C OMPOSITIO M ATHEMATICA
H ANS -G EORG R ÜCK
Abelian surfaces and jacobian varieties over finite fields
Compositio Mathematica, tome 76, n
o3 (1990), p. 351-366
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
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Abelian surfaces and Jacobian varieties
overfinite fields
HANS-GEORG
RÜCK
Fachbereich 9 Mathematik, Universität des Saarlands, D-6600 Saarbrücken, Federal Republic of Germany
© 1990 Kluwer Academic Publishers. Printed in the Netherlands.
Received 22 November 1988; accepted in revised form 12 January 1990
1. Introduction and results
Let C be a
projective non-singular
curve of genus 2 over a finite field kwith q (q
=p")
elements. The zetafunction of C is a rational function of the formwith
integers
al and a2. The Riemannhypothesis
for curves restricts these coefficients to|a1| 4q1/2
andla2l 6q.
Theproblem
is to find all al and a2 which occur in this way from curves over k of genus 2.The
theory
of abelian varieties over finite fields is animportant
tool to handlethis
problem.
Let A be an abelian
variety
over k. We denoteby fA(T)
the characteristicpolynomial
of the Frobeniusendomorphism
03C0A of A. Two abelian varieties A and B over k arek-isogenous
if andonly if fA(T)
isequal to fB(T).
Let J be the Jacobian
variety
of the curve C of genus2, then fJ(T) equals T 4
+a 1 T 3
+a2 T 2
+a 1 q T + q2, where
a1 and a2
are thecorresponding
coeffi-cients of the zetafunction of C.
Therefore the
problem of finding
all aland a2
can be divided into twoparts.
At first one has to determine the
polynomials fA(T)
for all the abelian varieties A over k of dimension 2. Then one has to decide whether agiven
abelianvariety
isk-isogenous
to the Jacobianvariety
of a curve.In this paper we
present
two theorems whichgive partial
solutions of theproblem.
Let
End,(A)
be thering
ofk-endomorphisms
of A. To avoid a lot of calculations forspecial
cases we restrict ourselves to those abelian varieties A whosealgebra End,(A) (x) Q
is a field. The same methods can be used to solvethe
problem
ingeneral (see
also[H-N]).
THEOREM 1.1. The set
of fA(T) for
all abelian varieties A over k(#
k = q =pn)
of
dimension 2 whosealgebra Endk(A) (D 0
is afield
isequal
to the setof
polynomials f (T)
=74
+al T3
+a2T2
+alqT + q2
where theintegers
aland a2 satisfy the following
conditions:(a) |a1| 4q1/2 and 2|a1|q1/2 - 2q a2 a21/4 + 2q, (b) ai - 4a2
+8q is
not asquare in 7L,
(c)
either(i) vp(al)
=0, vp(a2) n/2
and(a2
+2q)2 - 4qai
is not a square in7Lp
or(ii) vp(a2)
= 0 or(iii) vp(a1) n/2, vp(a2) n and f (T)
has no root in7Lp.
REMARKS.
(i)
Thecorresponding
theorem forelliptic
curves isgiven by
Waterhouse
[Wa].
(ii)
Theconditions lall 4ql/2
andvp(a1) n/2
inparts (a)
and(c) (iii)
of Theorem 1.1 have thefollowing
consequences:If n is odd and
p 7,
there areonly
the cases(ab a2) = (0, 0)
and(0, ±q).
Then f(T)
has no root in7Lp.
If n is even,
we just get (al, a2) = (0, 0), (0, - q)
and(I ql/2, q). f(T)
has no rootin
7Lp
if andonly
ifp ~
1 mod8, p ~
1 mod 12or p ~
1 mod 5respectively.
We want to decide whether an abelian
variety
of dimension 2 isisogenous
tothe Jacobian
variety
of a curve. If the field of definition isalgebraically closed,
this
problem
is solved in thefollowing
way:Every
abelianvariety
isisogenous
toa
principally polarized
one. Theprincipally polarized
abelianvariety
has apositive
divisor whose self-intersection number is 2. Thenby
a Theorem of Weil this divisor is either a curve of genus 2 or the sum of twopositive
divisors. In thefirst case the abelian
variety
isisogenous
to the Jacobianvariety
of that curve.If the field of definition is not
algebraically closed,
theproblem
is morecomplicated,
because theisogeny
mentioned above may not be defined over this field.We consider the case where the field of definition of the abelian
variety
is theprime
field7L/p7L.
THEOREM 1.2. Let A be an abelian
variety of dimension
2 over theprime field
k =
7L/p7L
whosealgebra Endk(A) ~ Q is a field,
and letfA(T)
=T4
+
a1T3
+a2T2
+a 1 pT + p2
be the characteristicpolynomial of the
Frobeniusendomorphism of
A. We denote theinteger ai - 4a2
+8p by
d.If (a) (a2
+2p)2 - 4pai is
not asquare in Q(d) and if (b) Z[a1/2
+d/2] is
equal to
thering of integers
inQ(d),
then A isk-isogenous
to the Jacobianvariety of a
curve over kof
genus 2.REMARKS.
