Constants of cyclotomic derivations

0.35 MB·English

by  Jean Moulin Ollagnier

Constants of cyclotomic derivations
Jean Moulin Ollagnier1 and Andrzej Nowicki2
1Laboratoire LIX, E´cole Polytechnique, F 91128 Palaiseau Cedex, France,
(e-mail : [email protected]).
2Nicolaus Copernicus University, Faculty of Mathematics and Computer Science,
87-100 Torun´, Poland, (e-mail: [email protected]).
Abstract
3 Let k[X] = k[x0,...,xn−1] and k[Y] = k[y0,...,yn−1] be the polynomial rings
1 in n > 3 variables over a field k of characteristic zero containing the n-th roots
0
of unity. Let d be the cyclotomic derivation of k[X], and let ∆ be the factorisable
2
n derivation of k[Y]associated withd, thatis, d(xj) = xj+1 and∆(yj) = yj(yj+1−yj)
a for all j ∈ Zn. We describe polynomial constants and rational constants of these
J derivations. We prove, among others, that the field of constants of d is a field of

6 rational functions over k in n−ϕ(n) variables, and that the ring of constants of d
2
is a polynomial ring if and only if n is a power of a prime. Moreover, we show that

] the ring of constants of ∆ is always equal to k[v], where v is the producty ···y ,
0 n−1
C
andwe describethe fieldof constants of ∆ in two cases: whenn is power of a prime,
A
and when n = pq.
.
h
t
a Key Words: Derivation; Cyclotomic polynomial; Darboux polynomial; Euler totient
m
function; Euler derivation; Factorisable derivation; Jouanolou derivation; Lotka-Volterra
[
derivation.

1
v 2000 Mathematics Subject Classification: Primary 12H05; Secondary 13N15.
1
5
2
6 Introduction
.
1
0
Throughout this paper n > 3 is an integer, k is a field of characteristic zero con-
3
1 taining the n-th roots of unity, and k[X] = k[x ,...,x ] and k[Y] = k[y ,...,y ]
0 n−1 0 n−1
:
v are polynomial rings over k in n variables. We denote by k(X) = k(x ,...,x ) and
0 n−1
i
X k(Y) = k(y0,...,yn−1) the fields of quotients of k[X] and k[Y], respectively. We fix the
r notations d and ∆ for the following two derivations, which we call cyclotomic derivations.
a
We denote by d the derivation of k[X] defined by
d(x ) = x , for j ∈ Z ,
j j+1 n
and we denote by ∆ the derivation of k[Y] defined by
∆(y ) = y (y −y ), for j ∈ Z .
j j j+1 j n
0 Corresponding author : Andrzej Nowicki, Nicolaus Copernicus University, Faculty of Mathematics
and Computer Science, ul. Chopina 12/18,87–100 Torun´, Poland. E-mail: [email protected].
We denote also by d and ∆ the unique extension of d to k(X) and the unique extension
of ∆ to k(Y), respectively. We will show that there are some important relations between
d and ∆. In this paper we study polynomial and rational constants of these derivations.
In general, if δ is a derivation of a commutative k-algebra A, then we denote by Aδ
the k-algebra of constants of δ, that is, Aδ = {a ∈ A; δ(a) = 0}. For a given derivation
δ of k[X], we are interested in some descriptions of k[X]δ and k(X)δ. However, we know
that such descriptions are usually difficult to obtain. Rings and fields of constants appear
in various classical problems; for details we refer to [5], [6], [27] and [25]. The mentioned
problems are already difficult for factorisable derivations. We say that a derivation δ :
k[X] → k[X] is factorisable if
n−1
δ(x ) = x a x
i i ij j
j=0
X
for all i ∈ Z , where each a belongs to k. Such factorisable derivations and factorisable
n ij
systems of ordinary differential equations were intensively studied from a long time; see
for example [8], [7], [23] and [26]. Our derivation ∆ is factorisable, and the derivation
d is monomial, that is, all the polynomials d(x ),...,d(x ) are monomials. With any
0 n−1
given monomial derivation δ of k[X] we may associate, using a special procedure, the
unique factorisable derivation D of k[Y] (see [16], [28], [22], for details), and then, very
often, the problem of descriptions of k[X]δ or k(X)δ reduces to the same problem for the
factorisable derivation D.
