J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
A
NDRZEJ
S
CHINZEL
On a decomposition of polynomials in several variables
Journal de Théorie des Nombres de Bordeaux, tome
14, n
o2 (2002),
p. 647-666
<http://www.numdam.org/item?id=JTNB_2002__14_2_647_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
On
adecomposition
of
polynomials
in several variables
par ANDRZEJ SCHINZEL
Dedscated to Michel France
RÉSUMÉ. On considère la
représentation
d’unpolynôme
aplu-sieurs variables comme une somme de
polynômes
à une variable en combinaisons linéaires des variables.ABSTRACT. One considers
representation
of apolynomial
insev-eral variables as the sum of values of univariate
polynomials
takenat linear combinations of the variables.
K. Oskolkov has called my attention to the
following
theorem used inthe
theory
ofpolynomial
approximation
(see
[6],
Lemma 1 andbelow,
Lemma4):
for every sequence of d + 1pairwise
linearly independent
vec-tors
[a,., , a,,2]
(l~/~~+l)
and everypolynomial
F ~C [Xl, X2]
ofdegree
d there existpolynomials
fILE
C [z]
(1
It d +1)
such thatHe has asked for a
generalization
and a refinement of this result. Thefollowing
theorem is astep
in this direction.Theorem 1. Let
n, d
bepositive
integers
and K afield
with char K = 0or char K > d. For every sequence
S"
(2 v
n)
of
subsetsof
K eachof cardinality
at least d -f- 1 there exist M =Cn n 1 1 J
vectors( n-1 )
[a,1 ,
2? - " ? E{I}
xS2
x... x8n
with thefollowing
property.
For everypolynomial
F EK~~1, ... ,
of degree
at most d there existpolyno-mials
fp
EK[z] (1
S J-t S
M)
such thatIt is not true that
polynomials
fp.
satisfying
(1)
exist for every sequence of vectors[a,,l,
... ,(1
/-tM)
such that each n of them arelinearly
independent.
See theexample
at the end of the paper.Let
P(n,
d,
K)
be the set of allpolynomials
F ExnJ
ofdegree
d. LetM(n, d, K)
be the least number M such that for every F EP(n,
d,
K)
(1)
holds for some sequence of vectors~a~l, ... ,
E K’~ and some se-quence ofpolynomials
f ~
EK[z]
( 1 ~a M)
if such sequences exist and00 otherwise. For an infinite field
K,
letm(n,
d,
K)
be the least number Msuch that for a Zariski open subset S of
P(n, d, K)
and for every F ES,
(1)
holds for some sequences of vectors andpolynomials
asbefore,
if such sequences exist and oo, otherwise. Theorem 1implies
Corollary.
oozf
andonly if
either n = 1 or char K = 0 orchar K > d.
If K is infinite
the sameequivalence
holdsfor
m (rt,
d,
K) .
The
problem
of determination ofm(n,
d,
K)
is related to theproblem,
much studied in the XIX thcentury
(see
[5],
for a modernaccount),
ofrepresentation
of ageneral
n-ary form ofdegree
d as the sum of powers oflinear forms. The two
problems
are notequivalent
even for Kalgebraically
closed,
since in our case neither F norfp
aresupposed homogeneous.
Theorem 2. For every
infinite field
K such that char K = 0 or char K > dwe have
For K =
Fq,
char K> d,
we haveIn
particular,
every n-ary form ofdegree
d over a field K of characteristic0 is
representable
as a linear combination of(’~~ d i l)
d-th powers of linearforms over K. This has been first
proved,
but notexplicitly
statedby
Ellison
[3].
Clearly
M(I,
d,
K)
=M(n,1, K)
=1 and oneeasily
provesTheorem 3.
2,
thenM(n, 2, K)
= nand if,
inaddition, K is
infinite,
thenm(n,
2,
K)
= n.Diaconis and Shahshahani asserted without a formal
proof
thatWe shall show .
