Another Proof of Ostrowski-Kolchin-Hardouin Theorem in Difference Algebra
◆自由論題＊研究ノート◆
Another Proof of OstrowskiKolchin-Hardouin Theorem in
Difference Algebra
差分代数における
Ostrowski-Kolchin-Hardouin 定理の別証明
Hiroshi Ogawara
Master's Program, Graduate School of Media and Governance, Keio University
小川原 弘士
慶應義塾大学大学院政策・メディア研究科修士課程
This paper gives another proof of an analog of Ostrowski-Kolchin theorem in difference algebra,
which was proved by Hardouin. Let K be a field of characteristic 0 and (L,τ) a difference extension
of a difference field (K,τ). Denote the invariant field of (L,τ) and that of (K,τ) by CL, C respectively.
Suppose C is an algebraically closed field. Suppose x1, . . . , xm, y1, . . . , yn are nonzero elements of L
satisfying τ(xi) = ui xi ,τ(yj) = yj + vj , where u1, . . . , um, v1, . . . , vn ∈ K. Then the analog states that
if x1, . . . , xm, y1, . . . , yn are algebraically dependent over KCL, there exists a nonzero element a ∈
K satisfying τ(a) = ( mi=1 uki i ) a for a nonzero element (ki) ∈ m or τ(a) = a + nj=1 ajvj for a nonzero
element (aj) ∈ C n.
本論文では Hardouin による Ostrowski-Kolchin の定理の差分化の別証明を与えた。K を標数 0
の体、(L,τ) を (K,τ) の差分拡大とする。(L,τ) と (K,τ) の不変体をそれぞれ CL, C と表記し、C は
代数閉体とする。L の 0 でない元 x1, . . . , xm, y1, . . . , yn が u1, . . . , um, v1, . . . , vn ∈ K に対して τ(xi)
= ui xi ,τ(yj) = yj + vj を満たすと仮定する。このとき差分化された定理は次のように記述される : x1,
. . . , xm, y1, . . . , yn が KCL 上代数的従属ならば、ある 0 でない元 a ∈ K が存在してτ (a)=( mi=1 uki i ) a
m
n
となる 0 でない (ki) ∈
が存在する、または τ(a) = a + nj=1 ajvj となる 0 でない (aj) ∈ C が存在
する。
Mathematics Subject Classification (2010). Primary 12H10; Secondary 65Q10
Keywords: Difference algebra, Linear difference equations, Module of differentials
KEIO SFC JOURNAL Vol.13 No.2 2013
99
自由論題
1　Introduction
2　Preliminaries
In [1], Hardouin has proved an analog of
Let A be an algebra over a ring R. There exist
Ostrowski-Kolchin theorem with one derivation
an A-module Ω called the module of differentials of A
operator [2] using difference Galois theory. The
over R and an R-linear derivation d : A → Ω called the
purpose of this paper is to give another proof in
universal R-linear derivation if for any A-module M and
difference algebra, in which we use module of
any R-linear derivation D : A → M there is a unique
differentials and its fundamental propositions instead.
To state our theorem we prepare some notions
A-module homomorphism f : Ω→ M such that D =
f 。d (cf. [4, pp.91-92]). The following propositions are
in difference algebra (cf. [4, pp.103 -115]). We always
well-known:
regard any ring (field) as a commutative ring (field)
with characteristic 0. Let K be a field and τK an
Proposition 1 (Rosenlicht [6]). Let L/K be a field
isomorphism from K to itself. We call the pair (K,τK )
extension and Ω its module of differentials with the
a difference field andτK the transforming operator of K.
universal K-linear derivation d. Then η1, . . . , ηr ∈
Let L be an extension of K which is also a difference
L are algebraically independent over K if and only
field with a transforming operator τL . We call (L,τL )
if dη1, . . . , dηr ∈Ω are linearly independent over L.
a difference extension of (K,τK) if τL｜K=τK. By CK, we
denote the invariant field of (K,τK), that is, the field of
Proposition 2 (Rosenlicht [6]). Let L/K be a field
r
extension and Ω its module of differentials with the
αi . Then we
universal K-linear derivation d. Suppose a1, . . . , ar ∈
invariant elements of τK . Forα=(α1 ,. . . , αr ) ∈ K
and k =(k1, . . . , kr) ∈
r
k
, we put α =
r
i=1
ki
shall show the following theorem:
K are linearly independent over
. If η,ζ1, . . . ,ζr ∈
L satisfy
Theorem　Let (L,τ) be a difference extension of a
r
difference field (K,τ) and the invariant field C = CK an
dη+
∑a
i
i=1
algebraically closed field. Suppose x1, . . . , xm, y1, . . . ,
dζi
= 0,
ζi
yn are nonzero elements of L satisfying
then dη = dζ1= . . . =dζr = 0.
