C OMPOSITIO M ATHEMATICA
D OMINIQUE D UVAL
Rational Puiseux expansions
Compositio Mathematica, tome 70, n
o2 (1989), p. 119-154
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119
Rational Puiseux expansions
DOMINIQUE
DUVALCompositio
© 1989 Kluwer Academic Publishers. Printed in the Netherlands.
Institut Fourier, B.P.74, F-38402 Saint-Martin-d’Hères Cedex, France Received 7 January 1988; accepted in revised form 21 September 1988
Abstract. The aim of this paper is to define and study "rational" Puiseux expansions of a plane
curve, and to describe a variant of Newton’s algorithm to compute them. The rational Puiseux
expansions of a curve give the same information as the classical ones, and in addition they give results of arithmetical nature about the curve. One major result is the easy determination of the residual field of the places of a curve over a non-algebraically closed field. This leads for example to a simple description of the real branches of curves defined over the real numbers.
My goal when 1 began studying Puiseux expansions simply was to implement Newton’s algorithm on a computer algebra system. It happens that the rational Puiseux expansions are
much easier to compute than the classical ones, mainly because less algebraic numbers are
needed.
Introduction
Let
F(X, Y)
denote apolynomial
with coefficients in a field K of characteris- tic 0. Let us assume that F is monic in Y andabsolutely irreducible, i.e.,
irreducible in
R[X, Y]
for anyalgebraic
closureÉk
of K. It is a classical factdue to Newton
[Ne],
that the roots of F(considered
as aY-polynomial)
arePuiseux series in
X, i.e.,
formal series inXile (for
somepositive integer e)
withcoefficients in
R.
For
example,
if K is the field Q of rationalnumbers,
the roots of theY-polynomial
are 6 Puiseux
series, beginning
as follows:120
In this case, the rational Puiseux
expansions
of Fgive
adescription
of theseroots
by pairs
of formal series with rational coefficients:In Section
1,
the notion of asystem
of rational Puiseuxexpansions
of Fover fM is
defined,
and it isproved
that iteasily gives
the classical Puiseuxexpansions
of F. These rational Puiseuxexpansions
of Fcorrespond bijec- tively
to the branches of theplane
curve ofequation F(x, y)
= 0 which passthrough
apoint (0, fi)
of the curve, i.e. to theplaces
of the fieldK(X)[Y]/
(F(X, Y))
which lie above theplace
Po ofR(X) corresponding
to 0 ER.
Butthe existence of
systems
of rational Puiseuxexpansions
of F over K is notproved
in this section.In the two
following sections,
we show that rational Puiseuxexpansions give
more information than the classical ones:Complete
factorization of F as aY-polynomial
is obtained over anyalgebraic
closure!K((X))
ofIK((X)),
and overIK((X)),
from classical Puiseuxexpansions
of F. It isproved
in Section 2 that from rational Puiseuxexpansions
of F is gotcomplete
factorization of F overK((X)), K((X)),
andK«X».
In section 3 are considered the
places
ofK(X)[Y]/(F(X, Y)).
We haveyet
noticed thatthey correspond bijectively
to the rational Puiseuxexpansions
of F. In
addition,
e two such
places
areconjugated
over K if andonly
if thecorresponding
rational Puiseux
expansions
areconjugated
overK,
and thus theplaces
ofthe field
K(X)[Y]I(F(X, Y)) correspond bijectively
to theconjugacy
classesover K of the rational Puiseux
expansions
ofF;
0 the
ramification
index of aplace
of eitherK(X)[Y]I(F(X, Y))
orK(X)[Y]/(F(X, Y))
is obtained aseasily
as in the classical case;e and the
residual field
of aplace
of!K(X)[Y]j(F(X, Y)) is,
up to K-isomorphism,
the extension field of K that isgenerated by
the coefficients of any of thecorresponding
rational Puiseuxexpansions.
In the
example above,
the curvehas 4 branches at
(0, 0)
overQ,
or over C. Two of them areunramified,
thetwo others have ramification index
2,
and each branchcorresponds
toexactly
oneplace
overQ,
with residual fieldQ,
since all coefficients arerational.
In Section 4 is described a variant of Newton’s
algorithm
forcomputing
Puiseux
expansions,
and it isproved
that this "rational" Newton’salgorithm computes
asystem
of rational Puiseuxexpansions
of F overK, thereby proving
its existence. Inaddition,
this rationalalgorithm
ischeaper
than the classical one because the coefficients of the series are in a smaller extension field of K.
The
implementation
of the rational Newtonalgorithm
has been per- formed on thecomputer algebra system Reduce, using
the D5 method(cf.