Theorem 4. For every
field
K such that either char K =0,
or char K > dand card K > 2d - 2 we have
In
particular,
everybinary
form F ofdegree
d over a field K ofcharac-teristic 0 is
representable
as a linear combination of d d-th powers of linearforms over
K,
whichslightly
improves
Theorem A of[4].
For K = C thiswas
proved by
Reznick[8].
The
following
theorem shows that the condition card K > 2d - 2 inTheorem 4 may be
superfluous.
Theorem 5. For every
field
K such thatchar K > d and card K d + 2
we have
Theorem 6. For every
algebraically
closedfield K, if
char K = 0 orchar K >
d,
thenThe
proof
of Theorem 1 is based on two lemmas.Lemma 1.
Letn 2::: 2~
(1
i~
be a subsetof K of cardinality
d+l.Then F = 0 is the
only polynomial
in~[~i,... ?
of degree
at most d ineach variable such that
F(ai, ~2,...,
= 0for
all(al, a2, ... ,
Proof.
See[1],
Lemma 2.2.Lemma 2. Let
for
each k =0,1, ... ,
n - 2 elementsof
K(0 ~
l
d)
be distinct and let
for
apositive integer q
(d
+1)wl
be the
expansion
of
q - 1 in base d + 1.Define
Then det 0.
Proof.
Let usput
in Lemma 1:Tï
={,Qi-i,c : ~ l
d} (1
i n -1).
By
the lemma theonly polynomial
F EK(xl, ... ,
xn-11
ofdegree
at mostd in each variable such that
Now,
all the vectors~lo, ... , ln_2~
E{O,
ll... ,
dln-1
can be orderedlex-icographically,
so that the vector~lo, ... , l~_2~
occupies
theposition
1 +n-2
E
+1 ) i
and then thesystem
ofequations
(3)
readsi=O
Also the
polynomial
F can be written asand
(3)
can be rewritten asThe fact that the
only
solution of thissystem
iscorresponding
to F =0,
implies
in view of(2)
thatBut then also det = det
,-~
0.Proof of
Theorem. 1. Let us choose in5~
distinct elements/3"_2,0, ... ,
(2 v n).
By
Lemma 27
hence the matrix B
consisting
of the rows r for which1 .
n
+ d -1
d is of rank
equal
to the number of such rows M= n + n - d 1 1 .
T herefo r e B has Mlinearly independent
columns 81, 82, ... , sM. Weput
Let
(note
that the multinomial coefficient isnon-zero).
For each 1 d we determine
( 1 ~
M)
from thesystem
ofequations
which can be rewritten as
By
the choice of sl, ... , sM the matrix of thissystem
has rankequal
to the number ofequations,
hence thesystem
is solvable for E K. We setand
(1)
follows from(6)
and(7).
Proof of Corollary.
In view of Theorem 1 it is sufficient to show thatM(n, d, K)
= oo if n > 1 andLet us consider an
arbitrary
polynomial F
of the form(6)
in which ad,O,...,o~
0 and ap-1,1,o,...,o
~
0. If K is infinite suchpolynomials
exist in every opensubset of
P (n, d, K) .
If(1)
holds,
then thepart
Fd-p
ofdegree
p of F satisfies
-which is
impossible,
sincexpl-l X2
occurs with a non-zero coefficient on theleft-hand
side,
but not on theright.
