C OMPOSITIO M ATHEMATICA
H AJIME K AJI
On the Gauss maps of space curves in characteristic p
Compositio Mathematica, tome 70, n
o2 (1989), p. 177-197
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177
On the Gauss maps of space
curvesin characteristic p
HAJIME KAJI
Compositio 177-197,
© 1989 Kluwer Academic Publishers. Printed in the Netherlands.
Department of Mathematics, School of Science and Engineering, Waseda University,
3-4-1 Ohkubo, Shinjuku, Tokyo 160, Japan
Received 16 May 1988; accepted 6 October 1988
§0.
IntroductionIn this
article,
we discuss the extensions of function fields definedby
Gaussmaps
of space
curves inpositive
characteristic p. As iswell-known,
if p =0,
then the Gauss maps of space curves are
always
birational onto itsimage.
However, if p > 0,
then this is not true.Precisely speaking,
our purpose is toinvestigate
the setsJf’, ag£
andK’imm
defined as follows: for a smooth connected
complete
curve C over analgebraically
closed field ofpositive
characteristic p, we define Jf’ as the set of subfields K’ ofK(C)
such that there exists amorphism
1 from C to someprojective
spaceP,
birational onto itsimage,
such thati (C)
is not a line inP and the extension of function fields defined
by
the Gauss map coincides withK(C)IK"; replacing
the word"morphism"
aboveby
"unramifiedmorphism"
and "closedimmersion",
we moreover defineK’un
andK’imm, respectively.
Our main results are
THEOREM 0.1:
(a) If
C is anordinary elliptic
curve, thenXî’..
contains anysubfield
K’of K(C)
such thatK(C)
isfinite, inseparable
over K’ and theseparable
closure
of
K’ inK(C)
is acyclic
extension over K’ withdegree
indivisibleby p.
(b) If
C is asupersingular elliptic
curve, thenotherwise.
(see
Section5)
178
THEOREM 0.2:
(a) If C
is a curveof
genusg 2,
then anarbitrary
element K’of K’imm
isof the form K(C)P’for
someinteger
1 0.(b) Moreover, for
aninteger
1 >0,
letbe a sequence
of Frobenius morphisms of C. Then, the following
conditionsare
equivalent:
(1) contrains K(C)P’;
(2)
there exists a rank 2 vector bundle E onC(p/)
such that&c(p)
is stableand éc
isisomorphic to
the bundleYc(Y) of principal
partsof Y o, f first
orderfor
some line bundle .P onC,
wheregc(p)
andEC
arethe
pull-backs of 0
toC(p)
andC, respectively.
(see
Corollaries 4.4 and6.2)
We shall show also the
following
results: K’(respectively, K’imm)
containsK(C)
if andonly if p ~ 2 (see Corollary 2.3);
for a proper subfield K’ ofK(C),
JT’ contains K’ if andonly
ifK(C)
isfinite, inseparable
over K’(see
Corollaries 2.2 and
3.4);
if C isrational,
then -’4"’ =K’imm (see Corollary 3.6); and, K’un = K’imm
for any C(see Corollary 4.3).
As a
corollary
to ourresults,
one can show that a smoothplane
curve hasseparable degree
1 over the dual curve via the Gauss map(see Corollary
4.5and
[16,
p.342];
compare with[24, Proposition 4.2]),
and that if C is aTango-Raynaud
curve(see,
forexample, [2]
or[27]),
thenK’imm
containsK(C)p (see Corollary 6.5).
The fundamental results in our
study
are Theorems2.1,
3.1 and4.1,
fromwhich we shall deduce all the results above.
§1.
Gauss mapsThroughout
thisarticle,
we shall work over analgebraically
closed field k ofpositive characteristic,
denotedby
p.Let C be a
smooth, connected, complete
curve defined over k. For amorphism
1 from C to aprojective
spaceP,
birational onto itsimage,
let Vbe a vector space
H0(P, OP(1)),
andP1C(i*OP(1))
the bundle ofprincipal
parts
ofi*OP(1)
of first order on C(see,
forexample, [14, IV, A], [19, §1] or
[25, §§2
and6]).
We have a natural mapwhich is
surjective
if andonly
if 1 is unramified. Theimage of al
is aquotient
bundle of V
Q9k (!Je
of rank 2.Let G be a Grassmann manifold
consisting
of 2-dimensionalquotient
spaces of V with the universal
quotient
bundleDEFINITION:
By
virtue of theuniversality
ofG,
one obtainsfrom a’
amorphism
We call this
morphism
the Gauss map of C in P via 1(see,
forexample, [3, 2, §4], [4, 1, (e)]
or[14, IV, B]).