Consider a derivation δ of k[X] given by δ(x ) = xs for j ∈ Z , where s is an integer.
j j+1 n
Such d is called a Jouanolou derivation ([10], [23], [16], [34]). The factorisable derivation
D, associated with this δ, is a derivation of k[Y] defined by D(y ) = y (sy −y ), for
j j j+1 j
j ∈ Z . We proved in [16] that if s > 2 and n > 3 is prime, then the field of constants
n
of δ is trivial, that is, k(X)δ = k. In 2003 H. Z˙ol a¸dek [34] proved the for s > 2, it is also
true for arbitrary n > 3; without the assumption that n is prime. The central role, in his
and our proofs, played some extra properties of the associated derivation D. Indeed, for
s > 2, the differential field (k(X),d) is a finite algebraic extension of (k(Y),δ).
OurcyclotomicderivationdistheJouanolouderivationwiths = 1, andthecyclotomic
derivation ∆ is the factorisable derivation of k[Y] associated with d. In this case s = 1,
the differential field (k(X),d) is no longer a finite algebraic extension of (k(Y),δ); the
relations between d and ∆ are thus more complicated.
We present some algebraic descriptions of the domains k[X]d, k[Y]∆, and the fields
k(X)d, k(Y)∆. Note that these rings are nontrivial. The cyclic determinant
x x ··· x
0 1 n−1
x x ··· x
w = (cid:12)(cid:12) n..−1 ..0 n..−2 (cid:12)(cid:12)
(cid:12) . . . (cid:12)
(cid:12) (cid:12)
(cid:12) x x ··· x (cid:12)
2 3 0
(cid:12) (cid:12)
(cid:12) (cid:12)
is a polynomial belonging to k[X]d(cid:12), and the product y y(cid:12)···y belongs to k[Y]∆. In
(cid:12) 0 1(cid:12) n−1
this paper we prove, among others, that k(X)d is a field of rational functions over k in
n−ϕ(n) variables, where ϕ is the Euler totient function (Theorem 2.9), and that k[X]d
is a polynomial ring over k if and only if n is a power of a prime (Theorem 3.7). The field
2
k(X)d is in fact the field of quotients of k[X]d (Proposition 2.5). We denote by ξ(n) the
sum n, where p runs through all prime divisors of n, and we prove that the number
p|n p
of the minimal set of generators of k[X]d is equal to ξ(n) if and only if n has at most
P
two prime divisors (Corollary 3.13). In particular, if n = piqj, where p 6= q are primes
and i,j are positive integers, then the minimal number of generators of k[X]d is equal to
ξ(n) = pi−1qj−1(p+q) (Corollary 3.11).
The ring of constants k[Y]∆ is always equal to k[v], where v = y y ...,y (Theo-
0 1 n−1
rem4.2)and, ifnisprime, thenk(Y)∆ = k(v)(Theorem5.6). Ifn = ps, wherepisaprime
and s > 2, then k(Y)∆ = k(v,f ,...,f ) with m = ps−1, where f ,...,f ∈ k(Y)
1 m−1 1 m−1
arehomogeneousrationalfunctionssuchthatv,f ,...,f arealgebraicallyindependent
1 m−1
over k (Theorem 7.1). A similar theorem we prove for n = pq (Theorem 7.5).
In our proofs we use classical properties of cyclotomic polynomials, and an important
role play some results ([11], [12], [32], [33] and others) on vanishing sums of roots of unity.
1 Notations and preparatory facts
We denote by Z the ring Z/nZ, and by Z∗ the multiplicative group of Z . The
n n n
indexes of the variables x ,...,x and y ,...,y are elements of Z . This means, in
0 n−1 0 n−1 n
particular, that if i,j are integers, then x = x ⇐⇒ i ≡ j (mod n). Throughout this
i j
paper ε is a primitive n-th root of unity, and we assume that ε ∈ k. The letters ̺ and τ
we book for two k-automorphisms of the field k(X) = k(x ,...,x ), defined by
0 n−1
̺(x ) = x , τ(x ) = εjx for all j ∈ Z .
j j+1 j j n
We denote by u ,u ,...,u the linear forms in k[X] = k[x ,...,x ], defined by
0 1 n−1 0 n−1
n−1
u = εj ix , for j ∈ Z .
