal classes of the base and
dealing
with vector(of length
classnumber)
valuedzeta functions. We will discuss these in the next section.
The main reference is
[G4]
where the reader can find anexposition
of variousresults and additional references. See also
[T2].
We will assume somebackground, especially
in the Section4,
on the relevanttheory
of Drinfeld modules of rank one.For
this,
see[Hl], [H2]
or the summary in[T3],
whose notation we will follow.Otherwise,
we have tried togive
a self-containedexposition
of the issues mostrelevant to this paper.
2. Zêta Values and Zeros
DEFINITIONS. Let K be a function field of one variable with field of constants
Fq
of characteristic p, oo be a
place
of K ofdegree b, K~
be thecompletion
of K at o0and A be the
ring
of elements of Khaving
nopole
outside oo. We take!1 to be thecompletion
of analgebraic
closure ofK~.
Let us also denoteby h
the class number ofK,
so that the class number of A is hb. Fix asign
function sgn :K*~~F*q03B4.
(In examples,
weusually
take sgn such that x, y, t have sgn1).
Now let S be aset of
representatives
ofF;6/F;.
LetA+ : = {a
e A :sgn(a) ~ S}. (This
way ofdealing
withsigns,
when b >1,
wassuggested
in[TI]
p.46).
Elements of sgn = 1 are called’positive’
or ’monic’.By deg
we will denote the residuedegree
over
Fq. By
conventiondeg(O)
= -~. LetAi := {a
E A :deg(a)
=il
andAi+ := {a
EA+ : deg(a)
=i}.
Let H be the Hilbert class field of Ai.e.,
themaximal abelian unramified extension of K in which oo
splits completely.
Recallthat K = H if hb = 1. Let L be a finite
separable
extension of K and letOL
= 0denote the
integral
closure of A in L.Now we define the relevant zeta functions.
For s E
Z,
define the ’absolute zeta function’:A
priori,
the second definition makes senseonly
forspositive,
as the terms thentend to zero
(and
also for s = 0 when((s)
=1),
but we will show below that itworks for for s a
negative integer
also and infact,
then03B6(s)
E A. The results on thespecial
values and the connection with Drinfeld modules mentioned in the paper later on,justifies
the use of elements rather than ideals even when the class number is more than one.Also,
we can look at a variant where a runsthrough
elements ofsome ideal of A or we can sum over all ideals
by letting s
to be amultiple
of classnumber and
letting
a’ to be thegenerator
of I’ withappropriate sign.
In the latter case, we have Eulerproduct
for such s, if 6 = 1.If L contains H
(note
that this is no restriction if hb =1,
e.g. for A =Fq[t]),
then it is known that the norm of an idéal 2’ of 0 is
principal.
Let NI denote thegenerator whose
sign
is in S. Let us now define the relative zeta functions in this situation.Put,
for s E ZThe same remark as above
applies
to the second definition.Finally,
we define the vector valued zetafunction,
whichgeneralizes
both def-initions above and works without
assuming
that L contains H. We will leave thesimple
task ofrelating
these definitions to the reader.Let
Cn (1 n h6)
be the ideal classes of A and choose an idealIn
inCn 1.
For s E
Z,
define the vectorZn (s, X) by defining
its nthcomponent ZO(s, X)n
via
(Here In
NI stands for thegenerator
of this ideal withsign
inS).
Again
the same remark as aboveapplies
to the second definition.Also note that Z
depends
verysimply
on the choice ofIn’s,
with components ofZ,
for different choicesof In’s, being
non-zero rationalmultiples
of each other.In
particular,
thequestions
of when it(i.e.,
all thecomponents) vanishes,
when itis rational or
algebraic
etc. areindependent
of such choice.REMARKS. Goss
[G4]
defines the zeta functions for abigger
spaceusing
abigger piece
of the character space to define ns . Since the results we are interested in dealonly
with sintegral,
for us X is an indeterminate and we dealonly
withintegral
s.Also note that Goss’ function on the
bigger
space has many more zeros. We leave the task ofrelating
our definitions to Goss’ to the reader. 0 We use X as a deformation parameter for otherwisediscretely
defined zetavalues and hence we define the order of
vanishing ords
of(, 03B6O
orZo
at s to bethe
corresponding
orderof vanishing of 03B6(s, X), 03B6O(s, X)
orZo (s, X)
at X = 1.This
procedure
isjustified by
the results described below. It is easy to see that the order ofvanishing
is the same for s and ps, as the characteristic is p.Now we state and prove a theorem which is
essentially
due to Lee[L1 ]. (See [T2]
for the relevant
history).