[D-D])
to handlealgebraic
numbers. This method does not use any fac- torizationalgorithm,
andactually
no suchalgorithm
is needed aslong
asone asks
only
for results overK.
For results overK,
factorizationalgorithms
may be needed
(for example
when K =Q),
but Sturm sequences aresufficient when K = R. These
points
are detailed onexamples
in section4,
too.
In Section
5,
we prove that the number ofelementary operations
overK that are needed for the
computation
of asystem
of rational Puiseuxexpansions
of F is"polynomial",
when the D5 method is used.Basic references are: Walker
[Wa]
forparameterizations
and classical Puiseuxexpansions,
Fulton[Fu]
for a moregeometric point
ofview,
andChevalley [Ch]
for arithmeticalpoints.
Contents
1. Classical and rational Puiseux
expansions
From now on, K denotes a field of zero
characteristics, K
analgebraic
closure of
K, F(X, Y)
a bivariatepolynomial
with coefficients inK,
monicin
Y,
irreducible inR[X, Y],
and M(resp. N)
thedegree
of F in X(resp.
inY),
with N > 0.Actually,
theseassumptions
may be muchweakened,
asexplained
in the conclusion.Generally,
F will be considered as aY-polynomial
with coefficients inIM[X]
where the
ai(X)’s
are inK[X],
theai,j’s
inK,
andby assumption aN(X)
= 1.As
usual,
the power series field in one variable T and with coefficients insome field L is denoted
by L((T)).
And the domain formedby
the elementsof
L((T))
ofnon-negative
T-orderby L[[T]].
For everypositive integer q,
aroot
Xllq
of X oforder q
is chosen in analgebraic
closure ofK(X),
in sucha way that
(X1/q1q2)q2
=XI/ql.
DEFINITION: The Puiseux series
field R
overK
is the union of all the fieldsK((X1/q)).
It is analgebraically
closedfield, by
Puiseux theorem[Wa, 4
thm3.1.].
DEFINITION: The
(classical)
Puiseuxexpansions
ofF(X, Y)
are the roots ofthe
Y-polynomial
F in the field R.Since F is irreducible in
R(X)[Y] (by
Gauss’lemma)
and since R isalgebraically
closed and of characteristic0,
theY-polynomial
F has Ndistinct classical Puiseux
expansions. They
are denotedIt will
always
be assumed thateach ék
is as small aspossible,
i.e.that ék
andthe nk,h
(for h 1)
have no common factorgreater
than 1.This ék
is calledthe
ramification
index of the seriesk .
Inaddition, yk
is inK[[X1/ek]]
becauseF has been assumed monic in Y.
Let us now recall the definition of the
parametrizations
of theplane
afiinecurve C of
equation F(x, y) =
0 overK, following
Walker[Wa, 4-2].
DEFINITION: A
parametrization
of the curve C is apair (x, y)
of elements ofK[[T]]
for some new variableT,
such thatF(x, y) =
0 inR[[T]J,
and x andare
not both inK.
The center of the
parametrization (x, )
is thepoint (xo, yo )
of the affineplane
such that xo(resp. Yo)
is the constant term of the series x(resp. ).
It is a
point
of the curve C. Theparametrization (x, )
is irreducible if there is nointeger
k > 1 such that both xand
are inK[[Tk]].
Andtwo
parametrizations (1, YI)
and(X2’ 2)
areequivalent
if there is somez E
K[[T]],
of T-order1,
such that2(T) = x1(z(T))
andY2 (T) = l (z(T)).
If such is the casethey
have the same center, but thereciprocal
isfalse.
In
addition,
thecoefficient field
of(x, )
over K is the extension field ofK
generated by
all the coefficients of the series x and.
It is a subfieldof K.
A branch of C is then defined as an
equivalence
class of irreducibleparametrizations
ofC,
and its center as the common center of the param- etrizations of the class. We say that a branch of C lies above apoint
xo ofK if
the center of the branch is(xo, yo)
for someyo in IK.
For each classical Puiseux
expansion
of
F,
letSince k=- vk(X1/ek),
it follows thatF(uk(X1/ek), vk(X1/ek)) =F(X, yk) =
0in R. The
pair (uk(T), vk(T))
is thus aparametrization
ofC,
which isirreducible
because ék
has been chosen minimal. We say that(uk(T), vk(T))
is the
parametrization
of Ccorresponding
to the Puiseuxexpansion Yk.
And that two classical Puiseuxexpansions
of F areequivalent
if theircorresponding parametrizations
areequivalent.
If 03B6k
denotes aprimitive
rootof unity of order ék
inÀ ,
the classical Puiseuxexpansions
of F that areequivalent
tok
are theAll of them have the same ramification index
ék .