In
particular,
theimage of al
isisomorphic
to thepull-back
of f2 toC,
denoted
by QC.
Weidentify
these bundles.If one considers G to be
consisting
of lines inP, then,
for ageneral point
x of
C,
theimage
of x in G under the Gauss map iscorresponding
to thetangent
line toi(C)
ati(x).
We
always
assume thati (C)
is not a line inP,
and denoteby
C* theimage
of C in G under the Gauss map, which is called the dual curve of C if P =
p2 (see,
forexample, [5, 1,
Exercise7.3]
or[14, 1, C]).
DEFINITION: For a curve
C,
we define Jf’ as the set of subfields K’ ofK(C)
as follows: there exists a
morphism
1 from C to someprojective
spaceP,
birational onto itsimage,
such thati (C)
is not a line in P and the extensionK(C)/K(C*)
definedby
the Gauss map coincides withK(C)/K’. Replacing
the word
"morphism"
aboveby
"unramifiedmorphism"
and "closedimmersion",
we moreover defineJÇi
andK’imm, respectively.
REMARK: It is well-known that
if p = 0,
then K’ ={K(C)}.
§2.
Genericprojections
Our first result is
THEOREM 2.1 : Let i : C - P be a
morphism
birational onto itsimage
asbefore,
let 03A0: P ~
P1
be aprojection of P from
ageneral point
inP,
and lety : C ~
Pi
be acomposition of
II with i. Let C* andC*
be theimages of
Cunder the Gauss maps via land ll’
respectively,
and letK(C*)s
andK(Ct)s
be the
separable
closuresof K(C*)
andK(C*1)
inK(C), respectively. If
dim P
3,
thenProof.-
Denoteby
P the centreof II,
let Gand G 1 be
Grassmann manifoldsof lines in P and
P1, respectively,
and let n:G - Gi
1 be a rational mapnaturally
inducedby
II.Then,
we have a commutativediagram
We note
that,
via the Plückerembeddings
of Gand G1 into
someprojective
spaces, n is
compatible
with a linearprojection
of theprojective
space, denotedby
A n.Writing 03C3(P)
for the subset of Gconsisting
of lines in Ppassing through P,
we see that the base locus of 03C0 isequal
to03C3(P),
whichcoincides with the centre of A 03C0. Choose a smooth
point x
ofC*,
and con-sider the embedded
tangent
spaceTx C*
to C* at x.Counting
the dimen-sions,
one proves thatTx C*
ando(()
do not meet for ageneral P,
whichimplies
that C* isseparable
overC* (see,
forexample, [28, §3]).
Thiscompletes
ourproof.
REMARK:
Similarly,
one can prove that ifdim P 4,
thenK(C)/K(C*) = K(C)/K(Ct). Moreover,
it can be shown that when dim P =3,
thisequality
does not hold if and
only
ifi(C)
isstrange
and not contained in anyplane
in
P,
wherei(C)
is called strange if there exists apoint
in P which lies onall the
tangent
lines at smoothpoints
ofi(C) (see,
forexample, [5, IV, §3]
or
[29, II]).
COROLLARY 2.2: For an element K’
of K’, if K(C)
isseparable
overK’,
thenK(C) =
K’.Proof.-
For amorphism
i: C ~ P associated toK’, using
Theorem2.1,
onecan reduce the
problem
to the case dim P = 2. In this case, the result isknown
(see,
forexample, [9, §9.4]
or[14,
p.310]).
COROLLARY 2.3: For a curve
C, the following
conditions areequivalent:
(1)
JT containsK(C);
(2) K’un contains K(C);
(3) K’imm
containsK(C);
(4) p ~
2.Pro of.
Theimplications (3)
~(2)
~(1)
are obvious.For a
morphism
1: C ~P,
birational onto itsimage,
we denoteby
m theintersection
multiplicity
of1 (C)
and ageneral tangent
line to1 (C)
at ageneral point.
We note that m + 1 isequal
to the third gap of the linear systemdefining
l. It follows from Theorem 2.1 and[6, Proposition 4.4]
thatK(C)/K(C*)
isseparable
if andonly
ifp =1=
2 and m = 2. This proves theimplication (1)
~(4).
Moreover,
it follows from the Riemann-Roch theorem that if 1 is a closed immersion definedby
acomplete
linearsystem
withsufficiently large degree,
then m = 2.