For anon-negative integer
k= 03C3kiqi,
with 0~ ki
q,we let
l(k) : = E ki,
hencel(k)
is the sum of base qdigits
of k.THEOREM 1. Let W be a
Fq -vector
spaceof
dimension d inside afield (or ring)
0
over Fq.
Letf
e 0 - W.If d
>l(k)/(q - 1),
thenEwcw(f
-f-w)k
= 0.Proof. (Compare [L1]).
Let 03C91,···03C9d be aFq -basis
of W. Then( f
+03C9)k
=( f
+B1 w
+ ··· +03B8d03C9d)k, 03B8i
eFq.
When youmultiply
out the kbrackets,
terms involve at most k of03B8i’s,
hence if d > k the sum in the theorem is zero, since we aresumming
oversome Oi
a term notinvolving
it and q = 0 in characteristic p.The next observation is that in characteristic p,
( a
-f-b)k
=11(aqi + bqi)ki,
hencethe sum is zero, if d >
l(k), by
the argument above.Finally,
note that¿BEFq (Jj
= 0unless q -
1divides j. Expanding the
sumabove by
the multinomialtheorem,
weare
summing multiples
ofproducts Bit... Bdd
and hence the sum is zero; because the sum of theexponents being l(k) (q - 1)d,
all theexponents
can not bemultiples
of q - 1. DFor the
applications,
note that{a
e A :deg(a) il
form suchFq
vectorspaces whose dimensions are
given by
the Riemann-Roch theorem.Similarly, Ai+
are made up of affine spaces as in the theorem. Hence
THEOREM 2
(Goss [G1]). For a negative integer s, 03B6(s) ~ A and 03B6((q - 1)s) =
0.
Proof:
Our remark above shows thatby
the Theorem1,
the sum over i is a finitesum, hence the first statement follows. Now if k
= -(q - 1)s,
thenwhere the first
equality
holds since¿OEFq 03B8q-
1 = -1 and the secondequal- ity
is seenby using
that the sum is finite andapplying
the Theorem 1 withW :=
ta
e A :deg(a) m}
for somelarge
m. 0REMARKS. Goss’
proof
of the firstpart
used a weaker version of the Theorem1,
with
d > k/(q - 1),
thesimpler
version d > k mentioned in theproof
would alsosuffice. Goss’
proof
of the second part involved an inductive argument. Theproof given
here was mentioned in[T2]
in thesimplest
case dealt there. It isinteresting
to note
(as
detailed in[T2])
thatthough
Carlitz and Lee almostproved this, nobody
before Goss seems to have dealt with
exactly
the zeta sum atnegative integer
as
such, though
Carlitz[Cl]
obtained results forpositive integers. (See below).
Since the zeta sums tum out to be finite and since a’
interpolates by
Fermat’s littletheorem,
we have[Gl] p-adic interpolation
for zeta values atnegative integers,
once we remove the Euler factor at the
prime p
of A. DLet us now recall what we know of the values of the Riemann zeta
functions
at
integers.
There is asimple
functionalequation relating
values at s and 1 - s.At s =
1,
we have apole.
At evenpositive
s, Euler showed that(( s)
is a rationalmultiple
of irl or rather(27ri)’.
For oddpositive
s, we knowonly
that((3)
isirrational
(Apery).
Atnegative integer s, (( s)
is arational,
which is zero, if s iseven and nonzero if s is odd.
In our context,
multiples of q-1
are considered even and the rest are consideredodd,
since the cardinalities of A* and Z* are q - 1 and 2respectively.