j i n
i=0
X(cid:0) (cid:1)
If r is an integer and n ∤ r, then the sum n−1(εr)j is equal to 0, and in the other case,
j=0
when n | r, this sum is equal to n. As a consequence of this fact we obtain, that
P
n−1
1
x = ε−i ju for all i ∈ Z .
i j n
n
j=0
X(cid:0) (cid:1)
Thus, k[X] = k[u ,...,u ], k(X) = k(u ,...,u ), and the forms u ,...,u are
0 n−1 0 n−1 0 n−1
algebraically independent over k. Moreover, it is easy to check the following equalities.
Lemma 1.1. τ(u ) = u , ̺(u ) = ε−ju for all j ∈ Z .
j j+1 j j n
For every sequence α = (α ,α ,...,α ), of integers, we denote by H (t) the poly-
0 1 n−1 α
nomial in Z[t] defined by
H (t) = α +α t1 +α t2 +···+α tn−1.
α 0 1 2 n−1
3
An important role in our paper play two subsets of Zn which we denote by G and
n
M . The first subset G is the set of all sequences α = (α ,...,α ) ∈ Zn such that
n n 0 n−1
α +α ε1+α ε2+···+α εn−1 = 0.Thesecond subset M istheset ofallsuch sequences
0 1 2 n−1 n
α = (α ,...,α ) which belong to G and the integers α ,...,α are nonnegative, that
0 n−1 n 0 n−1
is, they belong to the set of natural numbers N = {0,1,2,...}. Let us remember:
G = {α ∈ Zn; H (ε) = 0}, M = {α ∈ Nn; H (ε) = 0} = G ∩Nn.
n α n α n
If α,β ∈ G , then of course α±β ∈ G , and if α,β ∈ M , then α+β ∈ M . Thus G is
n n n n n
an abelian group, and M is an abelian monoid with zero 0 = (0,...,0).
n
The primitive n-th root ε is an algebraic element over Q, and its monic minimal
polynomial is equal to the n-th cyclotomic polynomial Φ (t). Recall (see for example:
n
[24], [13]) that Φ (t) is a monic irreducible polynomial with integer coefficients of degree
n
ϕ(n), where ϕ is the Euler totient function. This implies that we have the following
proposition.
Proposition 1.2. Let α ∈ Zn. Then α ∈ G if and only if there exists a polynomial
n
F(t) ∈ Z[t] such that H (t) = F(t)Φ (t).
α n
Put e = (1,0,0,...,0), e = (0,1,0,...,0), ..., e = (0,0,...,0,1), and let
0 1 n−1
e = n−1e = (1,1,...,1). Since n−1εi = 0, the element e belongs to M .
i=0 i i=0 n
ProPposition 1.3. If α ∈ G , thenPthere exist β,γ ∈ M such that α = β −γ.
n n
Proof. Let α = (α ,...,α ) ∈ G , and let r = min{α ,...,α }. If r > 0, then
0 n−1 n 0 n−1
α ∈ M and then α = β−γ, where β = α, γ = 0. Assume that r = −s, where 1 6 s ∈ N.
n
Put β = α+se and γ = se. Then β,γ ∈ M , and α = β −γ. (cid:3)
n
The monoid M has an order >. If α,β ∈ G , the we write α > β, if α−β ∈ Nn, that
n n
is, α > β ⇐⇒ there exists γ ∈ M such that α = β +γ. In particular, α > 0 for any
n
α ∈ M . It is clear that the relation > is reflexive, transitive and antisymmetric. Thus
n
M is a poset with respect to >.
n
Proposition 1.4. The poset M is artinian, that is, if α(1) > α(2) > α(3) > ... is a
n
sequence of elements from M , then there exists an integer s such that α(j) = α(j+1) for
n
all j > s.
Proof. Given an element α = (α ,...,α ) ∈ M , we put |α| = α +···+α .