Thus is obtained a
partition
of theset {y1, 2, ... , yN} in equivalence classes,
which are in one-to-onecorrespondence
with the branches of Clying
above 0.
Let g
be the number of theseclasses,
and assume that the number-ing
of the Puiseuxexpansions
is such thaty1, 2, ... , g
are in différent classes. It follows that124
DEFINITION: As
above,
letg
denote the number of branches of Clying
above0. A system
of rational
Puiseuxexpansions of
F over IM is a setof g pairwise non-equivalent
irreducibleparametrizations
ofC,
which is invariant under the action of the Galoisgroup
ofK
overK,
and suchthat,
for each
k, Xk
is a monomial03BBkTek with ek
> 0 and03BBk ~
0.The fact that such
systems
exist will beproved
in section 4. For agiven F,
there are differentsystems
of rational Puiseuxexpansions
of F: forexample
if F = Y2 - X there areIt must be noticed that the
parametrizations (uk (T ), vk(T)) (for
1
k g)
defined above from classical Puiseuxexpansions usually
do notform a
system
of rational Puiseuxexpansions of F,
becausethey
do not forman invariant set under the action of :
In the
example
of theintroduction,
theonly
classical Puiseuxexpansion equivalent
towhich is
conjugated
to(U3’ V3)
over K = Q.On the
contrary,
it is very easy to determine the classical Puiseuxexpansions
of F from asystem
of rational Puiseuxexpansions:
Let(for
1k g)
be asystem
of rational Puiseuxexpansions
of F overK,
let03B6k
denote aprimitive
rootof unity of order ek
inlÉ,
and03BB-1/ekk
a root of orderek of
Âk in R
for each k. Then:THEOREM 1: The classical Puiseux
expansions of
F are the seriesfor k =
1, 2,
...,g
and i =1, 2,
..., ek.The
ramification
indexof k(03B6ik03BB-k I/ek Xl/ek)
isexactly
ek.Pro of.
Consider some k between 1 andg.
ThenNow if we set X =
03BBTe, i.e.,
T =03B6ik03BB-k
IlekX1/ek
for somei,
weget
This proves that each of the
for 1
k g
and 1 iek
is a classical Puiseuxexpansion
of F.They
aremutually
distinct: for different k because the rational Puiseuxexpansions
arepairwise non-equivalent,
and for the same k and different i because the rational Puiseuxexpansions
are irreducible asparametrizations
of C. For the same reason, ek is the ramification index
of k(03B6ik03BB-1/ekkX1/ek)
forany i.
The fact that every classical Puiseux
expansion
of F is obtained in this waynow comes from the
identity
N =03A3gk=1ek,
and thus theorem 1 isproved.
Now, let g
denote the number of -orbits in agiven system
of rational Puiseuxexpansions
of F overK,
and suppose that(xk, k)
are in differentorbits for 1
k
g. For each k from 1 to g, let us denoteby Lk
thecoefficient field of
(Xk, k).
Because of the -invarianceproperty
of theset
{(k, k)}1kg,
the fieldLk
is a finite extension of K. Let.fk
denotethe dimension of
Lk
overK, i.e.,
the number of k’in {1, 2,
... ,gl
such that(k’, k’)
is in the -orbit of(k, k),
and let{03C31,
03C32,...,03C3fk}
be the set ofK-isomorphisms
fromLk into K.
Forevery j
from 1to fk,
let03BBk,j
=03C3j(03BBk)
and
Yk,J
=03C3j(k)
=03A3+~h=103C3j(03B1k,h)Tnk,h. By definition,
thegiven system
of rational Puiseuxexpansions
of F over K is made of thefor 1 k g and 1
j fk ,
and theorem 1 can beexpressed
as follows:THEOREM 1 bis: The classical Puiseux
expansions of
F are thefor
1k
g, 1j fk,
and 1 i ek. Theramification
indexof
theexpansion
above is ek.It follows that g =
03A3gk=1 fk
and thatIn the
following diagram,
the first linecorresponds
to the N classicalPuiseux
expansions
ofF,
the second line to theirg equivalence classes,
or toa
system
of rational Puiseuxexpansions
of F overK,
and the third line tothe g
orbits of thissystem
under the action of :2. Factorization of F over series fields
By
definition of the classical Puiseuxexpansions
ofF,
the factorization of F inR[Y]
isAnd if the classical Puiseux
expansions
aregrouped according
toequivalence,
a refinement of the above factorization is obtained:where
In
addition,
this factorization overK((X))
iscomplete
in the sense that eachFk(X, Y)
is irreducible inK((X))[Y]:
If the norm in the field extensionK((X1/ek))/K((X))
is extended topolynomials,
thenFk(X, Y)
is the norm of(Y - yk),
which is irreducible inK((X1/ek))[Y],
and it follows thatFk(X, Y)
is a power of some irreducible
polynomial
ofK((X))[Y] [Trl,
thm2.1].