Thus,
theimplication (4)
~(3)
follows fromCorollary
2.2.REMARK:
Combining
Theorem 2.1 with[6, Proposition 4.4],
one can prove that ifK(C)/K(C*)
is notseparable,
then itsinseparable degree
isequal
to m.REMARK:
Using
Theorem2.1,
one can prove that if the Gauss map of a curveC in P via
gives
aseparable
extensionK(C)/K(C*),
theni (C)
is notstrange
in P.§3.
Gauss maps and ruled surfacesThe
following plays
akey
rolethroughout
the rest of this article.THEOREM 3.1: For a
subfield
K’of K(C)
such thatK(C) is fi’nite, inseparable
over
K’, the following
conditions areequivalent:
(1)
K’ containsK’;
(2)
there exist a rank 2 vector bundle é on a curve C’ withK(C’)
= K’ anda
morphism
h: C ~P(&),
birational onto itsimage,
such that the extensionK(h(C))/K(C’) defined by
theprojection of P(é)
over C’ coincides withK(C)/K’.
This is true
for K’un (respectively, K’imm) if
one assumes moreover that h isunramified (respectively,
a closedimmersion)
in(2).
We first
verify
theimplication (1)
~(2).
LEMMA 3.2: Consider a trivial extension
(eo)
and aunique
non-trivial extension(el) of(!)c by 03A91C. Then, for
a line bundle Y on C, we have a natural extensionwhich coincides with
(eo) 0 Y if
thedegree of F
is divisibleby
the character- istic p.Otherwise,
the extension coincides with(03B51)
Q 2.For a
morphism
r C -P,
birational onto itsimage, combining
the Eulersequence on P with the extension
above,
we have a commutativediagram
with exact rows. Let C’ be the normalization of the
image
C* of C under theGauss map, and let
Co
be a section of the ruled surfaceP(9c)
over Cassociated to the
quotient i*OP(1)
of9c,
where9c
is theimage
ofal.
Wehave a commutative
diagram (see [ 11, §1])
where X is an
image
ofP(9c ) in P, and f is
a naturalmorphism
inducedby
the Gauss map.
Intuitively, Co
isconsisting
ofpoints
of contact onP(9c).
We see that the induced
morphism Co - P
above coincides with 1 viaC0 ~
C.Now,
let é be thepull-back DC’
and h thecomposition of f|C0
with theisomorphism
C -Co.
Since 1 is birational onto itsimage,
so ish,
and wehave
K(h(C))/K(C’)
=K(C)IK(C*). Moreover,
if i is unramified(respect- ively,
a closedimmersion),
then so is h. Thiscompletes
theproof
of(1)
~(2).
REMARK: The curve
i(C)
isstrange
if andonly
if themorphism P(QC’) ~ X
above is not finite. It follows from the
proof
of[18, Proposition 3]
that if iis
unramified,
theni(C)
is notstrange except
for the casewhen p
= 2 andi(C)
is a conic.The converse
(2)
=>(1)
follows fromLEMMA 3.3: Let lff be a rank 2 vector bundle on a curve
C’,
and let h: C -P(03B5)
be a
morphism
birational onto itsimage. If’h(C) is.finite, inseparable
over C’via the
projection of P(03B5),
then there exists amorphism
i: C ~P,
birationalonto its
image,
such that the extensionK(C)/K(C*) defined by
the Gauss map coincides withK(C)/K(C’).
Moreover,if h
isunramified (respectively,
a closedimmersion),
then so is 1.Proof:
Take asufhciently
veryample
line bundle Jt onC’,
andlet o
be anembedding
ofP(é)
into aprojective space P
as a scroll definedby
acomplete
linear
system
associated toOP(03B5) (1)
Q03C0*N,
where n is theprojection
ofP(é).
We define i to be acomposition of o
with h.Then,
theimage
of eachfibre of
P(03B5) under o
is a line inP, tangent
toi(C)
sinceh(C)
isinseparable
over
Ç’. Denoting by V
the vector spaceH0(P(03B5), (9P(8) (1) ~ 03C0*N),
wehave a commutative
diagram
where all the arrows are
surjective, Q
is determinedby
the upper arrow, andi is determined
by
the lower arrow. Let £f be aquotient
line bundle of(03C0h)*03B5
associated to h(see,
forexample, [5, II, Proposition 7.12]). Taking
the inverse
image by
h and amorphism P((03C0h)*03B5) ~ P(&), taking
thedirect
image by
theprojection
ofP((n 0 h)*é)
overC,
weget
where all the arrows are
surjective.