0 n−1 n 0 n−1
Observe that if α,β ∈ M andα > β, then |α| > |β|. Suppose that there exists aninfinite
n
sequence α(1) > α(2) > α(3) > ... of elements from M , and let s = α(1) . Then we have
n
an infinite sequence s > |α(2)| > |α(2)| > ··· > 0, of natural numbers; a contradiction. (cid:3)
(cid:12) (cid:12)
(cid:12) (cid:12)
Let α ∈ M . We say that α is a minimal element of M , if α 6= 0 and there is no
n n
β ∈ M such that β 6= 0 and β < α. Equivalently, α is a minimal element of M , if
n n
α 6= 0 and α is not a sum of two nonzero elements of M . It follows from Proposition 1.4
n
that for any 0 6= α ∈ M there exists a minimal element β such that β 6 α. Moreover,
n
every nonzero element of M is a finite sum of minimal elements.
n
4
Proposition 1.5. The set of all minimal elements of M is finite.
n
Proof. To deduce this result from Proposition 1.4, Dikson’s Lemma could be used :
in any subset N of Nn there exists a finite number of elements {e(1),··· ,e(s)} such that
N ⊆ e(j) +Nn .
It is simpler to use classical noetherian arguments. Consider the polynomial ring
S(cid:0) (cid:1)
R = Z[z ,...,z ]. If α = (α ,...,α ) is an element from M , then we denote
0 n−1 0 n−1 n
by zα the monomial zα0zα1 ···zαn−1. Let S be the set of all minimal elements of M ,
0 1 n1 n
and consider the ideal A of R generated by all elements of the form zα with α ∈ S.
Since R is noetherian, A is finitely generated; there exist α(1),...,α(r) ∈ S such that
A = zα(1), ..., zα(r) . Let α be an arbitrary element from S. Then zα ∈ A, and then
there(cid:16)exist j ∈ {1,...(cid:17),r} and γ ∈ Nn such that zα = zγ · zα(j) = zγ+α(j). This implies
that α = γ+α(j). Observe that γ = α−α(j) ∈ G ∩Nn, and G ∩Nn = M , so γ belongs
n n n
to M . But α is minimal, so γ = 0, and consequently α = α(j). This means that S is a
n
finite set equal to α(1),...,α(r) . (cid:3)
We denote by ζ(cid:8), the rotation(cid:9)of Zn given by
ζ(α) = (α ,α ,α ,...,α ),
n−1 0 1 n−2
for α = (α ,α ,...,α ) ∈ Zn. We have for example: ζ(e ) = e for all j ∈ Z , and
0 1 n−1 j j+1 n
ζ(e) = e. The mapping ζ : Zn → Zn is obviously an endomorphism of the Z-module Zn,
and is one-to-one and onto.
Lemma 1.6. Let α ∈ Zn. If α ∈ G , then ζ(α) ∈ G . If α ∈ M , then ζ(α) ∈ M .
n n n n
Moreover, α is a minimal element of M if and only if ζ(α) is a minimal element of M .
n n
Proof. Assume that α = (α ,...,α ) ∈ G . Then α +α ε+···+α εn−1 = 0.
0 n−1 n 0 1 n−1
Multiplying it by ε, we have 0 = α ε+α ε2+···+α εn. But εn = 1, so α +α ε+
0 1 n−1 n−1 0
α ε2 + ···+ α εn−2 = 0, and so ζ(α) ∈ G . This implies also, that if α ∈ M , then
1 n−2 n n
ζ(α) ∈ M .
n
Assume now that α is a minimal element of M and suppose that ζ(α) = β +γ, for
n
some β,γ ∈ M . Then we have α = ζn(α) = ζn−1(ζ(α)) = ζn−1(β)+ζn−1(γ) = β′ +γ′,
n
where β′ = ζn−1(β) and γ′ = ζn−1(γ) belong to M . Since α is minimal, β′ = 0 or
n
γ′ = 0, and then β = 0 or γ = 0. Thus if α is a minimal element of M , then ζ(α) is
n
also a minimal element of M . Moreover, if ζ(α) is minimal, then α is minimal, because
n
α = ζn−1(ζ(α)). (cid:3)
2 The derivation d and its constants
Let us recall that d : k[X] → k[X] is a derivation such that d(x ) = x , for j ∈ Z .
j j+1 n
Proposition 2.1. For each j ∈ Z , the equality d(u ) = ε−ju holds.
n j j
5
n−1 n−1 n
Proof. d(u ) = d (εj)ix = (εj)ix = (εj)i−1x
j i i+1 i
(cid:18)i=0 (cid:19) i=0 i=1
P P P
n n−1
= ε−j (εj)ix = ε−j (εj)ix = ε−ju . (cid:3)
i i j
i=1 i=0
P P
This means that d is a diagonal derivation of the polynomial ring k[U] = k[u ,...,u ]
0 n−1
which is equal to the ring k[X]. It is known (see for example [25]) that the algebra
of constants of every diagonal derivation of k[U] = k[X] is finitely generated over k.