ButFk(X, Y)
has distinct roots inSi,
and thus it is irreducible inR«X»[Y].
REMARK: If some classical Puiseux
expansion
of F is"finite",
sayk
EK[X1/ek], then Fk
ER[X, Y],
and since F isabsolutely
irreducible it follows that F =Fk
andg
= 1.Thus, complete
factorization of F over R and over!K((A"))
is obtainedfrom the classical Puiseux
expansions
of F. The next result proves that it can also be obtained from asystem
of rational Puiseuxexpansions
of F overK, together
with thecomplete
factorization of F overK((X)).
THEOREM 2: With the notations
of
section1,
thecomplete factorization of
F(as a Y-polynomial)
over the fieldsK((X)), K((X))
andSi,
isgiven by:
REMARK: The last
equality
can also be written:Pro of.
Theorem 1 bis proves that thecomplete
factorization of F over R isBut notations are such that a set of
g non-equivalent
classical Puiseuxexpansions
of F isgiven by
thek(03BB-1/ekk,jX1/ek) for 1 k g
and1
j fk,
and thus theFk,j’s
for 1k
g and 1j fk are exactly
the
Fk’s
for 1k g. So,
it has beenproved
above that thecomplete
factorization of F over
K((X))
isNow,
for agiven k
between 1and g,
if aj is theidentity
of03A3k,
thenFk(X, Y)
is the norm
of Fk,03C31(X, Y)
in the extensionLk((x))/K((X)).
Itfollows,
asabove,
thatFk(X, Y)
is a power of some irreduciblepolynomial
ofK((X))[Y],
and that it is itself irreducible inK((X))[Y]
since it has distinct roots in Si. Thus thecomplete
factorization of F overK((X))
isand theorem 2 is
proved.
Those différent factorizations of F can be visualized
by
thefollowing
diagram, where Fik,j
= Y -k,j(03B6ik03BB-1/ekk,jX1/ek),
and where the first line corre-sponds
to the factorization inR[Y],
the second line overK((X))[Y],
and thethird line over
K((X))[Y]:
3. The
places
ofK( C)
and ofK(C)
Since Fis irreducible over
K,
thequotient K(X)[Y]/(F(X, Y))
is an exten-sion field of
R(X) of degree N,
that will be denotedby K(C).
This notationcomes from the
interpretation
ofK(C)
as the function field of theplane
curve C of
equation F(x, y)
= 0. The extensionK(C)/K(X) corresponds
tothe
projection
of C on the "x-axis"[Fu, 6-3].
In the same way, since Fis also irreducible overK,
thequotient IK(X)[Y]j(F(X, Y))
is a fieldK(C),
exten-sion of
K(X)
ofdegree
N.The branches of C have been defined in section 1 as the
equivalence
classesof irreducible
parametrizations
of C. To eachparametrization (, )
of C areassociated two
subsets C and 03B2
ofK(C),
suchthat 6
is aring and 03B2
amaximal ideal of
(i. By definition, 03B2
is aplace
ofK(C) and (9
is thering of 03B2. They
may be defined as follows:Consider the
[K-algebra homomorphism
Notice that
Ker(g)
contains no non-zero element ofR[X]: If ~(G) =
0 withG E
K[X]
and G ~0,
then x would be a root of G inK, and
wouldbe a root of the univariate
polynomial F(x, Y)
ofR[Y]
ofpositive degree
N. But
then
would bein R
too, which contradicts the definition of aparametrization.
It is thus
possible
toextend ~
to aR-algebra homomorphism
still denotedby
03C8:Since
(x, )
is aparametrization
ofC,
thepolynomial
F is contained inKer(cp).
And since F is irreducible in theprinciple ring R(X)[Y],
weget Ker«p) = (F).
Thus is obtained an
injective !K-algebra homomorphism
Now, à and 03B2
are definedby
Two
parametrizations
of Cgives
thesame 6 and 13
if andonly
ifthey
areequivalent,
which proves that each branch of Ccorresponds
toexactly
oneplace
ofK(C).
Actually,
we do notget
in this way allplaces
ofrK(C).
We shouldhave,
in
addition,
consideredparametrizations
centered at the"points
atinfinity"
of C in order to
get
allplaces
oflÉ(C).