There isan
analogue
îr of 27ricoming
from thetheory
of Drinfeld modules. There is noreasonable functional
equation
known.(We
do have(( spn) == (( S )pn though
trivially).
At s =1,
we have nopole (the
zeta sumconverges),
but at evenpositive
s,
modeling
Euler’sproof
in thesetting
of thetheory
of Drinfeldmodules,
Carlitzfor A =
Fq[t] (where
in fact hedeveloped
the relevanttheory
from scratch and also identified Bernoulli numbers and factorialsoccurring
in the zeta value andproved
manyanalogies
forthem)
and Goss ingeneral (for b
=1,
but it is trivialto remove this restriction as noted on
[T1]
pa.46)
showed that03B6(s)/s
E H. Forodd
positive values,
there are results in[A1], [A-T], [Y1]
etc. Atnegative integer
s,
(( s)
isinteger
in A(by
the Theorem2),
which is zero, if s is even. Hence thequestion
arises whether it is non-zero if s is odd. That itis,
wasproved by
Gossfor A =
Fq [t] using
a recursion formula for the values. See[G1].
It is not knownin
general.
THEOREM 3. Let b = 1 and s be a
negative
oddinteger.
Then(( s)
is non-zeroif
A has a
degree
one rationalprime p (e.g. if
g = 0(this
reprovesFq[t] case)
orif q
is
large compared
to the genusof E)
orif
h = 1.(Together they
take careof
thegenus 1
case). In fact,
inthe first case, (( s)
is congruent to 1 mod p.Proof.
The first part followsby looking
at the zeta summodulo p
and notic-ing
that03A303B8~Fq
0-s =0,
hence theonly
contribution is 1 from the i = 0 term.Apart from g
=0,
there areonly
4 other A’s with hô = 1(these
have nodegree
1
primes)
andthey
are listed in[Hl]
p. 213.For q
=2,
there are no odd inte-gers. That leaves
only
twopossibilities: (a)
A =F3[X, y]/y2
=x3 -
x - 1 and(b)
A =F4[x, y]/y2
+ y =X3
+(3.
In both the cases, we look at contribution to the zeta sum for each i modulo x : i = 0gives 1,
i =1, 2, 3 give
0 and forhigher i’s,
with a modulo x we also getBa, B
eFq
and hence the contribution is zero asbefore. 0
REMARKS. In many
particular examples
of A we can settle thenon-vanishing by
this method. If we
ask, instead,
for oddnegative
s’s for which(( s)
isalways (i.e.,
for any
A)
non-zero, the Theorem1,
says that(( s)
=1,
ifl(-s)
q - 1.Again,
this can be
pushed
farther: Forexample,
ifl(-s) 2(q - 1) (and
s odd nega-tive as
before),
thenby
the Theorem 1 and theFq [t]
case of Theorem 3(by taking
t to be an element of A of the lowest
positive degree),
we see that03B6(s)
is nonzero. ~It should be mentioned that there is a close connection known between divis-
ibility properties
of Bernoulli numbersarising
from the Riemann zeta values and the class groupsof cyclotomic
fields. For similarresults,
due to Goss,Okada,
Shuand
Sinnott, coming
from the valuesof (
at bothpositive
andnegative integers,
see
[G4]
and[T5].
Let us now turn to the relative zeta functions.
THEOREM 4 (Goss [G4]). Foranegativeintegers, (0 (s)
EA.Infact, ZO(8)n
EA for
every n.Proof. (Goss [G4]).
The idea is todecompose
the sum inFq-vector
spaces anduse the Theorem 1 to show that for
large k,
the k-th term of the zeta sum is zero. Itis
enough
to show that for each ideal class C ofO,
the contribution from Z e C iszero. Let
J
E C. ThenIJ-1
=(i)
CJ-1.
Hence it is sufficient to show thatLet ~1, ... , oon be the infinite
places
of L with relative residuedegrees fl,
... ,fn
and thecorresponding embeddings
al, ... , anrespectively. Then,
writ-ing
thepolar
divisoras -(i)~ = 03A3 ml(~l),
we have k =deg N(i)
= 6E mi fi.