We find that thisdiagram
coincides withwhich is obtained from the commutative
diagram
under Lemma 3.2 for our v, and the extension of function fields definedby
the Gauss map of C via iis
equal
toK(C)/K(C’).
This
completes
theproof
of Theorem 3.1.COROLLARY 3.4: The set )f" contains any
subfield
K’of K(C)
such thatK(C) is.finite, inseparable
over K’.Proof.-
Let K’ be a subfield ofK(C)
asabove,
and let C’ be a curve withK(C’)
= K’.According
to Lemma 3.5below,
there is aprimitive element
184
of
K(C)
over K’.From 03C8
and amorphism
C ~ C’ definedby K(C)/K’,
weobtain a
morphism
Then,
we find that h is birational onto itsimage,
and the result follows from Theorem 3.1.LEMMA 3.5: An
arbitrary finite
extensionof function fzelds of dimension
1 overk is
simple.
Proof:
LetK/K’
be an extension asabove,
and letKs
be theseparable
closureof K’ in K. A
finite, separable
extensionK’s/K’
issimple.
On the otherhand,
an
arbitrary
intermediate field of thepurely inseparable
extensionK/KS
is ofthe form
Kp‘ for
someinteger
1 > 0(see,
forexample, [5, IV, Proposition 2.5]),
whose number is finite.So,
it follows from[10,
Theorem15,
p.55]
thatK/KS
issimple. Thus, according
to[10,
Theorem14,
p.54],
the extensionK/K’
issimple.
From now on, we shall focus our attention on
Y.n
andK’imm.
COROLLARY 3.6:
If C is a
rational curve, then f’ =K’imm.
Proof.-
For a proper subfield K’ inK’, let
C’ be a curve withK(C’) = K’,
and let h be a
graph morphism
of C ~ C’ definedby K(C)/K’:
h: C - C x
C’,
which is a closed immersion. The result follows from Theorem 3.1 and
Corollary
2.3.REMARK: A similar
argument
to the above can be found in[26, Example 2.13].
COROLLARY 3.7: For an element K’
of K’un (respectively, K’imm),
letKl
be anintermediate field of K(C)IK". If K(C)/K’1
is notseparable,
thenK’un (respect- ively, K’imm) contains K’1.
Proof.-
Let C’ be a curve withK(C’) =
K’.According
to Theorem3.1,
we have a vector bundle g ofrank
2 on C’ and an unramifiedmorphism
C -
P(03B5),
birational onto itsimage,
such thatK(h(C))jK(C’) = K(C)/K’.
Let
Cl
be a curve withK(C’1) = KI,
and let61
be apull-back
of éby
amorphism C’1 ~
C’ definedby K’1/K’. Using
h and amorphism
C ~C"
defined
by K(C)/K’1,
weget
an unramifiedmorphism
C ~P(03B51),
birationalonto its
image,
such thatK(h, (C))/K(C’1) = K(C)/K’1. Then,
the result forK’un follows
from Theorem 3.1. If h is a closedimmersion,
then so ishl.
Thiscompletes
theproof.
COROLLARY 3.8: For an element K’
of K’un (respectively, let Kl
be asubfield of
K’ such thatK’/K’1
isfinite, purely inseparable,
let C’ andC’1
becurves with
function fields
K’ andKï, respectively,
and letf :
C’ -+C,
be amorphism defined by K’/K’1. Then,
thefollowing
conditions areequivalent:
(1) (respectively., K’imm) contains K’1;
(2)
there exists a rank 2 vector bundle03B51
onCï
such thatfor
some vector bundle é on C’ associated to K’ as in Theorem 3.1.Proof.- (1)
~(2).
This follows from Theorem 3.1.(2)
=>(1).
Let h be themorphism
C -P(03B5)
associated to K’ as in Theorem3.1,
and lethl
be themorphism
h followedby
amorphism P(03B5) ~ P(03B51)
induced
from f. By
virtue of Theorem3.1,
it suffices to show thathl
isunramified,
birational onto itsimage (respectively,
a closedimmersion).
Since f is purely inseparable,
themorphism P(03B5) ~ P(,91)
isinjective. So,
it is sufficient to
verify
that atangent
vector toh(C)
inP(03B5)
at eachpoint of h(C)
is not vanished underP(03B5) ~ P( lC 1). But,
this is true since a fibreof
P(é)
over C’ istangent
toh(C),
andmapped isomorphically
intoP(,91) by P(03B5) ~ P( lC 1).