Proof of
Theorem 2. The dimension of the set of all n-arypolynomials
ofdegree
notexceeding
d andgreater
than e is(n!d) -
( e
On thed
where
f
= L
is at most d - e + n - 1 since the vectors a can beI=e+l
normalized
by
taking
the firstnon-vanishing
coordinateequal
to 1. Thisgives
the upper boundm(n
+ 6! 2013 1 -e)
for the dimension of the set of allm
polynomials
of the formE
f J1. (OoJ1.x)
and,
by
the definition ofm (n,
d,
K)
IA= 1
which
implies
the firstpart
of the theorem.In order to prove the second
part
let us observe that the number ofnor-malized vectors a is
q-1 ,
while the number of non-zeropolynomials
Hence we obtain at most
polynomials
of the formf (ax)
and at mostpolyno-m
mials of the
form 1:
On the otherhand,
the number of n-aryp.=1
polynomials
over1FQ
ofdegree
notexceeding
d andgreater
than e isBy
the definition ofM(n,
d,
thisgives
which
implies
the secondpart
of the theorem.Proof of
Theorem 3. LetFo,
theleading
quadratic
form ofF,
be of rank r.By
Lagrange’s
theorem there existlinearly independent
vectors~a~ 1, ... , apn]
in Kn( 1
pr)
such thatWe set au = 0 for r
ti
~
n and choose n - r vectors[0:1£1’...’
in Kn(1
:s; J..t
r)
such that Then there exist E Kand
(1)
follows with M = n .n
On the other
hand,
thepolynomial
F=
where 0 isclearly
v=1not
representable
in the form( 1 )
with M n.For the
proof
of Theorem 4 we needLemma 3. We have the
identity
where TZ is the i-th
fundamental
symmetric
functiort of
~1, ... , zd, To = 1.Proof .
See[7],
p. 333.Lemma 4. Let ap’
(1 ~ p,
d)
bearbitrary pairwise linearly independent
vectors in
K2.
If
char K = 0 or char K >d,
for
everypolynomial
F EK[XI,X2]
of degree
at most d - 1 there existpolynomials
f~,
EK[z]
such thatProof .
Since ce IA arepairwise
linearly independent
we may assume thateither
(i)
O:¡.¡.l =1,
c~2 are all distinct
(1 ~
d),
or(ii)
all =1,
a,2 are all distinct
(1 G ~c
d),
adl =0,
ad2 = 1.
Let now
(note
that the binomial coefficient isnon-zero).
In the case(i)
for eachsince the rank of the matrix of the coefficients
equals
the number of equa-tions. Then we setn-1
In the case
(ii)
for each I d we can solve for in K thesystem
ofequations
and then we set
Proof of
Theorem4.
In view of Theorem 3 we may assume d > 3. We shall prove first thatM(2, d, K) d.
Let F E
P(2,
d,
K)
and letFo
be thehighest homogeneous
part
of F.Supposing
that we haverepresented
Fo
in the form(1)
with M = d wemay assume that ap
(1 ~ JJ
d)
arepairwise
linearly independent
andthen
apply
Lemma 4 torepresent
F-Fo
in the form(1)
with the same a IA *Therefore,
it isenough
to find arepresentation
(1)
for Fhomogeneous
ofdegree
d.By
Lemma 1 there exist Cll, C21 in K such that 0.Replacing
Fby
F(cllxl
+C22~2)~
where c12, c22 are chosenin K so that CllC22 -
c12c21 ~
0 we may assume that the coefficient ofxd
in
F(xi, X2)
is non-zero. Let thenand let us consider the
polynomial
Since 0 the
polynomial
G is notidentically
0 and we have for each-Since card K > 2d -
2,
by
Lemma 1 there exist elements{31, ... , #d_2
of K such thatWe now
put
which makes sense, since
by
(9)
the denominator is non-zero.Again
by
(9)
we have,Qi ,-~
{3j
for 1 ij
d. HenceHowever,
by
(10)
and Lemma3,
Hence the
system
ofequations
We set
and obtain
(1)
from(8)
and(11).
It remains to show that
M(2, d, K) >
d. Let us consider theequation
In order to prove that it is
impossible
for every a E K it isclearly
sufficient to consider
fp
=bp.zd,
all =
1,
ap2 distinct.