(i)
The condition(a)
of Theorem 1.2 isequivalent
to the fact that the fieldEndk(A) ~ Q
is not a Galois extension of Q. This is an immediateconsequence of the
following properties:
The field
Endk(A) ~ Q
isequal
toQ(03C0A),
the minimalpolynomial
of nAis fA(T)
which
splits
overQ(d)
in aproduct (T2 - 03BB1 T
+p) (T2 - 03BB2 T
+p).
Theproduct
of thediscriminants Ai - 4p
and03BB22 - 4p equals (a2
+2p)2 - 4pa21.
(ii)
Thefollowing example
shows that the condition(a)
of Theorem 1.2 isimportant:
We takep = 2, a1 = 1, a2 = -1.
Then Theorem 1.1implies
that thepolynomial T4
+T3 - T2
+ 2T + 4 isequal to fA(T)
of an abelianvariety A
over
Z/2Z
of dimension 2. If we suppose that A isisogenous
to the Jacobianvariety
of a curveC,
then the zetafunctionexp(03A3~m=1 N m Tm/m)
of C isequal
toThe
integers Nm
are the number ofpoints
on C over extension fields ofZ/2Z;
inparticular
weget 0 N1 N2.
But in ourexample
we evaluateNi
1 = 4 andN2
= 2. Hence A is notisogenous
to a Jacobianvariety.
All theassumptions
ofTheorem 1.2
except
condition(a)
are fulfilled.(iii)
But on the other hand thisimportant
condition(a)
of Theorem 1.2 is not a necessary condition: The curvey2 = X5
+X4
+2X 2 - X
+ 1 overZ/5Z
hasa zetafunction with al = 3
and a2
= 4. For thisexample (a2
+2p)2 - 4pai
isequal
to 16.2.
Background
on Abelian VarietiesLet A be an abelian
variety
over the finite field kwith q
elements. We denoteby fA(T)
the characteristicpolynomial
of the Frobeniusendomorphism
nA of Aover k. This
polynomial
determines A up tok-isogeny,
i.e. two abelian varieties A and B arek-isogenous
if andonly if fA(T)
isequal to fB(T) ([T],
Theorem1).
The group of
k-homomorphisms
of abelian varieties A and B and thering
of k-endomorphisms
of A are denotedby Homk(A, B)
andEndk(A).
For aprime
number 1 which is different from the characteristic p of k one considers the Tate module
Tl(A)
of A.The Theorem
of
Tate([T],
MainTheorem)
states that the7L,-modules Homk(A, B)
QZl
andHomGal(k/k)(Tl(A), Tl(B))
arecanonically isomorphic.
Theelements of the second module are those
homomorphisms
çi of theZl-modules T (A)
andTl(B)
whichsatisfy
~lo(nA)l
=(03C0B)l °
~l where(nA),
and(nB),
are thecorresponding
Frobeniusendomorphisms
onTl(A)
andTl(B).
For the
prime number p
one considers the Dieudonné moduleM(A(p))
of thep-divisible
groupA(p) corresponding
to A([O],
Section3). M(A(p))
is a moduleover the
ring
a =W(k)[F, V],
whereW(k)
is thering
of Witt vectorsover k,
F isthe absolute Frobenius and V is the
Verschiebung (translation).
Analogous
to the 1-adic case there is a canonical(contravariant) isomorphism
from
Homk(A, B)
Q7Lp
ontoHoma(M(B(p)), M(A(p))).
If k is theprime
fieldZ/pZ,
thenW(k)
isequal
to7Lp
and Fis just
the Frobeniusendomorphism (03C0A)p
on
M(A(p)).
ThenHoma(M(B(p)), M(A)p)))
consists of thoseZp-homomorphisms
which commute with
(nA)p
and(nB)p’
This easy characterization is the reason for theassumption
on k in Theorem 1.2.A classification of
simple
abelian varieties over k isgiven by
theTheorem of
Honda-Tate
([T-H],
Théorème1):
The
assignment A H ?LA gives
a one-to-onecorrespondence
betweensimple
abelian varieties over k
(up
toisogenies)
and Weil numbers(these
arealgebraic integers
whoseconjugates
have absolutecomplex
valueq’ /2 ). Endk(A) (D 0
is a divisionalgebra
with centerQ(03C0A),
itsplits
at each finiteplace prime
to p, it does notsplit
at any realplace
and its invariant at aplace
p over p isgiven by
The dimension of A is
equal
toFrom this we
get immediately
that the invariant at pequals vp(fi(0))/vp(q)
wherefi(T)
is the irreducible monic divisorof fA(T)
overZp
whichcorresponds
to p.