§4.
Calculation on ruled surfaces The third of our fundamental results isTHEOREM 4.1 : Let t: C - P be a
morphism
birational onto itsimage,
let g andg’
be the generaof
C andC’, respectively,
and denoteby pa(f(C0))
thearithmetic genus
of f (Co),
with the same notations asbefore. If
1 isunramified,
then we have
186
We denote
by d
thedegree
ofi*OP(1),
thatis,
thedegree
ofi(C)
inP,
andby
d* thedegree
of C* in G via the Plückerembedding,
where C* is theimage
of C in G under the Gauss map. We write n for thedegree
of theextension
K(C)/K(C*),
thatis,
thedegree
of the Gauss map of C. We denoteby v
themorphism P(DC’) ~ X,
andby
y thedegree
of the cokernel of the mapa 1.
One can
easily
prove thefollowing (see,
forexample, [ 11, §1], [19, §3]
or[25, §3]).
LEMMA 4.2: With the same notations as
before,
we haveTo prove Theorem
4.1, using
ageneric projection
ofP,
we may assume that¡P = ¡p2.
Writing
D for theimage f(C0),
we firststudy
the numerical class of D in the ruled surfaceP(DC’).
For ageneral line
L inP2,
setin
P(QC’). Then,
H is a section ofP(QC’)
over C’. Since X coincides with1fD2,
in
particular,
it hasdegree 1,
it follows from Lemma 4.2 that v hasdegree
d*.
Thus,
we have(H2) =
d*. Sincevlf(Co)
is birational onto itsimage,
wehave
(D. H) =
d. On the otherhand, denoting by
F the numerical class ofa fibre of the
projection
ofP(QC’)
overC’,
we obtain(D. F) = n
from theproof
of(1)
~(2)
in Theorem 3 .1.Therefore, using
Lemma4.2,
we find thatDenoting by K
the numerical class of a canonical divisor ofP(9c, ),
we seeNow,
it follows from theadjunction
formula thatSo,
we obtainby
Lemma 4.2 thatWe here have
pa(D) g g’, n 1,
and y = 0 since i is unramified.Therefore,
the result follows from theequality
above. Thiscompletes
theproof
of Theorem 4.1.COROLLARY 4.3:
%:n = K’imm.
Proof.-
It follows from Theorem 4.1 thatf (Co)
issmooth,
so that theunramified
morphism
h in Theorem 3.1 is a closed immersion.COROLLARY 4.4:
If C has
genusg 2,
then anarbitrary
element K’of )fun
is
of
theform K(C)P’for
someinteger
1 0.Proof.-
Use Hurwitz’s formula(see,
forexample, [5, IV, §2]).
COROLLARY 4.5: A smooth
plane
curve hasseparable degree
1 over the dualcurve via the Gauss map.
Proof.-
Ifg 2,
then the result followsdirectly
fromCorollary
4.4. Ifg =
0,
then the resultclearly
follows. We hence assume that g =1,
so thatd = 3. We obtain from Theorem 4.1 that
d* a 3,
and the result follows fromCorollary
2.2 and Lemma 4.2.REMARK: At the last
part
of theproof above,
we do not need Theorem 4.1 because wealways have n
d.REMARK: This result answers a
question
in[16,
p.342],
andimproves [24, Proposition 4.2].
§5. Elliptic
curvesThis section is devoted to
elliptic
curves.THEOREM 5.1: Let C be an
ordinary elliptic
curve in characteristic p, and let K’ be asubfield of K(C)
such thatK(C) is finite, inseparable
over K’.If
theseparable
closureof
K’ inK(C)
iscyclic
over K’ and theseparable degree of
K(C)/K’
is not divisibleby
p, thenK’imm contains
K’.Proof.-
Let C’ be a curve withK(C’)
= K’.By
virtue of Theorem3.1,
it issufficient to show that there exist a rank 2 vector bundle g on C’ and a closed immersion h: C -
P(é)
such thatK(h(C))/K(C’)
=K(C)/K’.
Writing s and q respectively
for theseparable
andinseparable degree
ofK(C)/K’,
we have an exact sequence of group schemesTaking
thedual,
weget
n
Since s and q are
coprime,
one may choose asingle
element 2 of C’ whichgenerates
the kernel above.Now,
weput
We note that there exist
only
two sections ofP(é)
over C’ such that the self-intersection number isequal
to zero. It followsand we have a commutative
diagram
where f is
amorphism
inducedby K(C)/K’.