Therefore, k[X]d is finitely generated over k. We would like to describe a minimal set of
generators of the ring k[X]d, and a minimal set of generators of the field k(X)d.
If α = (α ,...,α ) ∈ Zn, then we denote by uα the rational monomial uα0···uαn−1.
0 n−1 0 n−1
Recall (see the previous section) that H (t) is the polynomial a +a t1 +···+a tn−1
α 0 1 n−1
belonging to Z[t]. As a consequence of Proposition 2.1 we obtain
Proposition 2.2. d(uα) = H (ε−1)uα for all α ∈ Zn.
α
Note that ε−1 is also a primitive n-th root of unity. Hence, by Proposition 1.2, we
have the equivalence H (ε−1) = 0 ⇐⇒ H (ε) = 0, and so, by the previous proposition,
α α
we see that if α ∈ Zn, then d(uα) = 0 ⇐⇒ α ∈ G , and if α ∈ Nn, then d(uα) = 0 ⇐⇒
n
α ∈ M . Moreover, if F = b uα(1) +···+b uα(r), where b ,...,b ∈ k and α(1),...,α(r)
n 1 r 1 r
are pairwise distinct elements of Nn, then d(F) = 0 if and only if d b uα(i) = 0 for
i
every i = 1,...,r. Hence, k[X]d is generated over k by all elements of(cid:16)the form(cid:17) uα with
α ∈ M . We know (see the previous section), that every nonzero element of M is a
n n
finite sum of minimal elements of M . Thus we have the following next proposition.
n
Proposition 2.3. The ring of constants k[X]d is generated over k by all the elements of
the form uβ, where β is a minimal element of the monoid M .
n
In the next section we will prove some additional facts on the minimal number of
generators of the ring k[X]d. Now, let us look at the field k(X)d.
Proposition 2.4. The field of constants k(X)d is generated over k by all elements of the
form uγ with γ ∈ G .
n
Proof. Let L be the subfield of k(X) generated over k by all elements of the form
uγ with γ ∈ G . It is clear that L ⊆ k(X)d. We will prove the reverse inclusion. Assume
n
that 0 6= f ∈ k(X)d. Since k(X) = k(U), we have f = A/B, where A,B are coprime
polynomials in k[U]. Put
A = a uα, B = b uβ,
α β
αX∈S1 βX∈S2
where all a , b are nonzero elements of k, and S , S are some subsets of Nn. Since
α β 1 2
d(f) = 0, we have the equality Ad(B) = d(A)B. But A,B are relatively prime, so
d(A) = λA, d(B) = ΛB for some λ ∈ k[U]. Comparing degrees, we see that λ ∈ k.
Moreover, by Proposition 2.2, we deduce that d(uα) = λuα for all α ∈ S , and also
1
6
d(uβ) = λuβ for all β ∈ S . This implies that if δ ,δ ∈ S ∪ S , then d uδ1−δ2 = 0.
2 1 2 1 2
In fact, d uδ1−δ2 = d uuδδ12 = u21δ2 d(uδ1)uδ2 −uδ1d(uδ2 = u21δ2 λuδ1uδ2 −(cid:0)λuδ1uδ2(cid:1) = 0.
This mean(cid:0)s, tha(cid:1)t if δ1(cid:16),δ2 ∈(cid:17) S1 ∪ S(cid:0)2, then δ1 − δ2 ∈ Gn(cid:1)Fix an (cid:0)element δ from S1(cid:1)∪ S2.
Then all α−δ, β −δ belong to G , and we have
n
A a uα u−δ a uα a uα−δ
α α α
f = = = = ,
B b uβ u−δ b uβ b uβ−δ
P β P β P β
and hence, f ∈ L. (cid:3) P P P
Let us recall (see Proposition 1.3) that every element of the group G is a difference
n
of two elements from the monoid M . Using this fact and the previous propositions we
n
obtain
Proposition 2.5. The field k(X)d is the field of quotients of the ring k[X]d.