The
key
for aproof
of Theorem 1.2 is a Theoremof
Weil([W],
Satz2):
Let A be an abelian
variety
of dimension 2. Consider apositive
divisor D on Awhose self-intersection number
(D, D)
is 2. Then this divisor is either a curve of genus 2 or the sum of twopositive
divisors. In the first case A isisomorphic
tothe Jacobian
variety
ofD,
in the second case A is theproduct
of twoelliptic
curves.
The self-intersection number of a divisor D on A can be evaluated
by
theRiemann-Roch Theorem
([M], 111.16):
Let
(DD
be thehomomorphism
from A to the dual abelianvariety Â,
whichassigns
to x E A thepoint 03A6D(X) corresponding
to the divisor class ofT*xD -
D.Let
x(D)
be the Euler characteristic of the line bundle associated to D. Thesequantities satisfy: deg (DD
=X(D)2
and 2 ·X(D) = (D, D).
Divisors on A are
closely
related with"skew-symmetric" homomorphisms
from A to
Â:
For each
prime number l (l ~ p)
there is anon-degenerate
bilinearmapping
ofZl-modules e’: Tl(A)
xTl(Â) ~ Zl satisfying
thefollowing properties (c.f. [M], IV.20):
Let 9 be a
homomorphism
from A to B with dualhomomorphism : À - Â,
then
eBl(~lx, y)
=e’(x, ly).
If A is defined over the finite field k
with q elements,
then the Frobeniusendomorphisms (nA)1
and(nÂ)1
onTl(H)
andTl(Â)
arecompatible with, e’ by
theformula
et«nA),x, (03C0Â)ly)
= q.eAl(x, y).
To eachhomomorphism
~: A ~Â
onecan associate a bilinear form
E~l: Tl(A)
xTl(A)
~Zl by Er(x, y)
=et(x, ~ly).
Hence a
homomorphism
~: A ~Â
is defined over k if andonly
ifE~l((03C0A)lx, (03C0A)ly)
= q ·Eî (x, y).
Conversely
letE,
be a bilinear form onTl(A)
which satisfiesEl((03C0A)lx, (03C0A)ly)
= q ·E,(x, y),
then there is aZ,-homomorphism
~l fromTl(A)
toTl(Â)
which commutes with
(03C0A)l
and(03C0Â)l
such thatEl(x, y)
=eAl(x, ~ly). By
theTheorem of Tate 91 can be
approximated by k-homomorphisms
from A toÂ.
The
corresponding
statements are true for theprime
number p if one takes the Dieudonné modulesM(A(p))
andM(Â(p»,-
for details we refer to[O],
Section 3.These considerations can be used to construct a divisor on an abelian
variety:
The bilinear form
ET
of ahomomorphism ~ = 03A6D
isskew-symmetric.
Conversely
let ~: A --+ A be ak-homomorphism
such that thecorresponding E~l
is
skew-symmetric (skew-symmetric
modulolN),
then the divisor D =(1
x~)*P
where P is a Poincaré divisor on A x
A
satisfies 2 ’ ç =03A6D (2 ’
~ ~03A6D
modlN).
For a
proof we
refer to[M],
IV.20,
Theorem 2 and[O], Proposition
1.15.There is a Poincaré divisor which is defined
over k,
therefore the divisor D has also k as a field of definition.We recall that a divisor D is defined
over k,
if D can begiven locally by equations
with coefficients in k. Another characterization is thefollowing:
D isdefined over k if and
only
ifQ(D)
= D for eachautomorphism a
of the Galois group of k(c.f. [R],
Lemma4, especially
the laststep
in theproof).
3. Proof of Theorem 1.1
The theorem of Honda-Tate
(c.f. §2)
shows that the set ofpolynomials fA(T) belonging
to abelian varieties A over k of dimension2,
whosealgebra Endk(A) (D 0
is afield,
isequal
to the set of irreducible monicpolynomials f(T) of degree 4,
whose roots are Weil numbers and where in additionvp(fi(0))/n
is aninteger
for each irreducible monic divisorfi(T) of f (T) over Op.
LEMMA 3.1.
f(T) is
an irreducible monicpolynomial of degree
4 inZ[T],
whoseroots are Weil
numbers, if
andonly if f(T)
=T4
+al T3 + a2T2
+alqT + q2
with
integers
al and a2satisfying:
Proof. Let f (T)
be an irreduciblepolynomial of degree
4 inZ[T],
whose rootsare Weil numbers. Since the roots are
permuted by complex conjugation, they
are of the form 03C01,
q/03C01,
n2,q/n2’ Hence f (T)
=T4
+a1 T3
+a2T2
+ai q T + q2
with
integers
aand
a2; weget lall 4ql/2.
The elements
À1
= nI +q/03C01
andÀ2
= 03C02 +q/n2
are distinct real numbers and zeros ofT2
+a 1 T + a2 - 2q,
whose discriminanta i - 4a2
+8q
ispositive;
this
yields
a2ai/4
+2q. From f(T) = (T2 - 03BB1 T + q)(T2 - À2 T + q)
we seethat this discriminant is not a square in
7L,
since otherwisef (T)
would bereducible in
7L[T].