We consider a constant section
Co
ofP(éc)
over C which comes fromneither of the two sections of
P(é)
mentionedabove,
and let us prove thatf|C0
is a closed immersion.Let C" be the normalization of the
image f(C0). Using
a basechange by
the
naturally
inducedmorphisms
C ~ C" ~C’,
weget
a commutativediagram
where we denote
by C"0
a section ofP(03B5C")
over C"naturally
definedby
thebase
change. Then,
we find thatCo
is apull-back
ofCJ
and the self- intersection number ofCJ
is alsoequal
to zero.Now,
supposethat f|C0
is not birational. It follows that C - C" shouldnot be birational.
Therefore, Oc,,
is not trivial andCJ
comes from either of the two sections ofP(03B5) specified above,
so doesCo.
This is acontradiction,
and it follows that
flco
is birational.It remains to show
that f(C0)
is smooth inP(03B5). Computing
its arithmetic genus, one caneasily
deduce this result.THEOREM 5.2:
If
C is asupersingular elliptic
curve in characteristic p, thenotherwise.
Proof:
We show the inclusion ~.By
Theorem3.1,
we see that the conversefollows from the existence of suitable vector
bundles,
which can be verifiedsimultaneously
below(use,
forexample, [1]).
Take an element K’ of
K’imm,
and let r C - P be a closed immersionassociated to K’. We
employ
the same notations as before.Case:
p ~
2If K(C) = K’,
then there isnothing
to prove. We may assume thatK(C)
isnot
separable
over K’.Let
Ci,
C’ and C" be curves with function fieldsK(C)p,
K’ andK’l/p, respectively.
Weget
a commutativediagram
Since is
unramified, using
Lemmas 3.2 and4.2,
we haveand
Co
is a constant section ofP(9c)
over C via anisomorphism P(QC) ~
¡pl X C.
I claim that
QC1 is indecomposable. Suppose
that9c,
isdecomposed.
SinceCI
issupersingular,
a line bundle Y onCI
withLC ~ i*OP(1)
isuniquely
determined.
Thus, LC1
should be of the form,!£ffi2
for some line bundle £f Thisimplies
that the mapis not birational onto its
image. Then,
we find a contradictionbecause flco
is birational onto its
image.
It follows that
9c,
isindecomposable.
1 claim moreover that2c,
has evendegree. Suppose
that1-9c,
has odddegree.
Itclearly
follows that.2c,,
shouldhave also odd
degree.
Since9c
is a direct sum of two line bundles with thesame
degree, 9c,,
must beindecomposable. Therefore, LC’
andLC"
are bothindecomposable
ifthey
have odddegree.
Since C - C" isisomorphic
toCi -
C’ as a finite cover of abstract curves,9c
isindecomposable
if andonly
if so is9c, (see,
forexample, [1, II]).
This is a contradiction.Since C" is
supersingular,
we havefor some line bundle uH on C".
If K(C)/K’ is
notpurely inseparable
ofdegree p, again
this contradicts thebirationality of f|C0
because the mapshould not be birational onto its
image.
Thiscompletes
theproof.
Case: p = 2
According
to Corollaries 2.2 and2.3, K(C)
isalways inseparable
over K’.Since i is
unramified,
we have an exact sequenceIt follows from Lemma 3.2 that
(8) splits
if andonly
if d is even.By
the similar way to the former case, let us consider the curvesCl , C’,
C" and the commutative
diagram
above.Subcase:
(03B5) splits
1 first claim that
22Cl
isindecomposable
and has evendegree.
Theindecompos-
ability
ofLC1
followsby
the similar way to the former case. If22Cl
has odd191
degree, then, according
to[23,
Theorem2.16], 2,
should beindecomposable.
This is a contradiction.
Next,
1 claim thatK(C1)/K(C’)
is either trivial or notseparable. Suppose
the
contrary.
SinceCi
issupersingular,
thedegree
ofCi -
C’ should not bedivisible
by p,
thatis,
odd. SinceQcl
isindecomposable
with evendegree,
sois
9c, . Thus, LC"
should be of the form N~2 for some line bundle N on C"because C" is
supersingular,
and this contradicts thebirationality of f|C0
asabove.