τ(xi) = ui xi ,
Proposition 3 (Kubota [3]). Suppose (L, τ) is a
τ(yj) = yj + vj ,
difference extension of a difference field (K, τ). Let
Ω be the module of differentials of L/K with the
where u1, . . . , um, v1, . . . , vn ∈ K. If x1, . . . , xm, y1, . . . ,
universal K-linear derivation d. Then there exists an
yn are algebraically dependent over KCL, then there
additive mapping τ* : Ω→Ω such that
exists a nonzero element a ∈ K satisfying
τ(a) =
(
m
i=1
τ*(ηdζ) = τ(η)d(τ(ζ)) (η,ζ∈ L).
)
uiki a (1)
for a nonzero element (ki) ∈
3　Proof of Theorem
m
or
From the assumption, we may suppose that L is
finitely generated over K. In fact, there is a nonzero
n
∑a v (2)
τ(a) = a +
polynomial F∈ KCL [X1, . . . , Xm, Y1, . . . , Yn] satisfying
j j
j=1
n
for a nonzero element (aj) ∈ C .
F(x1, . . . , xm, y1, . . . , yn) = 0.
100
Another Proof of Ostrowski-Kolchin-Hardouin Theorem in Difference Algebra
Let L' be an extension over K generated by x1, . .
r
∑
. , xm, y1, . . . , yn and the elements of CL being in the
coefficients of F. Then we haveτ(L' ) ⊂ L' , so that
cj
j=1
dzj
=
zj
r
m'
∑∑
cj
j=1
nij
i=1
dxi
=
xi
m'
∑a
i
i=1
dxi
= 0.
xi
(L',τ)/(K,τ) is a difference extension. Furthermore,
we only have to prove our theorem in case v1, . . . , vn
From Proposition 2, each zj is algebraic over KCL.
are linearly independent over C.
Take some zj of them such that not all n1j , . . . , nm'j
First, suppose that x1, . . . , xm are algebraically
are zero. Considering its minimal polynomial, we can
dependent over KCL and take the minimal number
take a nonzero element z ∈ KCL satisfying
m' such that x1, . . . , xm' are algebraically dependent
over KC L. Let Ω be the module of differentials of
τ(z) = u(r n ) z,
z
(3)
ij
L/KCL with the universal KCL-linear derivation d :
L →Ω . Then there is a nontrivial equation of linear
where rz is a positive integer and (rznij) = (rz n1j , . . . ,
dependence over L,
rz nm'j).
Next, suppose that x1, . . . , xm are algebraically
m'
∑a
i
i=1
independent over KC L. Take the minimal number
dxi
= 0,
xi
n' such that x1, . . . , xm, y1, . . . , yn' are algebraically
dependent over KCL. There is a nontrivial equation of
where a i ∈ L and a m' = 1. Applying the additive
linear dependence over L,
mappingτ* of Proposition 3 to this equation, we have
m
∑
m'
∑
dxi
τ(ai)
= 0.
xi
i=1
ai
i=1
n'
dxi
+
xi
∑b dy = 0,
h
h
h=1
where ai, bh ∈ L and bn' =1. Applyingτ* to this equaHence we get byτ(am′) = am′= 1,
tion, we have
m
m'-1
∑
dxi
(τ(ai)− ai)
= 0.
xi
∑
i=1
τ(ai)
i=1
n'
∑τ(b )dy = 0.
dxi
+
xi
h
h
h=1
Hence ai, bh are included in CL. Take c1, . . . , cr ∈ CL,
Since dx1, . . . , dxm' − 1 are linearly independent over
nij ∈
L from Proposition 1, it follows thatτ(a i) = a i for
above. Then we get
each i. Hence every ai is a member of CL. There are
r
∑
elements c1, . . . , cr ∈ CL such that they are linearly
independent over
and zj ∈ L following the same procedure as
cj
j=1
and satisfy
(∑ )
n'
dzj
+d
zj
bh yh = 0.