It does not matterhere,
because weare
only
interested in theplaces
of(C)
that "lie above" po(cf. below),
andall of them are finite since F is monic in Y.
In the case of F =
Y,
we haveR(C) = R(X).
The curve C is thex-axis,
and has
only
one branchlying
above 0. One of itsparametrizations
is(0, 0)
=(T, 0),
and thecorresponding ring
andplace
arerespectively
K[X] and p0 = XK[X].
In the
general
case, aplace 03B2
offM(C)
lies above theplace Po
ofR(X)
if03B2
nR[X]
=p0. Those places correspond
to the branches of Clying
above0. We have seen that such a branch of C is the
equivalence
class of e classical Puiseuxexpansions
ofF,
each with ramification index e. Thisinteger e
iscalled the
ramification
index of thecorresponding place 03B2.
The
places
ofK(C)
are now defined as theintersections 03B2
nK(C)
of theplaces
ofn«C)
withK(C).
In this situation too, we saythat 13
lies abovei3
nK(C).
Let03B21
and03B22 be
twoplaces
ofR(C) lying
above theplace Po
of
K(X).
It is not easy todecide,
from classical Puiseuxexpansions
ofF,
whether03B21
and03B22 lie
above the sameplace
ofK(C).
But we shall see intheorem 3
that,
on thecontrary,
it is veryeasily
done from anysystem
of rational Puiseuxexpansions
of F over K.If i3
is aplace
ofK(C), 03B2
aplace
of-K(C) lying
above03B2,
and if(9
= 6
nK(C) where 6
is thering
of03B2,
then L =W1i3
is a field. It is afinite extension of
K,
called the residualfield
of03B2,
and itsdegree f
is theresidual
degree of 03B2.
Thisinteger f is
alsoequal
to the number ofplaces 03B2
of
K(C) lying
above03B2. If F = Y,
then po =XK[X]
is aplace
ofK(X);
theonly place
offM(C) lying
above po ispo,
and the residual fieldof po
is K.Now, when i3
is aplace
ofK(C) lying above po,
we shall see that Land f are easily
obtained from a set of rational Puiseux
expansions
of F overK,
which is nottrue for classical Puiseux
expansions.
Let us denote
by 03B21, 03B22,..., 03B2g*,
theplaces
ofK(C) lying
above theplace
po ofK(X).
For 1k g*,
letet
be the ramificationindex, fk*
theresidual
degree,
andL*k
the residual fieldof 03B2k.
Theintegers et and f*k
arerelated to N
by
the formulaThe
following
result proves that the *’s are useless:THEOREM 3: For k =
1, 2,...,
g, letek , fk
andLk
bedefined
as in section2, from
a set{(xk, yk)}1kg of representatives of
the -orbits in a systemof
rational Puiseux
expansions of
F over K.131 For k =
1, 2, ... , g*,
letet, fk*
andL*k
bedefined
asabove, from
theplaces {03B2k}1kg* of K(C) lying
above Po.Then
g*
= g, and(up
toreordering
theplaces 03B2k) e* k = ek , fk* = fk ,
andL*k
isK-isomorphic
toL k for
k =1, 2,
... , g.Proof.- (using Chevalley’s
book[Ch]).
It has been
proved
in theorem 2 that thecomplete
factorization of F inK((X))[Y]
is F =03A0gk=1 Fk.
Itfollows, by
the Chinese remaindertheorem,
that the
K((X))-algebras
are
isomorphic,
and that eachK((X))[Y]/(Fk)
is a field.On the other
hand, K((X))[Y]/(F(X, Y))
isequal
to thepo-adic
com-pletion K(C) ~K(X) K((X))
ofK(C),
it is thusisomorphic
to theproduct
ofthe
03B2k-adic completions
ofK(C)
for k = 1 tog*,
which are fields.This proves that
g*
= g, and that theplaces 03B2k
can be numbered in sucha way that
K((X))[Y]/(Fk)
is the03B2k-adic completion
ofK(C).
The fact that
Fk
is theproduct of fk
irreducible factors of the samedegree
ek in
K((X))[Y]
proves that thereare fk places
of!K(C) lying
above03B2k,
eachof them with ramification index ek. This means
that fk
is the residualdegree
of
03B2k and ek
its ramificationindex,
i.e.that f*k = fk and e*k
= ek .It remains to prove that the coefficient field
[lk of (k, k)
isK-isomorphic
to the residual field
L*k
of03B2k.
Since we havejust proved
thatthey
both havethe same
degree
overK,
weonly
have toget
aninjective K-algebra
homo-morphism
fromL*k
intoLk.