The formulas
IÂ1-Â21
+lÂl + 03BB2|
=2·max{|03BB1|, |03BB2|} 4q1/2, IÂI-Â212=
ai - 4a2
+8q
and|03BB1
+Â21
=lall give
theinequality
a2 >2lallql/2 - 2q.
Conversely let f(T)
=T 4
+a1T3 + a2T2
+a1qT + q2
be apolynomial
withproperties (a)
and(b).
Theinequalities
in(a)
show that the elements03BB1 = -al/2
+(ai/4 -
a2 +2q) 1/2
and03BB2 = -al/2 - (ai/4 -
a2 +2q)1/2
aredistinct real numbers
with |03BB1|, IÂ21 2ql/2.
We alsoget
thedecomposition f(T)
=(T2 - Â1 T + q)(T 2 - 03BB2T+ q).
Let ni,
q/03C0i
be the zeros ofT2 - 03BBiT +
q, then easy calculations prove thatInil | = |q/03C01| = q1/2 .
Hence the rootsof f(T)
are Weil numbers.f(T)
is irreducible inZ[T],
becauseai - 4a2
+8q
is not a square in7L. D
LEMMA 3.2. Letf(T)
=T4
+al T3
+a2T2
+alqT + q2
be apolynomial
in7L[T].
Thenvp(fi(O»In
is aninteger for
each irreducible monicdivisor fi(T) of f(T) over Q p if
andonly if
oneof
thefollowing
conditions issatisfied:
Proof.
We consider the Newtonpolygon
of thepolynomial f(T) over Qp.
Thispolygon
is the lower convexenvelop
of thepoints (0, 2n), (1, vp(al)
+n), (2, vp(a2)), (3, vp(a1))
and(4, 0),
and itsshape
leads to adecomposition
off(T)
over
Op (see
e.g.[We], §3).
Wedistinguish
three cases(each
casecorresponds
toa formal
isogeny type
of thep-divisible
group of an abelianvariety
of dimension2,
c.f.[Ma]).
Case 1. The
point (3, vp(a1))
is a vertex of thepolygon:
This is
equivalent
tovp(al) n/2
andvp(a2)
>2vp(al).
Weget
thedecomposition f(T) = (T - rxl)(T - a2)(T - a3)(T - a4)
withvp(al)
=vp(al)
andvp(a4)
= n -vp(al).
In addition thepolynomials
T - al,(T - a2)(T - a3)
andT - a4 are
in Qp[T].
Hence all the numbersvp(fi(0))/n
areintegers
if andonly
ifvp(al)
=0, vp(a2) n/2, (T - a2)(T - a3)
is irreducible overOp.
The latter isequivalent
to condition(i),
becausef(T)
isequal
to aproduct (T2 - 03BB1T + q)(T2 - Â2T+ q)
with(03BB21 - 4q)(Â2 - 4q) = (a2
+2q)2 - 4qa21 (c.f.
the
proof
of Lemma3.1).
Case 2. The
point (3, vp(a1))
is not a vertex, but(2, vp(a2))
is a vertex of thepolygon:
This is
equivalent
tovp(a2) n
and2vp(a1) vp(a2).
Then
f(T)
can bedecomposed
in(T - al)(T - rx2)(T - a3)(T - a4)
withvp(a1)
=vp(a2)
=vp(a2)/2
andvp(a3)
=vp(a4)
= n -vp(a2)/2.
The twoquadratic
polynomials (T - oc,)(T- rx2)
and(T - a3)(T - a4)
arein Qp[T].
Here we
get
that all the numbersvp(fi(O»In
areintegers
if andonly
ifvp(a2)
= 0.Case 3. The Newton
polygon of f(T)
is astraight
line:This is
equivalent
tovp(a1) n/2
andvp(a2)
n. Weget f(T) = (T - al) (T - a2)(T - a3)(T - a4)
withvp(ai)
=n/2.
In this case the additional condition isthat f(T)
has no root inZp.
DLemma 3.1 and Lemma 3.2
give
aproof of Theorem
1.1.4. Some Lemmas
The basic idea of the
proof
of Theorem 1.2 is to construct apositive
divisor onan abelian
variety
with self-intersection number 2 which is defined over the field of definition of the abelianvariety.
Wepresent
some lemmas which will lead to the construction of such a divisor.From now on let k denote the
prime
fieldZ/pZ.
The lemmas are true for an
arbitrary
finitefield,
whenever theprime number p
is not
involved,
but for sake ofsimplicity
we do notdistinguish
these cases.LEMMA 4.1. Let A be an abelian
variety
over kof
dimension g. For agiven
basison the Tate module
Tl(A) (the
Dieudonné moduleM(A(p)) if
1 =p)
letFI
be thematrix
of
the Frobeniusendomorphism (03C0A)l.