Now we will prove that k(X)d is a field of rational functions over k, and its transcen-
dental degree over k is equal to n−ϕ(n), where ϕ is the Euler totient function. For this
aim look at the cyclotomic polynomial Φ (t). Assume that
n
Φ (t) = c +c t+···+c tϕ(n).
n 0 1 ϕ(n)
All the coefficients c ,...,c are integers, and a = a = 1. Put m = n−ϕ(n) and
0 ϕ(n) 0 ϕ(n)
γ = c ,c ,...,c , 0,...,0 .
0 0 1 ϕ(n)
(cid:16) m−1 (cid:17)
Note that γ0 ∈ Zn, and Hγ0(t) = Φn(t). Consider th|e e{lzem}ents γ0,γ1,...,γm−1 defined by
γ = ζj(γ ), for j = 0,1,...,m−1.
j 0
Observe that H (t) = Φ (t) · tj for all j ∈ {0,...,m − 1}. Since Φ (ε) = 0, we have
γj n n
H (ε) = 0, and so, the elements γ ,...,γ belong to G .
γj 0 m−1 n
Lemma 2.6. The elements γ ,...,γ generate the group G .
0 m−1 n
Proof. Let α ∈ G . It follows from Proposition 1.2, that H (t) = F(t)Φ (t), for
n α n
some F(t) ∈ Z[t]. Then obviously degF(t) < m. Put F(t) = b + b1 + ···+ b tm−1,
0 t m−1
with b ,...,b ∈ Z. Then we have
0 m−1
H (t) = b (Φ (t)t0)+b (Φ (t)t1)+···+b (Φ (t)tm−1)
α 0 n 1 n m−1 n
= b H (t)+···+b H (t),
0 γ0 m−1 γm−1
and this implies that α = b γ +b γ +···+b γ . (cid:3)
0 0 1 1 m−1 m−1
Consider now the rational monomials w ,...,w defined by
0 m−1
w = uγj = uc0 uc1 uc2 ···ucϕ(n)
j 0+j 1+j 2+j ϕ(n)+j
for j = 0,1,...,m−1, where m = n−ϕ(n). Each w is a rational monomial with respect
j
to u ,...,u of the same degree equals to Φ (1) = c +c +···+c . It is known (see
0 n−1 n 0 1 ϕ(n)
for example [13]) that Φ (1) = p if n is power of a prime number p, and Φ (1) = 1 in all
n n
other cases. As each u is a homogeneous polynomial in k[X] of degree 1, we have:
j
7
Proposition 2.7. The elements w ,...,w are homogeneous rational functions with
0 m−1
respect to variables x ,...,x , of the same degree r. If n is a power of a prime number p,
0 n−1
then r = p, and r = 1 in all other cases.
As an immediate consequence of Lemma 2.6 and Proposition 2.4, we obtain the equal-
ity k(X)d = k(w ,...,w ).
0 n−1
Lemma 2.8. The elements w ,...,w are algebraically independent over k.
0 m−1
Proof. Let A be the n×m Jacobi matrix [a ], where a = ∂wj for i = 0,1,...,n−1,
ij ij ∂ui
j = 0,1,...,m−1. It is enough to show that rank(A) = m (see for example [9]). Observe
that ∂w0 = c uc0−1uc1···ucϕ(n) 6= 0 (because c = 1), and ∂wj = 0 for j > 1. Moreover,
∂u0 0 0 1 ϕ(n) 0 ∂u0
∂w1 6= 0and ∂wj = 0forj > 2,andingeneral, ∂wi 6= 0and ∂uj = 0foralli,j = 0,...,m−1
∂u1 ∂u1 ∂ui ∂ui
with j > i. This means, that the upper m×m matrix of A is a triangular matrix with a
nonzero determinant. Therefore, rank(A) = m. (cid:3)
Thus, we proved the following theorem.
Theorem 2.9. The field of constants k(X)d is a field of rational functions over k and
its transcendental degree over k is equal to m = n − ϕ(n), where ϕ is the Euler totient
function. More precisely,
k(X)d = k w ,...,w ,
0 m−1
(cid:16) (cid:17)
where the elements w ,...,w are as above.