From now on, aK-algebra homomorphism
isalways
one that preserves 1.Thus,
sinceL*k
is afield,
weonly
need aK-algebra homomorphism
fromL*k
in[lk.
This isgiven by
thefollowing
lemma:
LEMMA:
Let 03B2
be aplace of [K(C)
and L* itsresidual field.
Let(, )
be aparametrization of
anyplace 03B2 of K(C) lying
above03B2,
and let L be thecoefficient field of (x, ).
Then there is a
K-algebra homomorphism from
L* into IL.Pro of.
Let us come back to the[K-algebra homomorphism
132
The
restriction 03C8 of ~
toK[X, Y]
is aK-algebra homomorphism
with values inL[[T]].
As for ç, it induces aK-algebra homomorphism
and if O is the
ring of i3
then’F«9)
cL[[T]].
On the other
hand,
theapplication
. such that
is a
K-algebra homomorphism.
And since theK-algebra homomorphism 039803A8|o
from O to Lhas 03B2
in itskernel,
it induces aK-algebra
homo-morphism
fromW1i3 =
L* to L. This proves thelemma,
and theorem 3.REMARK: The fact that the -orbits of rational Puiseux
expansions
of F arein one-to-one
correspondence
with theplaces
ofK(C) lying
above po issimply
a different statement of theorem 2.In the
following diagram,
ramification indices are on the firstline, places
of
K(C) lying
abovepo
are on the secondline,
andplaces
ofK(C) lying
above po on the third line:
In
addition,
theorem 3 proves that the residual field of aplace
is K-iso-morphic
to the coefficient fieldof any
rational Puiseuxexpansion
of F in thecorresponding
-orbit.Among
the consequences of thisproperty
are:0 The fact that the coefficient field of the rational Puiseux
expansion
ofa
given
branch of C is "the smallest one" among the coefficient fields of theparametrizations
of this branch(by
the lemmaabove).
It will be seen insection 4 that the
computations
can be made in this field.2022 When K is the field R of real
numbers,
since a branch of acomplex
curve
(defined by
a realpolynomial)
is real if andonly
if its residual fieldsis the field of
reals,
weget
thefollowing
result:COROLLARY: Let
F(X, Y) be
a bivariatepolynomial
with realcoefficients,
and{(k, k)}k a
systemof rational
Puiseuxexpansions of F
over R.Then for
eachk, the
branchof
Cparametrized by (xk, k)
is realif and only if
thecoefficient of xk
and everycoefficient of k
are real numbers.This
corollary
leads to asimple
determination of the real branches ofC,
since it can be decided whether every coefficientof xk
andof k
is real witha finite computation (cf.
section4).
EXAMPLES: Let us come back to F =
(X2
+Y2)3 - 4X2Y2
over K = Q.We have seen that F has a
system
of rational Puiseuxexpansions
over Qconsisting
of 4expansions,
each with coefficient field Q. It follows thatQ(C) (resp. Q(C))
has 4places lying
abovePo (resp.
abovep.).
Now,
let F =(2X2
+Y2)3 - 8X2 y2
over Q. It also has asystem
of 4 rational Puiseuxexpansions
overQ, namely
It follows that
Q)(C)
still has 4places lying
above po, while the fieldQ(C)
has
only
3places lying above po :
the two lastexpansions
areconjugated
overQ and thus
they correspond
toonly
oneplace
ofQ(C),
with ramification index 2 and residualdegree
2.The two
following examples, though
verysimple, point
out the basic ideasof the
algorithm:
Let F = y2 + X over Q. The classical Puiseux
expansions
of F areand it has a
system
of rational Puiseuxexpansions consisting
of theunique pair
The coefficient field is
Q(i)
in the classical case, andsimply Q
in the rationalcase, and thus the
only
branch of center(0, 0)
of thisparabola
is real. Insome sense, the coefficient
(2013 1) of xl
causes thedisparition
of the irrational coefficient i.But it is not
always
sosimple.
Anexample
of a moregeneral
situation is F = Y7 -9X5
over Q. The classical Puiseuxexpansions
of F are the134
where
v, (T) = 91/7 TS
and(,
is aprimitive
rootof unity
of order 7in Q.
Asystem
of rational Puiseuxexpansions
of F is made of theunique expansion
which
implies
that C hasonly
one branch centered at(0, 0)
and that this branch is real.Actually
in thisexample
theexponent 1/7
of 9in v1
is"replaced" by - 2,
which is the inverse of 7 modulo5,
and this "mistake"is corrected
by
the coefficient9-3 of 1 (notice
that - 2 x 7 + 3 x 5 =1).