Comparing
the coefficients of on both sides of(12)
we obtainThe determinant of this
system
is(1 ~
d)
andby
(12)
, hence
b~ -
0a contradiction. This
argument
is valid without theassumption
onFor the
proof
of Theorem 5 we needLemma 5. Let al, ... , ak be distinct elements
of
Pp,
k > p - 3. ThenProof .
If k = p - 1 we use theidentity
If k = p - 2 we argue
by
induction.For j
= 0 the statement istrue,
forhence, by
inductionIf k = p - 3 we argue
again by
induction.If j
= 0 the statement is true. Ifk
> j
> 1 we have theidentity
hence,
by
inductionProof of
Theorem 5.By
the last statement in theproof
of Theorem 4 wehave
M(2,
d,
K) >
d,
thus it is remains to prove the reverseinequality.
Let F EP(2,
d,
K).
By
Lemma 4 we may assume that F ishomogeneous.
Letand consider first card K = p = d + 1.
Let us assume first that the
mapping
Pp
-Fp
given by t
Hf (t)
=p-i ,
E
ap-l-it’
is notinjective.
Then there exist r, s such thatr #
s andi=O
f (r)
=f (s),
hence71= I
Setting
a12 =0,
~a22, ... ,
ap_2,2~ _ ~ ~
jr, s}
we haveby
Lemma 5hence,
by
(14),
Since det
(~2)
0,
this suffices forsolvability
overFp
of thesystem
of
equations
-Then we obtain from
(13) - (15)
thatAssume now that the
mapping Pp -+
IFP
given by
t ~f (t)
isinjective.
We shall consider three casesIn the case
(i),
letso that
Setting
a12 =0,
{a22, ... ,
ap_1,2} _ ~
B Irl
we haveby
Lemma 5hence, by
(16),
Since det
(2)
0,
this suffices forsolvability
overIFp
of thesystem
of
equations
Then we obtain from
(13)
and(17)
thatIn the case
(ii),
letso that
Setting
all =0,
{a21, ... ,
B {r}
we haveby
Lemma 5hence, by
(18),
and, by
Lemma3,
Since det
(~1)
0_ jGp-1=1=
0,
this suffices forsolvability
overFp
of thesystem
of
equations
In the case
(iii),
sincewe have ao =
Op-i. Hence the first and the last row of the determinant
are
equal
and the determinant vanishes.Since
0,
this suffices forsolvability
overFp
of thei
system
ofequations
Then we obtain from
(13)
and(20)
Consider now the case, where
Again,
let us assume first that themapping F; --+
IF~
given by
t ~f (t)
=is not
injective.
Then there exist r, sE PP
such thathence
Setting
we haveby
Lemma 5and,
by
Lemma3,
Since det
(~2)
o~jp-3 7~
0,
this suffices forsolvability
overFp
of thesystem
>of
equations
-Then we obtain from
(13)
and(22)
thatAssume now that the
mapping Pp
-IFp
given by
t 1-4f (t)
isinjective.
We shall consider two casesIn the case
(iv)
let 0 =f (r),
r E so thatSetting
we haveby
Lemma 5hence,
by
(23)
Since det
(C~,,2)
Ojp-2 0
0,
this suffces forsolvability
over1Fp
of thesystem
of
equations
Then we obtain from
(13)
and(25)
thatIn the case
(v) t
->f (t)
is abijective mapping
ofF; onto F;.
If themapping
t H
t f (t)
had the sameproperty,
we should obtainwhich is
impossible.
Hence there exist r, sE IF;
such thatr 0 s and
Setting
we have
by
Lemma 5hence, by
(26), (24)
holds and we conclude theargument
as in the case(iv).
Theproof
of Theorem 5 iscomplete.
Proof
of
Theorem 6. We shall prove first thathence on
taking
e = d + 1 - u we obtain from Theorem 2which
gives
(27).