We will show that parts of
Sk corresponding
to a fixed choice of{ml}
also vanish.Namely,
let D:= 03A3 ml(~l),
then we will show that if thedeg(D)
islarge,
then03A3N(i)s
=0,
where the sum is over S : ={i
EJ-1-{0}/units, -(i)~
=D}.
Once we fix a
polar divisor,
theambiguity
in i isonly through
a finitefield.
Hence, given
iE S,
fix arepresentative r(i) by demanding
that thesign
of03C31(r(i))
is1,
for some choice of thesign
inLOOI.
£ :=tj
EJ-1 : (j)+
03A3(ml-1)(~l) 0}
acts on Sby translation: j
or(i)
=T(i) +j.
This is an actionbecause as k is assumed
large,
this translation does notchange
the topdegree
andhence
polar
divisor or asign. Further,
if(jl
+r(i))/(j2
+r(i)
is aunit,
then sinceits
polar
divisor is zero, it has to be in a finitefield,
but then the choice of thesign
makes it 1 and hence
ji
=j2.
In otherwords,
there is noisotropy.
Now we show that the contribution of each orbit of S under this action is zero. In other
words,
we have to show thatE N (j
+r( i)) -s
=0,
wherenow the sum is
over j running though £,
which is aFq-vector
space. NowN( j
+r(i))-S
= yfl ul (j
+r(i))-S, where g
is the root ofunity
to take care ofthe
sign,
and can be taken out of the sum, as it is the same for allj.
Hence aslight
modification of the Theorem
1,
withTI(fl
+w)
in theplace
off
+ 03C9(see [G4] ),
we see that the sum is zero, when k is
large.
This finishes theproof.
DTHEOREM 5.
Foranegativeintegers, (0«q- 1)s)
=O.Infact, ZO((q-1)s)n
=0 for
every n.Proof. By
theprevious theorem,
the sum over idefining
the zeta value isfinite,
so the
proof
followsexactly
as in the Theorem2,
the fact that some norms getrepeated being
irrelevant. DREMARKS.
(a)
LetHl
be the class field of Kcorresponding
to K*tZ
xU+,
where t is an uniformizer at oo with
sgn(t)
= 1 andU+
is thesubgroup
of the idele groupconsisting
of those ideles with unitcomponents
at the finiteplaces
and withcomponent at o0 of sgn = 1
(see [T3]
for the altemate definition and summary of basicproperties
ofHl,
wejust
mention thatNi
1 = H when 6 =1).
In the case Lcontains
Hl,
this theorem follows from the results of Goss[G4]
described in thenext
section,
which use L-functions.(b)
Theproof
of the Theorem 4 can be used to getbounds,
similar to thoseobtained
by
the Theorem1,
on i for which the ith term of(0
can be nonzero. Forthe abelian
extensions, using
L-functionfactorization,
much better bounds followimmediately
from the Theorem 1.3. Orders of
Vanishing
Now we tum to the
question
of the orderof vanishing.
Goss[G4]
gave a very nice method to find a lower bound for orders ofvanishing,
when L containsHl
and mentioned as an openquestion
whetherthey
are exact. We will show thatthey
arenot and indeed that the
patterns
of extravanishing
arequite surprising
in terms ofthe established
analogies.
The main idea of Goss is to tum the
similarity
in the definition of our zeta function with the classical one into a double congruenceformula, using
the Teich-muller
character,
and to use theknowledge
of the classical L function Euler factors to understand the order ofvanishing.
This is done as follows.(See
the references mentioned in the introduction for the facts recalled from thetheory
of Drinfeldmodules).
Let p
be aprime
of A and let W be the Wittring
ofA/s.