It may be noticed
that {(3T7, 3T5)}
is anothersystem
of rational Puiseuxexpansions
of F over Q. It issimpler,
but cannot be obtaineddirectly by
thealgorithm
of Section 4.4. Rational Newton
algorithm
The aim of this section is to describe
simultaneously
the classical and rational Newtonalgorithms,
as two différent versions of aunique algorithm.
This
algorithm
consists in twoparts, respectively
called thesingular
andthe
regular part.
0 The
singular
part is the one we are interested in: thispart gives
theinteger g,
and for each k from 1 tog,
itgives k, ek, fk, Lk,
andRk
coefficients of the seriesk,
for somepositive integer Rk .
0 The
regular
part issimpler
and does notdepend
on the chosenversion,
but
computations
areperformed
in a smaller field in the rational case. Thispart gives
as many terms as wanted for thek’s.
If alarge
number of termsis
required,
thealgorithm of Kung
and Traub[K-T]
is recommended in both versions.4.1. Notations
and formulae
A K-term is defined as a list 03C4 =
(q,
m,03B2)
of fourelements,
where q andm are
coprime integers, q
ispositive, and y and fi
are non-zero elementsof K.
A finite K-expansion
is a finite sequence(LI’
T2, ...,LM)
of K-terms such that mh > 0 forh
2(where
Lh =(qh,
03BCh, mh,Ph»»
AK-expansion
is thena sequence
(usually infinite)
03C0 =(rl
i2 , ..., 9 ’rh, -...)
ofK-terms,
such thatmh > 0 for
h 2,
and there exists aninteger
R such thatqh = 1, 03BCh = 1, and Ph
EK(03BC1, fil,
03BC2,03B22, ...,
03BCR,P R) for
every h > R.To each
K-expansion n
is associated apair P(n) = (x, j)
of elements ofK[[T]],
where x is a non-constantmonomial,
in thefollowing
way:Let 03C0 =
(ïi,
03C42,..., 03C4h,... ) and 03C4h = (qh,
llh, mh ,03B2h).
Consider newsymbols Xo, Xl ,
... ,XR
andYo , Y1,..., Yh ,
... and the relationsThe elimination of
Xl , X2,
...,XR-1
and ofY1, Y2 , ... , YM-1
for any M Rgives
two relationswhere Oth
does notdepend
of M. Theexpression of 03BB, e,
of ther1h’S
and thenh’s
iseasily obtained,
but is notsimple.
It isgiven
below for the sake ofcompleteness,
but it will note be used. Letso
that q’ h =
1 and ThenIt follows that
And from :
comes
and thus:
for
136
Now,
thepair P(n) = (ac, )
is definedby
The result of Newton’s
algorithm (described below)
defines a set ofK-expansions {03C0k}k.
In the classical version of thealgorithm, k
varies from 1 to N and theP(03C0k)’s
are theparametrizations (uk, Vk)’S
of section1,
associated to the classical Puiseux series of F. In the rational case, we shall prove that k varies from 1to g
and that theP(03C0k ’s
form asystem
of rational Puiseuxexpansions
of F over K.4.2.
Description of
thealgorithm
In the
description
of thealgorithm,
finite ordered sets arerepresented by
lists(... ) -
thus theempty
set is( )
-, and(a, ,
a2 , ... ,ak )
u(bl, b2 , ... , bl) represents (a, ,
a2,..., ak,b1, b2,..., bl).
Comments are in italic. And L stands for any finite extension of K.The Newton
polygon
of apolynomial F(X, Y) = 03A3Ni=003A3Mj=0ai,jXjYi
isdefined as the lower
part
of the convex hull of the set ofpoints (i, j )
in theplane
such that a¡,j is non-zero[Wa, 4].
We shall use the
following
five"primitive" algorithms:
Algorithm
COEFFICIENT(F, i, j).
In: F E
L[X, Y]; i, j non-negative integers;
Out: the coefficient of Xi Y’ in F.
Algorithm
BEZOUT(q, m).
In: two
coprime integers q
and m,with q
>0;
Out:
(u, v),
where u and v areintegers
such that uq + vm - 1.Algorithm SQUARE-FREE (03A6).
In: 0 E
L[Z];
Out: A finite set of lists
(W, r)
with 03A8 EL[Z] and r
apositive integer,
suchthat each ’ is
square-free
and (D =03A0(03A8,r) 03A8r
and the ’V’s arepairwise
coprime.
137 It is well known that such a
decomposition
can be obtained with derivations andgcd’s computations,
without anyfactorization algorithm [D-S-T].
Algorithm
NEW-POLYNOMIAL(F, i, 1).