Suppose
M is anon-singular (2g
x2g)-matrix
withcoefficients
inZl
such that thecoefficients of M -1FFIM
are also inZl.
Then there are ak-isogeny
~: B ~ A(9: A B if
1 =p) of degree lvz(detM)
and a basisof Tl(B)(M(B(p)) if
1 =p)
suchthat
has
the matrix M. And the matrixM-1FIM corresponds
to the Frobeniusendomorphism (n,),.
Proof.
We consider the finitesubgroup U
= M ·Tl(A)/lvl(det M) · %(A),
ifl ~
p.Otherwise let U be the finite commutative group scheme
corresponding
to thea-module
M(A(p»)/M. M(A(p)) (c.f. [0],
Theorem3.14).
Since the coefficients ofM -1 1FlM
are inZ,,
the canonicalisogenies
A ~A/ U
andA/U -
A are definedover k. The
isogeny A/U ~
A(A - A/U
if l =p)
fulfills theproperties
of thelemma. D
With the next lemma we can construct a divisor on an abelian
variety
with agiven
local behaviour.LEMMA 4.2. Let A be an abelian
variety
over kof
dimension 2. We assume that the characteristicpolynomial fA(T)
is irreducible over Z.Suppose
that S isa finite
set
of prime
numbers. Then we get ak-isogeny
B ~ A where Bsatisfies
thefollowing properties:
There are a divisor D on B which isdefined
over k and a k-homomorphism
9: B ~Û
whosedegree
isprime
toS,
such that03A6D
= 2 · ~.Proof.
We may assume that 2 is contained in S.(i) Suppose
that B is an abelianvariety
over k with thefollowing properties:
For each 1 ES there is a
Zl-homomorphism fl: Tl(B)
~Tl() (M((p)) ~
M(B(p))
if 1= p)
with(a) fz 0 (03C0B)l = (03C0)l 0 fz (fp o (03C0)p = (03C0B)p o fp
if 1 =p),
(b)
the bilinear form(x, y) ~ eBl(x, fly)
isskew-symmetric, (c) det(/i)
is a unit inZl.
Then
by
the Theorem of Tate(c.f. §2)
weget
for each 1 E S ak-homomorphism
(1): B
such that~(l)l = f mod l2.
The bilinear formE9(’)
isskew-symmetric
modulo
l2.
Thus there is a divisorD(l)
which is defined over k such that 2 ·~(l)
~03A6D(l) mod l2 (c.f. 92).
Now we take
D is defined over k and satisfies
Since 2
E S,
we see that there is ak-homomorphism
~: B ~ B with(D,
= 2 · ~. In addition thedegree
of 9 isprime
to S.(ii)
We have to find an abelianvariety
B andZl-homomorphisms fl
withproperties (a), (b)
and(c):
In terms of matrices we have to find matrices
MI
with coefficients inZl
with(c.f.
theproperties
ofeBl)
(à) Fr MIFI = p · Ml,
whereF,
is the matrix to(03C0B)l,
(6) MI
isskew-symmetric, (ê)
detMl
is a unit inZ,.
If
F,
onTl(A) (M(A(p))
if l =p)
has a"simple" form,
we can take B = A andgive MI directly.
Otherwise wechange
A to an abelianvariety
B over kby applying
Lemma
4.1,
this movesF,
to a matrix of the formM -1 ’F,M.
Since thedegree
ofthe
isogeny
A - B in Lemma 4.1 is a powerof l,
the Tate modules(Dieudonné module)
at otherprime
numbers are notchanged.
Weapply
thisprocedure
foreach
l E S,
hence weget
an abelianvariety B
over k of dimension2,
where foreach l E S the matrix
F,
has a"simple"
form.We
distinguish
several cases whichdepend
on the factorizationof fA(T)
overZI.
In each case wegive
a"simple"
form forF,
and a matrixMI
withproperties
(à), (b)
and(ê).
The details are easy toverify.
Case 1.
Case
2.
Case
3.
Case 4.
We assume that
Case 5.
This proves the lemma.
The next lemma de scribes the
decomposition
into irreduciblecomponents
of the Tate module and of a bilinear form on it.Let A be an abelian
variety
over k. For aprime
number 1 we consider thefactorization
f, (T) = 03A0ifi(T)
overZl
with irreduciblepolynomials fi(T).
For
each f (T)
we define(03C4fi)(T) = (1/fi(0))fi(p/T)Tdeg fi(T).
Since all the rootsof fA(T)
are Weilnumbers, (03C4fi)(T)
is also an irreducible divisorof fA (T).
It is not difficult to prove:
LEMMA 4.3. Let 1
(1
~p)
be aprime
number which does not divide the discriminantof fA(T).
There is a
decomposition of the
Tate moduleTl(A) = ~i Tl(A)(fi)
as aZl [(03C0A)l] -
module where each
Tl(A)(fi)
isisomorphic
toZl[T]/fi(T).