The identificationW/pW ÉÉ A/s provides
us with the Teichmuller character w :(A/s)*~W*
satisfying wk(a
modp) = (ak
modp)
mod p.Now let
039B&
denote thep-torsion
of the rank one,sgn-normalised
Drinfeldmodule p
of generic
characteristic. Let L containHI
and let G be the Galois group ofL(039B&)
over L. Then G can bethought
of as asubgroup
of(A/&)*
and hencew can be
thought
of as a W-valued character of G. LetL(03C9-s, u)
eW(u)
be theclassical L-series of Artin and Weil in u :=
q-sm,
where m is the extensiondegree
of the field of constants of L over
Fq.
Let
S~ := {~j}
denote the set of the infiniteplaces
of L and letGjdenote
theGalois group of
Loo) (039B&)
overLoo).
ThenGj
CF*q.
Given s, letSS
CS~
be thesubset of the infinite
places
at which w-’ is an unramified character of G.Then SS
does not
depend
on p. PutTHEOREM 6
(Goss [G4]).
Let L containHl
and let s be anegative integer.
Then(o (s, X)
EA[X].
Sketch
of
theproof. Tracing through
thedefinitions,
the property of w men- tioned abovegives
the double congruenceformula L (w -’, Xm)mod
p =03B6O(s, X) mod p
forinfinitely
many p. But as the L function is known to be apolynomial,
the result follows. 0
This
immediately gives
thefollowing
lower bound for the order ofvanishing.
THEOREM 7
(Goss [G4]).
Let L containHl
and let s be anegative integer,
thenthe order
of vanishing of 03B6O (s)
is at leastREMARKS. Note that
VS depends
on sonly through
its value modulo q - 1 andVs [L : KI.
This isanalogous
to theproperties
of the exact orders ofvanishing
in the classical case, but not in our case as we will see below. D EXAMPLES:
(i)
L = K =Hl: (This
ispossible,
when h03B4 =1). Then Vs
= 0 or1, according
as s is odd or even. Gossproved
that for A = O =Fq[t],
the order ofvanishing
isVS.
(ii)
L istotally
real extension ofdegree
d of Ifcontaining Hl, i.e.,
oosplits completely
in L:VS
is 0 or daccording
as whether s is odd or even.(Note
that thebounds Vs
in(i)
and(ii)
are inanalogy
with the orders ofvanishing
in the classicalcase mentioned in the
introduction).
(iii)
L =K(039B&):Vs
=(qdeg(&) - 1)/(q - 1)
for all s.(iv)
L =Fqn(t)
and K =Fq(t) : VS
= 0 is s is odd and for s evenV,
=pk,
ifn = pkl,(l,p) = 1.
(v)
L =F5(-5)
and K =F5 (t): Vs
is 1 or 0according
as whether 2(not 4)
does or does not divide s. ~
First few calculations we made on situations
(ii)
to(v)
showedequality
ofthe bounds with the exact orders of
vanishing
in theexamples. (We
detail the dataafterwards).
But in thefollowing
situation of(i) (which
can bethought
of asspecial
case of
(ii) also),
we found that the bound is not exact:Let (a)
A =F2[x,y]/y2+y
=x3+x+1, (b) A
=F2[x,y]/y2+y = x5+x3+1.
In both these
situations,
hô = 1 and hence K =Hl.
For small s wecomputed 03B6(s, X) by
hand as follows: The genus of K is 1 and 2 for(a)
and(b) respectively.
Hence
by
the bounds obtained from the Theorem 2 and the Riemann-Roch theorem shows that for(a)
and(b), (( -1, X)
=1+0*X+(x+(x+1))*X2
=1+X2
=(1
+X)2.
Hence the order ofvanishing
is 2 for s = -2n. Similarsimple
handcalculation showed that the order of
vanishing is 1 if 2n ~ -s 9 for (a)
and is 1for s = -7 and 2 for other s
with -s 9
for(b).
As the pattern, if indeed there should be one, was not clear to us, more calcula- tions
(we
list the resultsbelow)
wereperformed, using
the method and bounds describedabove,
on macsymaby
Jennifer Johnson forexample (b),
first for- s fi 43
not divisibleby
2(without
loss ofgenerality).