In: F E
L[X, Y];
a K-term i =(q,
J1, m,fi);
aninteger 1 ;
Out:
X-1F(03BCXq, Xm(03B2
+Y))
EL(03BC, 03B2)(X)[Y].
This
algorithm
willalways
be used with valuesof
the parameters such that the result is inL(03BC, 03B2)[X, Y].
Algorithm
POLYGON(F, I).
In: F E
L[X, Y];
I = 1 or2;
Out: A finite set of lists
(q,
m,1, 03A6) with q,
m, 1integers, q
and mcoprime,
q >
0,
and (D EL[Z].
Those lists are in one-to-one
correspondence
with the segmentsof
the Newtonpolygon of F if I
=1,
andonly
with the segments withnegative slope if I
= 2.The
correspondence
is thefollowing:
Thesegment à corresponding
to thelist
(q,
m,1, 03A6)
is on the lineqj
+ mi = 1 in the(i, j)-plane,
and03A6 =
03A3(i,j)~0394ai,jZ(i=i0)/q
whereio
is the smallest valueof
i such that there is apoint (i, j)
on A. It iseasily
seen that (D is inL[Z]
and notonly
inL(Z’/").
Let us now come to the
description
of thealgorithm
NEWTON. Thesubalgorithm
SINGULARcomputes
thesingular part
of theexpansions,
with the
help
of SINGULAR-TERM whichcomputes
a set of terms. The result of SINGULAR is then sent to thesub-algorithm REGULAR,
whichcomputes
theregular part
of theexpansions,
and isindependent
ofthe chosen version. The différence between both versions comes from the SINGULAR-TERM
sub-algorithm.
Algorithm
NEWTON(version, F, H).
In: version has one of the two values classical or
rational ;
H is apositive integer;
F is anabsolutely
irreduciblepolynomial
ofK[X, Y],
monic and ofpositive degree
in Y. In order tosimplify
thedescription
of thealgorithm,
we assume in addition that F has no "finite" Puiseux
expansion (cf.
Section2).
If such is not the case,during
thecomputation
is met a Newtonpoly-
nomial with no
point
on thej-axis,
which means that thecorresponding expansion
iscompletely
obtained.See the conclusion
of
this paperfor
looserassumptions
on F.Out: a finite set of finite
K-expansions,
each of themcontaining
at least Hterms.
Sub-algorithms: REGULAR,
SINGULAR.begin
return REGULAR
(SINGULAR (version, F, K, ( )), H )
end.
138
Algorithm
REGULAR(S, H).
In: S is a finite set of
pairs {(03C0k, Fk)}1kB
where for eachk,
nk is afinite
K-expansion, Fk ~K[X, Y],
withFk(0, 0) = 0, (Fk)’Y(0, 0) ~ 0
andFk(X, 0) ~ 0;
H is apositive integer;
Out: A set
{03C0’k}1kB
of finiteK-expansions,
such that eachni begins
with03C0k and contains at least H K-terms.
Sub-algorithms: COEFFICIENT,
NEW-POLYNOMIAL.begin
R -
( );
results will begrouped
in the set R.for each
(n, F)
in S dowhile cardinal
(03C0)
H dom -
Min {j/COEFFICIENT (F, 0, j)
:001;
03B2
~ -COEFFICIENT(F, 0, m)/COEFFICIENT (F, 1, 0)
03C4 ~
(1, 1,
m,03B2);
7t ~ 7C U
(i)
F ~ NEW-POLYNOMIAL
(F,
i,m);
R - R w
(03C0);
return R end.
Algorithm
SINGULAR(version, F, L, 7r).
In : version as in
NEWTON;
F EL[X, Y];
03C0 is a finiteK-expansion ;
Out: a finite set of
pairs (nl, F1)
where 03C01 is a finiteK-expansion beginning by
n, and FI~L1[X, Y]
for some extension L1 of L.Sub-algorithms : SINGULAR-TERM, NEW-POLYNOMIAL,
SIN-GULAR.
This
algorithm
is recursive.begin
S -
( ); results of the algorithm
will begrouped
in the set S.if 03C0 =
( )
then I ~ 1 elseI ~ 2;
for each
(i, 1, r)
in SINGULAR-TERM(version, F, L, I )
do03C01 E- 7T U
(r);
Fl - NEW-POLYNOMIAL
(F,
T,1);
L1 ~ the extension of L
generated by
theelements J1 and 03B2
of i;if r = 1 then S - S u
((03C01, FI»
else S - S u SINGULAR
(version, FI, LI, 03C01),
return S end.
Algorithm
SINGULAR-TERM(version, F, L, I).
In: version, F and
L,
as inSINGULAR;
I = 1 or2;
I =