LetEl: Tl(A)
xTl(A) ~ Zl
be askew-symmetric
bilinearform
whichsatisfies El((03C0A)lx, (03C0A)ly) = P · El(x, y)
and det
El ~
0. Then we get:(i) El | Tl(A)(fi)
xTl(A)(fj)
= 0 iffj(T) ~ (03C4fi)(T),
(ii) El 1 Tl(A)(fi)
xTl(A)(03C4fi)
= lmi -El(fi)
wheremi
0 andEl(fi)
is a bilinear formover
Zl
onTl(A)(fi)
xTl(A)(03C4fi)
withvl (det El(fi))
= 0.(iii) vl(det El)
=Li
mi.deg,h(T).
~We want to use Lemma 4.3 to
change
agiven
divisor. Let D be a divisor on anabelian
variety
A and lett/J:
B - A be anisogeny,
then the definition of the dualisogeny : Â ~ Ê yields
that03BF03A6D03BF03C8 = 03A603C8*D. With
the next lemma we findsuitable
isogenies
for thatprocedure.
LEMMA 4.4. Let A be an abelian
variety
over k. We consider a divisor D on Awhich is
defined
over k and ak-homomorphism
~: A ~Â
with 2 wp =03A6D.
For a
prime
number 1(l
~p)
which does not divide the discriminantof fA(T)
wefix
an irreducibledivisor fi(T) of fA(T)
overZl.
Then there is ak-isogeny 03C8:
B ~ Awhose
degree
is a power of 1 such that thebilinear forms ET
andE°~°03C8l satisfy:
(i) El
° ~11 | Tl(B)(fj)
xTl(B)(03C4fj) ~ ET | Tl(A)(fj)
xTl(A)(03C4fj), if fj(T) ~ fi(T)
andfj (T)
~(03C4fi)(T).
(ii) E03BF~03BF03C8l | Tl(B)(fi)
xTl(B)(03C4fi) la ET | Tl(A)(fi)
xTl(A)(03C4fi),
where ô = 1if (03C4fi)(T) ~ fi(T)
and ô = 2if (03C4fi)(T)
=fi(T).
Proof.
In view of thedecomposition Tl(A) = EB j Tl(A)(fj)
of Lemma 4.3 weconsider the
diagonal
matrix M which de scribes theidentity
onTl(A)(fj) (if fj(T) ~ fi(T))
and themultiplication by
1 onTl(A)(fi).
If
F,
is a matrix of(03C0A)l respecting
thedecomposition
ofTl(A),
weget
M-1 F,M
=Fi .
Therefore Lemma 4.1yields
ak-isogeny t/J:
B ~ A such thatt/J,
isgiven by
M.Let
Me
be the matrix of the bilinear formE9,
thenMT MT M corresponds
toE03BF~03BF03C8l.
The evaluation of the matrixmultiplication
proves the lemma. ri With Lemma 4.4 wetry
tochange
ak-homomorphism
(p: A ~Â by
a k-isogeny tjJ:
B - A such that the new bilinear formE°~°03C8l
has the same value l’on each
part T,(B)(h)
xTl(B)(03C4fi) (c.f.
Lemma4.3).
If wesucceed,
then03BF
~ 03BFis
of the form lm.
fp,
wherefp:
B -is
ak-homomorphism
whosedegree
isprime
to1.
Only
those divisorsfi(T)
offA(T)
with(03C4fi)(T) =fi(T) might
cause sometrouble,
in all other cases we can take ô = 1 in Lemma 4.4. For two-dimensional abelian varieties acase-by-case
examination of thepossible decomposition
offA(T) (like
in theproof
of Lemma4.2) yields
thefollowing:
LEMMA 4.5. Let A be an abelian
variety
over kof dimension 2,
and let 9: A -Â
be a
k-homomorphism
with 2 · qJ =03A6D for
a divisor D on A. For aprime
number 1(l
~p)
which does not divide the discriminantof fA(T)
we consider the decom-position
in irreduciblefactors fA (T) = 03A0ifi(T)
overZl.
If (i)
thedecomposition
isdifferentfrom (T2 - 03BB1 T + p)(T2 - 22 T + p)
orif (ii)
the
decomposition
is(T2 - 21
T +p)(T2 - 22 T
+p)
and in addition the matrixof Er
isof
theform
such that
v,(a)
-vl(f)
is an eveninteger,
then we get ak-isogeny 03C8:
B -A,
whosedegree
is a powerof l,
and ak-homomorphism fp:
B ~Ê,
whosedegree
isprime
tol,
with
03BF~°03C8
=lm. fp.
DWe want to
give
anexample:
Suppose fA(T) = (T2 -
03BBT +p)(T - a)(T - (p/a))
is thedecomposition
inirreducible factors. Then there is a basis of
Tl(A)
such that the Frobeniusendomorphism (nA)l
isgiven by
Let
be the matrix of ~: A ~ A.