This showed that within the range ofcomputation
the order ofvanishing
at s is 2 or 1depending
on whetherl(-s) 2 or
not. From this we could see thepattern
and prove thefollowing
theorem.
THEOREM 8.
If
b =1,
q = 2 and K ishyperelliptic,
then the orderof vanishing of (( 8)
atnegative integer
s is 2if l(-s)
g, where g is the genusof
KProof.
Let x E A be an element ofdegree
2 and let n be the first odd non-gap for A at oo. Then the genus g of K is seen to be(n - 1)/2.
LetSA(i)
denotethe coefficient of
X in (A (s, X).
Asimple application
of the Theorem 1 and the Riemann-Roch theorem shows that for i >1 (- s)
-I- g,SA(i)
= 0. Hencesince
1( -s) fi
g =(n - 1)/2.
On the otherhand,
as x hasdegree
1 inFq[x],
we have
SA(2i)
=6’p (z),
for0
i(n - 1)/2. Further, by
the Theorem1, SFq[x](i) = 0, if i > (n - 1)/2,
as(n - 1)/2 l(-s). Hence, 03B6A(s,X) =
03B6Fq[x] (s, X2).
Henceby
theexample (i) above,
the order ofvanishing
is 2 = 2 * 1as
required.
~REMARKS:
(i)
We do not need a restriction on the class number ofIiC,
so apart from(a)
and(b)
this falls outside the scope of the theoremgiving
the lower bounds.(ii)
Let A be of genus g, with b =1,
and with the gap structure at oo sothat 1(iqoo)
=i + 1, for 1
irand1((g+r)00)
=r + 1, rq g + r.
Then a
straightforward generalization
of theproof
of the theorem abovegives
thevanishing order q (the
lower boundVs
is1)
at even s such thatl(-s)/(q - 1)
r.For q >
2,
thesimplest
suchexample
would mean03B4 = 1, q = 3, r = 1, g = 5
andl(300)
=2, 1(6oo) -
3. I amobliged
to J. F. Voloch forproviding
thefollowing
concrete
example
of this situation:x2(x - 1)3 - y2
+yx 3+ y3 + y5
= 0 with o0being (1,0).
Infact,
for thisexample l(700)
=3,
so we can take r = 2. Othervariants of this
phenomenon
are alsopossible.
(iii) Analogies
between the number fields and the function fields areusually
the strongest for A =Fq[y].
But even in that case, here is anexample
of extra vanish-ing,
when A =F4[y]
and O =F4[x, y]/y2
+ y =x3
+(3:
Orderof vanishing
of03B6O
at s = -1 is 2(as
can be verifiedby
directcomputation,
forexample, by using
the remark
(b) following
the Theorem5)
whereas the lower bound of the Theorem7 is 1. D
We refer to the last section for some issues raised
by
this theorem.We do not know yet whether in these cases, the order of
vanishing
isVs, except
for the extra
vanishing exceptional
situation mentioned in the theorem. This is true in thecomputational
range and we have thefollowing partial
result(which
can
again
bepushed
furtherby
theproofs along
the sameline),
which shows that situation ingeneral
canonly
be worse as far as the pattern of zeros is concerned.PROPOSITION: For the
example (a) above,
the orderof vanishing of 03B6(s)
is 1(= Vs), if l(-s)
=2 or if -s - 0, 3, 5 or 6 mod
7.Proof.
We know that(( s)
= 0by
the Theorem 2. Hence we haveSince the genus is
1, by
the Riemann-Roch theorem and the Theorem1,
we seethat in the first case, the terms of the last sum vanish if i >
1,
and hence the sumis 1 + x-s +
( x
+1) -s ~
0 as claimed. For i >1,
the terms of the second sum are all zero modulo y(as they
can bedecomposed
as sums overpairs
b and b +y),
hence the sum is congruent to 1 + x-S +
(x
+1)-s
modulo y, which takes care ofthe second case. o
We end this section
by listing
thecomputational
results mentioned above. Let0,
denote the orderof vanishing
of the relevant zeta function at s. Note that in(c), (d)
and(e) below,
it agrees withVs .