0 m−1
Now we will describe all constants of d which are homogeneous rational functions
of degree zero. Let us recall that a nonzero polynomial F is homogeneous of degree
r, if all its monomials are of the same degree r. We assume that the zero polynomial
is homogeneous of arbitrary degree. Homogeneous polynomials are also homogeneous
rational functions, which (in characteristic zero) are defined in the following way. Let
f = f(x ,...,x ) ∈ k(X) We say that f is homogeneous of degree s ∈ Z, if in the field
0 n−1
k(t,x ,...,x ) the equality f(tx ,tx ,...,tx ) = ts ·f(x ,...,x ) holds It is easy
0 n−1 0 1 n−1 0 n−1
to prove (see for example [25] Proposition 2.1.3) the following equivalent formulations of
homogeneous rational functions.
Proposition 2.10. Let F,G be nonzero coprime polynomials in k[X] and let f = F/G.
Let s ∈ Z. The following conditions are equivalent.
(1) The rational function f is homogeneous of degree s.
(2) The polynomials F, G are homogeneous of degrees p and q, respectively, where
s = p−q.
(3) x ∂f +···+x ∂f = sf.
0∂x0 n−1∂xn−1
Equality (3)iscalledtheEuler formula. Inthispaperwe denotebyE theEuler derivation
ofk(X), thatis, E isaderivationofk(X)defined byE(x ) = x forallj ∈ Z . Asusually,
j j n
we denote by k(X)E the field of constants of E. Observe that, by Proposition 2.10, a
rational function f ∈ k(X) belongs to k(X)E if and only if f is homogeneous of degree
zero. In particular, the set of all homogeneous rational functions of degree zero is a
8
subfield of k(X). It is obvious that the quotients x1,..., xn−1 belong to k(X)E, and they
x0 x0
are algebraically independent over k. Moreover, k(X)E = k(x1,..., xn−1). Therefore,
x0 x0
k(X)E is a field of rational functions over k, and its transcendence degree over k is equal
ton−1. Put q = xj+1 forallj ∈ Z . Inparticular, q = x0 . Theelements q ,...,q
j xj n n−1 xn−1 0 n−1
belong to k(X)E and moreover, xj = q q ···q for j = 1,...,n−1. Thus we have the
x0 0 1 j−1
following equality.
Proposition 2.11. k(X)E = k x1, x2,..., xn−1, x0 .
x0 x1 xn−2 xn−1
(cid:16) (cid:17)
Now consider the field k(X)d,E = k(X)d ∩k(X)E.
Lemma 2.12. Let d ,d : k(X) → k(X) be two derivations. Assume that K(X)d1 =
1 2
k(c,b ,...,b ), where c,b ,...,b are algebraically independent over k elements from k(X)
1 s 1 s
such that d (b ) = ··· = d (b ) = 0 and d (c) 6= 0. Then k(X)d1 ∩k(X)d2 = k(b ,...,b ).
2 1 2 s 2 1 s
Proof. Put L = k(b ,...,b ). Observe that k(X)d1 = L(c), and c is transcendental
1 s
over L. Let 0 6= f ∈ k(X)d1 ∩ k(X)d2. Then f = F(c), where F(t),G(t) are coprime
G(c)
polynomials in L[t]. We have: d (F(c)) = F′(c)d (c), d (G(c)) = G′(c)d (c), where
2 2 2 2
F′(t),G′(t) are derivatives of F(t),G(t), respectively. Since d (f) = 0, we have
2
0 = d (F(c))G(c)−d (G(c))F(c) = F′(c)G(c)−G′(c)F(c) d (c),
2 2 2
(cid:16) (cid:17)
and so, (F′G − G′F)(c) = 0, because d (c) 6= 0. Since c is transcendental over L, we
2
obtain the equality F′(t)G(t) = G′(t)F(t) in L[t], which implies that F(t) divides F′(t)
and G(t) divides G′(t) (because F(t),G(t) are relatively prime), and comparing degrees
we deduce that F′(t) = G′(t) = 0, that is, F(t) ∈ L and G(t) ∈ L. Thus the elements
F(c), G(c) belong to L and so, f = F(c) belongs to L. Therefore, k(X)d1 ∩k(X)d2 ⊆ L.