h=1
n'
bhyh is algebraic over KCL. There is also an
Hence ∑h=1
element w ∈ KCL satisfying
r
ai =
∑n c (n ∈
ij j
ij
).
j=1
n'
∑r b v
τ(w) = w −
Then not all nij are zero. Putting zj =
m'
i=1
w h h
(4)
h=1
xinij , we
have
for some positive integer rw.
We can embed CL into the field of formal power
KEIO SFC JOURNAL Vol.13 No.2 2013
101
自由論題
series C((t)) over C as a field, since C L is finitely
∞
generated over C. We see C L and K are linearly
w =
μ=q
disjoint over C, and so are K and C((t)). In fact,
suppose a1, . . . , ar ∈ CL are linearly dependent over
∞
∑β
tμ ,
hμ
μ=q
whereγμ ∈ K,βhμ∈ C and βn'0 = 1. From (4), we see
K. If r = 1, clearly a1 is linearly dependent over C.
∞
∑
Assume that a1, . . . , ar-1 are linearly independent over
∞
γμtμ−
μ=q
∑( ∑r β
∞
r
μ=q
∑
ki ai = 0.
i=1
∞
n'
∑ ∑ ∑β t
τ(γμ)tμ =
μ=q
K. There are k1, . . . , kr ∈ K with kr = 1 such that
∑
γμtμ , bh =
w
h=1
h=1
hμ
μ
=
μ=q
)
n'
γμ−
rwvh
v t μ.
hμ h
Since v1, . . . , vn' are linearly independent over C from
Applying τ to this, we have
the assumption, we see γ0 ≠ 0. Therefore γ0 is a
n'
rwβh0 vh, the
nonzero element satisfying τ
（γ0）= γ0−∑
h=1
r
∑τ(k )a = 0.
i
form of (2).
i
i=1
Hence we obtainτ(ki) = ki , so that ki ∈ C. This means
References
a 1, . . . , a r are linearly dependent over C. Next,
[1]
suppose k1, . . . , kr ∈ K are linearly dependent over
∞
C((t)). Then there are formal power series ∑
c tv, .
v=p 1v
crv t ∈ C((t)) which make a nontrivial equation
. . ,∑
v=p
[2]
of linear dependence,
[3]
∞
v
∞
r
∑(∑k c
∑∑
ki
i=1
civ tv =
v=p
∞
r
v=p
i=1
i iv
)
[4]
tv = 0.
[5]
r
k i c iv = 0 for all v. Since some c iv is a
So we get ∑
i=1
[6]
nonzero element, k1, . . . , kr are linearly dependent
over C.
〔受付日　2013. 7. 19〕
〔採録日　2013. 10. 30〕
Hence there is an embedding from KC L into
the field of formal power series K((t)) over K as a
difference field defining τ
= t in K((t)). The above z
can be described in K((t)) as
z =
∞
∑α t
μ
μ
(αμ∈ K, αp ≠ 0).
μ=p
From (3), we see
∞
∑
μ=p
∞
∑u(r n )α t
τ(αμ)tμ =
z
ij
μ
μ
Hardouin, C., Hypertranscendance des systèmes aux
différences diagonaux, Compos. Math., 144, No.3, 2008,
pp.565-581.
Kolchin, E. R., Algebraic groups and algebraic dependence,
Amer. J. Math., 90, No.4, 1968, pp.1151-1164.
Kubota, K. K., On the algebraic independence of
holomorphic solutions of certain functional equations and
their values, Math. Ann., 227, No.1, 1977, pp.9-50.
Levin, A., Difference Algebra, Springer Science+Business
Media B.V., 2008.
Matsumura, H., Commutative Ring Theory, Cambridge
Univ. Press, 1986.
Rosenlicht, M., On Liouville's theory of elementary
functions, Pacific J. Math., 65, No.2, 1976, pp.485-492.
.
μ=p
Therefore we obtain τ(αp ) = u(rz nij)αp, the form of (1).
We also put
102