In order to show that
we notice that
Let us consider
independent
variables whereand the matrix B = where
1~~
being
determinedby
theinequality
Let
Bv
be the minor of the matrix B obtainedby omitting
the v-th column and D be the discriminant of thepolynomial
Polynomials
Bo
and D in the variables aid are notidentically
zero.In order to see that 0 let us order all variables aid
linearly assuming
aki if either ~ + l or
2 -+- j
=k + 1 and j
l. Then allproducts
of aid are ordered
lexicographically.
Theproduct
occuring
in theexpansion
ofBo
precedes
in thelexicographic
order any other term in thisexpansion,
hence it does not cancel and 0. On the other hand-
Now,
the discriminant of thepolynomial
is not
identically
0 as a functionof t"
(it
is different from 0 fortv
= 0 forv m - p,
tm-p
=1),
henceDo
= 0implies
analgebraic dependence
overthe
prime
field II of K between(1 v m -
p).
LetWe assert that for
This is
obviously
true for i m - 1. Assume that it is true for all ij,
where
m j m -f- 1 - 1.
Since,
by
the Cramerformulae,
for p =.i_; - _ , ...
and all a’s
occuring
on the left-hand side have the second index at mostit follows that E Q
(28)
iscomplete.
But thenand the inductive
proof
ofimplies
while the number of
independent
variables1, m j m ~- k - 1)
equals
We now assert that if for a
polynomial
we have
BoD =1=
0,
then there existsuch that
Indeed, using
the notation introduced earlier we take forthe m distinct zeros of the
polynomial
Now for each 1 d we solve the
system
ofequations
for in K and assert that the solution satisfies the
larger
system
The
proof
isby
induction onj.
We assume that(32)
is true forall j
i,
i > m and obtain
However the sum on the
right-hand
side of(33)
coincides with the sum onthe left-hand side of
(29)
onputting
there k=1-m+l,
a =i-rrz+~2~+1
(kt1).
Henceby
(29)
which proves
(32).
we obtain
(31)
from(30)
and(32).
Example.
Each three of the vectors~1, 0, 0~,
(0,1, 0~, [0,0,1],
(3,1,1~,
[1, 3, 1], (3, 3, 2~
arelinearly independent
overQ,
nevertheless for allpoly-nomials
fi
EQ[z]
(1
-i6)
we haveIndeed,
it isenough
to consider the casefi
=biz2 (1 i G 6).
Assuming
the
equality
in(33)
we obtaincomparing
the coefficients of XIX2, XlX3 andwhich is
impossible,
sinceI conclude
by expressing
my thanks to U. Zannier for a remarkhelpful
in theproof
of Theorem 5.Note added in
proof.
M. Kula has checked thatM(2,d,K) =
d in thesimplest
cases not coveredby
Theorems 4 and 5: d = 7 or8,
K =References
[1] N. ALON, M. B. NATHANSON, I. RUZSA, The polynomial method and restricted sums of
congruence classes. J. Number Theory 56 (1996), 404-417.
[2] P. DIACONIS, M. SHAHSHAHANI, On nonlinear functions of linear combinations. SIAM J. Sci. Stat. Comput. 5 (1984), 175-191.
[3] W. J. ELLISON, A ’Waring’s problem’ for homogeneous forms. Proc. Cambridge Philos. Soc. 65 (1969), 663-672.
[4] U. HELMKE, Waring’s problem for binary forms. J. Pure Appl. Algebra 80 (1992), 29-45.
[5] A. IARROBINO, Inverse system of a symbolic power II. The Waring problem for forms. J.
Algebra 174 (1995), 1091-1110.
[6] B. F. LOGAN, L. A. SHEPP, Optimal reconstruction of a function from its projections. Duke Math. J. 42 (1975), 645-659.
[7] T. Mum, A treatise on the theory of determinants. Dover, 1960.
[8] B. REZNICK, Sums of powers of complex linear forms, unpublished manuscript of 1992.
Andrzej SCHINZEL
Institute of Mathematics Polish Academy of Sciences 00-950 Warsaw
Poland