Following
Lemma 4.4 we takeand find a
k-isogeny 03C8:
B - A withHence Í/1 0 ~03BF03C8
= l ·Ç
where thedegree
of(’p: B --+ Ê
is not divisibleby
1.On the other hand if the difference
vj(a) - vl( f )
in Lemma4.5(ii)
isodd,
thereis no way to find a
k-homomorphism (’p
with thisprocedure.
Before we canapply
Lemma
4.5,
we have tochange
this difference to an even numberby
a differentmethod.
LEMMA 4.6. Let A be an abelian
variety
over k with an irreduciblepolynomial fA(T)
over Z. We consider a divisor D on A over k and ak-homomorphism
qJ: A ~Â
with
2 · ~
=(D,. Let 03B5 ~ Endk(A),
then we get:(i)
Then there is a divisorDe
on A over k with 2. qJ 0 B =03A6D03B5 if
andonly if BE Q(nA
+(p/03C0A)).
(ii) If e E 0(7[A
+(p/03C0A)),
then the Euler characteristicx(D,) of De
isequal
toN Q(1tA
+p/03C0A)/Q(03B5) · x(D)·
Proof. (i)
Letl(l ~ p)
be aprime
number. There exists a divisorD,
on A over kwith
03A6D03B5
= 2 · qJ 0 e if andonly
if the bilinear formE~03BF03B5l
isskew-symmetric (c.f. §2).
The
endomorphism
e is apolynomial 9(nA)
in nA,because fA(T)
is irreducibleover Z. Since
E9
isskew-symmetric
and defined over kby assumption,
the formE~03BF03B5l
is alsoskew-symmetric
if andonly
ifg(nA)
=g(p/nA).
The latter condition isequivalent
to 03B5 ~Q(03C0A
+(p/03C0A)).
(ii)
Since thedegree
of 03B5 isequal
to the normNQ(03C0A)/Q(03B5),
the Riemann-Roch Theoremimplies ~(D03B5)2 = (NQ(03C0A+p/03C0A)/Q(03B5)·X(D))2.
The two functionsX(D,)
andNQ(03C0A + p/03C0A)/Q(03B5)
arehomogenous polynomials
in e. Thisyields
the formula for~(D03B5).
Let
l (l
~p)
be aprime
number where thepolynomial fA(T)
has the decom-position (T2 - À1 T + p)(T2 - Â2T+ p)
overZl.
The differencev1(a) - vl( f ) (c.f.
Lemma
4.5(ii))
of ak-homomorphism
~: A -Â
may be odd.If we find an
endomorphism 03B5 ~ Q(03C0A
+(p/03C0A))
whose norm isexactly
divis-ible
by
an odd power ofl,
then qJ 0 E satisfies condition(ii)
of Lemma4.5,
i.e. thecorresponding
difference is even. Of course thedegree
of 9 - e should not containnew
prime
numbers where Lemma 4.5 is notapplicable.
The existence of E isguaranteed by
the next lemma.LEMMA 4.7. Consider an abelian
variety
A whichsatisfies
theassumptions of
Theorem 1.2. Let S be
afinite
setof prime
numbers and in addition let 1 be aprime
number(l
~2,
1 ~ p,le S)
such that thedecomposition of fA(T)
in irreduciblefactors
overZl
isequal
to(T2 - 03BB1
T +p)(T2 - À2 T + p).
Then there is an
endomorphism 03B5 ~ Endk(A)
n0(n,
+(p/nA))
whose normNQ(03C0A + p/03C0A)/Q(03B5)
isprime
to S andequal
to 1.(03A0i pi) · a2, where
a is aninteger
and where pi areprime
numbers suchthat fA(T)
has a root inZpI.
Proof.
Like in Theorem 1.2 we denote theinteger a’ - 4a2
+8p by d,
let inaddition e =
(a2
+2p)2 - 4pai.
Then0(n,
+(p/nA))
isequal
toQ(d). By
assumption (a)
of Theorem 1.2 theinteger
e is not a square inQ(d).
Aconsequence of
assumption (b)
is that thering
ofintegers
inQ(d)
isequal
toEndk(A)
n0(nA
+(p/03C0A)).
In view of these remarks the
problem
can be reduced to find an element inQ(d)
whose norm is 1.(03A0i Pi).
Let
do
and eo be thesquarefree parts
of thepositive integers
d and e.If a
prime
number pi satisfies(d0 pi) = 1,
thenT
isequal
toSince e =
(03BB21 - 4p)(03BB22 - 4p),
the additional condition(e0 pi) = -1 yields
thatone of the two
quadratic polynomials splits
over7Lpz.
Theprime
number lsatisfies
(d0 l) = (e0 l) =1.
Let
do
= q1 ··· qr be thedecomposition
inprime
factors. Since e is not asquare in
Q(d),
eo is différent from 1 anddo.