G(c)
The reverse inclusion is obvious. (cid:3)
Let us return to the rational functions w ,...,w . We know (see Proposition 2.7)
0 m−1
that they are homogeneous of the same degree. Put: d = d, d = E, c = w and b = wj
1 2 0 j w0
for j = 1,...,m − 1, Then, as a consequence of Lemma 2.12. we obtain the following
proposition.
Proposition 2.13. k(X)d,E = k w1,..., wm−1 .
w0 w0
(cid:16) (cid:17)
Since w ,...,w are algebraically independent over k (see Lemma 2.8), the quotients
0 m−1
w1,..., wm−1 are also algebraically independent over k. Thus, k(X)d,E is a field of rational
w0 w0
functions and its transcendental degree over k is equal to n − ϕ(n) − 1, where ϕ is the
Euler totient function. In particular, if n is prime, then n−ϕ(n)−1 = 0 and we obtain:
Corollary 2.14. k(X)d,E = k ⇐⇒ n is a prime number.
9
3 Numbers of minimal elements
Let F be the set of all the minimal elements of the monoid M , and denote by ν(n)
n
the cardinality of F. We know, by Proposition 1.5, that ν(n) < ∞. We also know (see
Proposition 2.3) that the ring k[X]d is generated over k by all the elements of the form uβ,
where β ∈ F. But k[X] is equal to the polynomial ring k[U] = k[u ,...,u ], so k[X]d
0 n−1
is generated over k by a finite set of monomials with respect to the variables u ,...,u .
0 n−1
It is clear that if β,γ are distinct elements from F, then uβ ∤ uγ and uγ ∤ uβ. This
implies that no monomial uβ,β ∈ F belongs to the algebra generated by other uγ,γ ∈
F,uγ ∤ uβ. Thus, uβ; β ∈ F is a minimal set of generators of k[X]d.
Moreover, uβ; β ∈ F is a set on generators of k[X]d with the minimal number of
(cid:8) (cid:9)
elements according to the following proposition.
(cid:8) (cid:9)
Proposition 3.1. Let f ,...,f be polynomials in k[X]. If k[X]d = k[f ,...,f ], then
1 s 1 s
s > ν(n).
Proof. As the uβ are monomials in the u’s, they constitute a Gro¨bner base for the
ideal I generated in k[X] by k[X]d. This basis is minimal for any admissible order, for
example the lexicographical one.
Making a head reduction of the f , a new head-reduced system of generators appears,
i
maybe with less than s elements. Thus, without loss of generality, we can suppose that
the system (f ,...,f ) is head-reduced, which means that the leading monomial of one f
1 s i
does not belong to the multiplicative monoid generated by the other leading monomials.
The leading monomials of the various f are uα for some α ∈ M .
i n
The exponents α are minimal in the sub-monoid they generate, but this sub-monoid
has to be M itself. (cid:3)
n
In this section we prove, among others, that k[X]d is a polynomial ring over k if and
only if n is a power of a prime number. Moreover, we present some additional properties
of thenumber ν(n), which areconsequences ofknown results onvanishing sums ofrootsof
unity; see for example [12], [30], [32] and [33], where many interesting facts and references
on this subject can be found.
We denote by ξ(n) the sum n, where p runs through all prime divisors of n. Note
p|n p
that if a,b are positive coprime integers, then ξ(ab) = aξ(b)+ξ(a)b.
P
First we show that the computation of ν(n) can be reduced to the case when n is
square-free. For this aim let us denote by n the largest square-free factor of n, and by n′
0
the integer n/n . Then ϕ(n) = n′ϕ(n ), Φ (t) = Φ tn′ (see for example [24]), and
0 0 n n0
ξ(n) = n′ξ(n ).
0
(cid:0) (cid:1)
Assume now that n = mc, where m > 2, c > 2 are integers. For a given sequence
γ = (γ ,...,γ ) ∈ Zm, consider the sequence
0 m−1
γ = γ ,0,...,0,γ ,0,...,0,...,γ ,0,...,0 .
0 1 m−1
(cid:16) c−1 c−1 c−1 (cid:17)
This sequence is an element o|f Z{zn, }and it|is{zeas}y to prove t|he{fzoll}owing lemma.
10