Now, if K(C1) = K(C’),
thenK(C)
ispurely inseparable of degree p
overK’. We hence consider the case when
Ci -
C’ is notseparable. Then,
onecan take a curve
C2
betweenCi
and C’ withK(C2) = K(C)p2,
and C"’between C and C" with
K(C"’) = Ktl/p2.
We obtain a commutativediagram
1 claim that
fLC2
has odddegree. Suppose
thatfLC2
has evendegree. Then, 2c,
should bedecomposed
sinceCl
issupersingular.
This is a contradiction.Therefore, 2c,
isindecomposable
with odddegree
andLC’’’
is of the form N°2 for some line bundle X on C"’.If K(C)/K’
is notpurely inseparable
ofdegree p2, then
we find a contra-diction
by
the similar way to the lastpart
of the former case. Thiscompletes
our
proof.
Subcase:
(03B5)
does notsplit
We see that
Co
is aunique
section ofP(9c)
over C such that the self- intersection number isequal
to zero, and9c
isindecomposable
with evendegree.
1 claim that
LC1
has odddegree. Suppose
thatJcl
has evendegree. Then, 2c,
should have a non-trivial extensionwith some line bundle Y on
CI
because-2c,
isindecomposable.
We find thatthe
pull-back
of(03B51)
to C must coincide with(E). Therefore, similarly
asabove,
this contradicts thebirationality
offlco.
Hence, 9c,
and9c,,
areindecomposable
with odd and evendegree, respect- ively,
and we haveK(CI) = K(C’) since f|C0 is
birational onto itsimage.
Thus, K(C)/K’
ispurely inseparable
ofdegree
p.REMARK: In the latter subcase
above,
we have not used theassumption
thatC is
supersingular.
§6.
Curves ofhigher
genusThe main purpose of this section is to prove
THEOREM 6.1: Let C be a curve in characteristic p with genus
g 2,
letf :
C ~ C’be a Frobenius
morphism of C,
and let 2 be a line bundle on C such that thedegree of P1C(L)
is divisibleby
p.Then, the following
conditions areequivalent:
(1) 4i"mm
containsK(C)P;
(2)
there exists a stable vector bundle éof
rank 2 on C’ such thatCombining
Theorem 6.1 withCorollary 3.8,
weget
COROLLARY 6.2: Let C be as above. For an
integer
1 >0,
letbe a sequence
of Frobenius morphisms of
C.Then, the following
conditions areequivalent:
(1) K’imm contains K(C)P’;
(2)
there exists a vector bundle éof rank
2 onc(p/)
such that03B5C(p)
is stable andéc
~&b(2) for
some line bundle 2 on C.To prove Theorem
6.1,
let us use Theorem 3.1. For a rank 2 vector bundletff on
C’,
togive
amorphism
h : C -P(03B5),
it isequivalent
togive
aquotient
line bundle 2 of
f*03B5,
and there is a commutativediagram
where
P(L)
is a section ofP(f*03B5)
over C.Then,
we haveLEMMA 6.3: The
following
conditions areequivalent:
(1)
h isbirational
onto itsimage;
(2)
thequotient
line bundle 2of f*é
does notcome from
anyquotient
linebundle
of
03B5.Pro of.
SinceP(L)
ispurely inseparable
ofdegree
p overC’,
one of the twocases below occurs:
P(L)
ispurely inseparable
ofdegree
p overh(C),
andK(h(C))
=K(C’),
soh(C)
is a section ofP(03B5)
overC’;
h is birational onto itsimage,
andK(h(C))/K(C’)
ispurely inseparable
ofdegree
p. Thiscompletes
theproof.
PROPOSITION 6.4: Let C be a curve in characteristic p, let
f :
C ~ C’ be aFrobenius
morphism of
C, and let 2 be a line bundle on C such that thedegree of P1C(L)
isdivisible by
p.Then, the following
conditions areequivalent:
(1) K’imm contains K(C)P;
(2)
there exists a vector bundle 03B5of
rank 2 on C’ such thatand the natural
quotient
line bundle Yof P1C(L)
does notcome from
anyquotient
line bundleofg.
Proof:
We first show theimplication (1)
~(2)
in caseof p =1=
2. For a closedimmersion 1: C - P associated to
K(C)p
inK’imm, we
haveIn
particular,
L andi*OP(1)
have the samedegree
modulo psince p ~
2.It follows from Lemma 3.2 that
where
i*OP(1)
is the dual ofi*OP(1). Thus,
there exists a vector bundle lffon C’ such that
P1C(L) ~ f*03B5,
and it follows from Lemma 6.3(1)
=>(2)
that é has the
required property.