If eo is not a divis or of
do,
we choose aprime
number pi with thefollowing
conditions:
Since these conditions are
independent,
there areinfinitely
many p1satisfying
them. We
set do =
1 and(e0 p1) = - 1. Using
the Hilbertsymbols
it is notdifficult to show that
l · p1
is norm of an element inQ(d).
If eo
is a divisorof do,
we may assume that eo = q1 ··· qs with s r. Now wechoose
prime
numbers PI and p, with the conditions:There are
infinitely
many p 1 and p,, andthey satisfy
The calculation of the Hilbert
symbols
shows that l · p1 · p, is norm of an elementin
Q(Jd).
05. Proof of Theorem 1.2
Let
again
k be theprime
fieldwith p
elements.Step
1. We choose a finite set S ofprime
numberscontaining 2,
p and all the divisors of the discriminant offA(T).
Applying
Lemma 4.2 there is ak-isogeny Al
~A,
a divisorD 1 on A over k
and a
k-homomorphism ~1:A1 ~ Â1,
such that 2 · ç 1 =(D,, and deg
91 is prime
to S.
We may assume that
x(D 1 )
>0,
because otherwise wechange
çi 1by
anendomorphism
e whose norm isnegative (c.f.
Lemma4.6).
Step
2. Eachprime
divisorof deg
(pi does not divide the discriminantof fA(T).
The
only prime
numbers 1 which do notsatisfy
theassumptions
of Lemma4.5 are those
where fA(T)
has thedecomposition (T2 - 03BB1 T + p)(t2 - 03BB2T + p)
over
Zl
and wherevl(deg ~1) ~ 2 mod 4 (i.e.
the differencevl(a) - vl(f)
inLemma
4.5(ii)
isodd).
Foreach 1; (i
=1,
... ,r)
with these conditions we findby
Lemma 4.7 anendomorphism 03B5i ~ Endk(A1)
n0(nA
+(p/03C0A))
with normNQ(03C0A + p/03C0A)/Q(03B5i) = li · (03A0ipij) · a2i, where fA(T)
has a zero inZpiJ.
Weapply
Lemma 4.6 and
get
ak-homomorphism (Pl
= ~1 03BF £ 03BF ··· 03BF £r and a divisorD 1 on
A over k
with 2 ·1
=03A61,
such thatdeg (Pl contains only prime
numbers for which Lemma 4.5 isapplicable.
Step
3.Let 1,
be aprime
divisorof deg (P 1. By
Lemma 4.5 there is ak-isogeny t/Jl:A2 --+Ab
whosedegree
is a power ofl1,
such that1 03BF 1 03BF 03C81
=lm11 · ~2,
where the
degree
of thek-homomorphism
qJ2:A2 --+ Â2 is
not divisibleby li.
In addition 2 · qJ2 =
03A6D2,
whereD2
is a divisor onA2
over k withZ(D2)
> 0(c.f.
§2
and[M],
III.12,
Theorem2).
Since
deg 03C8 1 and deg §i
1 are powersof 11, nothing
ischanged
at otherprime
numbers. Then we take the next
prime divisor l2
ofdeg
~2 andrepeat
the sameprocedure.
Finally
weget
ak-isogeny 03C8:
B -A1 (03C8 = 03C8s 03BF···03BF 03C81), a
divisor E on B over kand a
k-homomorphism
~: B ~B, such
that 2 · ç(DE
anddeg
9 = 1.Step
4. Since 2 · ç =(DE,
there is a divisorE
on B with 2 ·~
E and 9= 03A6 ([M],
IV.23,
Theorem3).
With the Riemann-Roch Theorem we evaluate:(E, E) = 2 · ~() =
2.Unfortunately
the divisorE
may not be defined over k.Without loss of
generality
we assume that E ispositive (either
E or - E isample,
becausex(E) = 1).
Let 03C3 be the
generator
of the Galois groupGal(Kjk).
Themapping
Gal(k/k) ~ Pic0(B) =
with03C3m ~ 03C3m -
is a1-cocycle,
because2 · (03C3 - ) ~ 03C3E - E ~ 0.
Since
H1(Gal(k/k), Ê)
= 1(c.f. [L]),
there is ab ~
with03C3 - = - 03C3.
We choose b E B with= T*b -É.
Let D be the divisor
Tt E,
then weget
03C3D ~ D.This
implies
6D= D,
because D ispositive
withx(D)
= 1. Hence D is definedover k and
(D, D)
= 2(c.f. §2).
In view of Weil’s Theorem
(c.f. §2)
D is either a curve of genus 2 or the sum of twoelliptic
curves.In the second case the abelian
variety A
isisogenous
over thequadratic
extension of k to the
product
of twoelliptic
curves. But then thereducibility
ofthe characteristic
polynomial
ofn2 yields
that theinteger (a2
+2p)2 - 4paf
is asquare in Z.
Hence the
assumptions
of Theorem 1.2imply
that A isk-isogenous
to theJacobian
variety
of the curve D.References
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