(a)
For theexample (a) above,
for 0 -s32, 0 s
is 2 ifl(-s)
1 and 1otherwise.
(b)
For theexample (b) above,
and for A =F2[s, y]/y2+y=x7+x+1
andfor A =
F2[x, y]/y2 + y = x9 + x + 1 respectively,
for 0 -s65,
snot divisible
by
2(without
loss ofgenerality), Os
is 2 ifl(-s)
2 or 3 or 4respectively
and 1 otherwise.(c)
A =F3[x, y]/y2
=X3 _ x -
1.OS
for 0 -s 28 is 1 if s is even and 0 otherwise.(d)
A =F3[t]
with O =F3[t, y]/y2 = t3 - t -
1 and O =F3[t] respectively,
for 0 -s
31, OS is 1.
(e)
A =F3[t]
with (1 =F9[t],
for 0 -s179, Os
is 1 if s is even and 0otherwise.
4. Values at Positive
Integers
We now look at the values of the zeta functions at the
positive integers.
The results obtained areanalogues
of the classical results with the basebeing Q (with
27rias the relevant
period)
or animaginary quadratic
field(with
the relevantperiod being
theperiod
ofelliptic
curve withcomplex multiplication by
the field. In this second case, the similar results for the absolute and relative zeta values may beknown,
but we do not know a suitablereference).
For the necessarybackground
on Drinfeld modules and the
explicit
class fieldtheory
based on them see thereferences mentioned in the introduction. Notice that since there is no functional
equation
knownlinking
the values atpositive
andnegative integers,
these seemto be two distinct theories. Let 7r be a fundamental
period
of asgn-normalized
rank one Drinfeld A module p with the lattice
corresponding
to pbeing
3tA.(This
determines -k up to
multiplication by
an element ofF*q).
This is ananalogue
of 203C0i.As an
analogue
of Euler’s theorem wehave,
THEOREM 9. Let s be a
positive
even(i.e.,
amultiple of
q -1) integer,
then(A(s)/ifs
EH1.
REMARK. Much more
precise
version of this theorem wasproved by
Carlitz for the case A =Fq [t], using (what
is now understoodas) Fq[t]
Drinfeld module. Gossproved
the Theorem 9 under the additionalhypothesis 6
= 1using
the Drinfeld A modules ingeneral,
withessentially
the same method. The full Theorem 9 wasproved
in[Tl],
with theonly
additionalingredient
in theproof being
thehandling
of the
signs
as in the Section 2. Also note that if we use aperiod
of a Drinfeldmodule defined over H
instead,
the ratiobelongs
to H.THEOREM 10.
(Goss [G4]).
Let p be asgn-normalized
rank oneDrinfeld
modulewith
corresponding
latticeif plp,
whereIp
is an idealof
A. Let J be an idealof A, Ap,j
be the J-torsionof
p and let a EJ-1I03C1.
Thenfor
s apositive integer,
wehave
Proof.
We willdrop
thesubscripts
p in theproof.
Lete(z)
be theexponential
function
corresponding
to p. It has coefficients inHl. Also, e(03B103C0)
is a J-torsionpoint
of p and hencebelongs
toHl (Aj ). (For
a nontrivialJ,
this field is the same asJ( (AJ), by
the class field theoretic resultsof Hayes,
see p. 245 of[T3]).
Hence, if we write1/e(z
+a3t) = L cnzn,
then cn EHl(AJ).
Nowe(z)
= z11(l - z/(03C0i)),
where the
product
runs over i e I -{0}. Taking
thelogarithmic derivative,
ifa e I,
so that a + i~ 0,
we getWe get a similar
equality
if wemultiply
both sidesby z
to get rid of thepole
at0, if a E I.
Comparison
of the coefficients now proves the theorem. DProof of the
Theorem 9. Thespecial
case a = 0 and J = I = A of theprevious
theorem shows that
03C0-s 03A3
a-’ cHi,
where a runsthrough
the nonzero elementsof A.
By looking
at thesigns,
we see that the sum is zero, if s is odd andis -03B6(s),
if s is even. D