Next,
we consider the case p = 2. For a closed immersion 1: C - P such that thedegree
of1 (C)
is even(respectively, odd),
it follows from Corollaries 2.2 and 2.3 that the extensionK(C)/K(C*)
definedby
the Gauss map via iis not
separable. Using Corollary 3.7,
we obtain a closed immersionli: C -
Pi
such that thedegree
ofi(C)
is even(respectively, odd)
and theextension defined
by
the Gauss mapvia i is purely inseparable of degree p.
In
particular,
thisimplies
the condition(1)
in caseof p -
2.Now, apply
theargument
in caseof p ~
2 above to this i1. We obtain therequired 03B5
in(2)
when the
degree
of fil is even(respectively, odd). Thus,
we haveproved
thatboth
(1)
and(2)
arealways
satisfied when p = 2.Finally,
we show theimplication (2)
~(1).
For the naturalquotient
linebundle Y
of f*03B5 via f*03B5 ~ P1C(L), by
virtue of Lemma 6.3(2)
~(1),
wehave a
morphism
h: C -P(é)
such thatK(h(C))/K(C’)
ispurely inseparable
of
degree
p.Computing
the arithmetic genus ofh(C),
we see thath(C)
issmooth,
so that h is a closed immersion.Then,
the result follows from Theorem 3.1.We here consider a
special
case of this situation.DEFINITION: For a curve C with a Frobenius
morphism f :
C -C’,
if C’ hasa line bundle .K such that:
(1) f*N ~ 03A91C,
and(2) f * : H1(C’, NV) ~
Hl(C,f*NV)
is notinjective,
then C is called a
Tango-Raynaud
curve.COROLLARY 6.5:
If C is a Tango-Raynaud
curve in characteristic p, thenK’imm
contains
K(C)p.
Proof.-
It follows from theassumption
that there is a non-zeroelement 03BE
ofH1(C’, Xv)
for some line bundle .K on C’ such thatf*(03BE) =
0 inHl
(C, f*NV ).
Take 03B5 to be the extension of Xby (9c,
determinedby 03BE,
andfil to
be f*OC’
in the situation above. The result follows fromProposition
6.4.
REMARK: If a line bundle N on C’ with
deg f* % = 2g - 2
satisfies the condition(2),
then wehave f*N ~ 03A91C.
REMARK: Consider a curve C of
genus g 2,
and denoteby n(C) Tango’s
invariant of C in
[32,
Definition11]. Then,
C is aTango-Raynaud
curve inour sense if and
only
ifand our definition above is
equivalent
to theordinary
one(see,
forexample,
[2]
or[27]).
195 REMARK: Neither rational nor
ordinary elliptic
curve C is aTango-Raynaud
curve. On the other
hand,
asupersingular elliptic
curve C is aTango- Raynaud
curve, because(9c, enjoys
therequired properties.
REMARK:
Any
curve C in characteristic 2 is aTango-Raynaud
curve if C isneither rational nor
ordinary elliptic,
because the cokernel of a natural mapOC’ ~ f*(9c
is a line bundlehaving
therequired properties (see,
forexample, [2, Exemple i),
p.81] or [20, 2.2]).
Now,
we go back to theproof
of Theorem 6.1. For a rank 2 vector bundle IF on a curve C and aquotient
line bundle Y ofIF,
weput
Moreover,
weput
where £f’s are
quotient
line bundles of F. The values(F)
is called thestability degree
of F because of the fact that F is stable if andonly
ifs(F)
> 0.LEMMA 6.6: For a rank 2 vector bundle F on a curve,
if s(F) 0,
then aquotient
line bundle .pof 5’
withs(F, L)
0 isuniquely
determined.Pro of. Taking
aquotient
line bundle L of F withs(F, L) 0,
we denoteby
-0 the kernel of F ~ L. Let L’ be anarbitrary quotient
line bundle of F withs(F, L’) 0,
and denoteby aV’
the kernel of f7 -+ L’.Then,
we have thatdeg
-SY >deg Y,
anddeg N’ deg
L’. It follows thatdeg
aV >deg L’,
anddeg
aV’ >deg
Y.Thus,
we see that £f = £f’ as aquotient
line bundle of F.For a line bundle Y on a curve C with genus g, we have
s(P1C(L), L) = - (2g - 2). Moreover, if g 2,
then it follows from Lemma 6.6 thatREMARK: One can say
that 91:(2)
is the farthest from a stable bundle on Ceven if