◆研究ノート◆

Linear Ordinary Differential
Equations and Fermat Equations
線形常微分方程式とフェルマ方程式
Keiji Nishioka

Professor, Faculty of Environment and Information Studies, Keio University
西岡 啓二
慶應義塾大学環境情報学部教授

A theorem analogous to Picard's theorem on representation of a plane algebraic curve of genus
greater than 1 with meromorphic functions will be proved. Its enunciation will be done for elements
in a Fuchsian extension defined in this note instead of considering for meromorphic functions. As
seen straightforwardly, a differential extension generated with solutions of linear ordinary differential
equations turns out to be Fuchsian, hence the theorem deduces a corollary that solutions of linear
ordinary differential equations substaintially satisfy no Fermat equations.
種数＞１の平面代数曲線の有理型関数による表現に関するピカールの定理の類似が証明される。
命題は有理型関数に対してではなく、このノートで定義されるフックス拡大の要素に対して記述される。線形常微分方程式の解で生成される微分拡大はフックス拡大であり、この定理は系として、線形常微分方程式の解はフェルマ方程式を本質的に満足しないという結果を導く。
Keywords: Linear Ordinary Differential Equation, Fuchsian Extension, Prime Divisor, Picard'sTheorem, Picard-Vessiot Extension, Strongly Normal Extension

126Linear Ordinary Differential Equations and Fermat Equations

as a required one.

1 Fuchsian extensions

A differential algebraic function ﬁeld of one,

Picard s theorem [P] states: Suppose twodimension R/K which is Fuchsian has no movable

meromorphic functions u, v sataisfy the irreducible

singularities in the sense of Matsuda [M], namely

algebraic relation of degree mbeing a differential extension of which every valuation

F(u, v) = 0.

ring is stable under D. In fact let P be a prime divisor─of R/K. There is an element u ∈ R /K with its pole
only at P. By deﬁnition we see P ∈Π , hence DOP ⊂OP, which indicates the assertion. The converse canThen the genus of the algebraic relation must be 0

be readily seen.

or 1. Single-valuedness has been a little investigatedin the theory of differential algebra, for instance [M]A differential extension R/K is said to depend

and [B]. This paper attempts to propose a concept

rationally on arbitrary constants if there exists a

relating to algebraic differential equations with nodifferential extension L/K such that R and L are free

movable algebraic branches and apply it to show an,analogy of Picard s theorem.

over K and LR=LCLR holds [N]. If this is the case R/K

Let K be a differential ﬁeld of characteristic 0 withdeﬁned as follows. Let c1, c2,..., cn be a transcendence

differentiation D and let CR denote the constant ﬁeldbase of CLR over CL and Π deﬁne the set consisting ofof a differential extension R/K. In the following we

restrictions to R of prime divisors of LR/Li (1 < i< n),set C=CK and assume it to be algebraically closed. A

where Li = L(c1, ... , ci-1,ci+1, ... , cn). Noting that LR/Liprime divisor of a ﬁnitely generated ﬁeld extenstion
R/K in this paper means an equivalence class of

has no movable singularities in Matsuda's sense, foru ∈ R /─K ⊂ LR ─/L for some i there is a prime divisor P

discrete valuation of rank 1 of R/K.

of LR/Li where u has a pole. The restriction Pi / isis Fuchsian with a set of prime divisors Π , which isiRour prime divisor.

A differential extension R/K which is ﬁnitely
generated as a ﬁeld extension will be said to be

Any strongly normal extension R/K is Fuchsian

Fuchsian if there is a set Π of prime divisors of R/K

since it depends rationally on arbitrary constants

such that each valuation ring OP of P ∈Πis stable under─ has athe differentiation D and every element of R /K

viewing the interpretation of [BB]: R/K is bypolar prime divisor in Π .

× R provided K
is generated with constants over 1 ○× KR
deﬁnition strogly normal if the quotient ﬁeld of R ○being algebraically closed in R .

Here we will discuss some basic properties ofFuchsian extensions.

2 Theorem

If R/K is Fuchsian and S is an intermediate
differential ﬁeld between R and K, then S/K is

In the sequel we assume that K is algebraically

Fuchsian. In fact let Π be the required set of prime

closed in R and C=CK. Let R/K be a differential ﬁeld

divisors of R/K in the deﬁnition. Then clearly D(OPextension. Then its differential module Ω R/K has a CR -∩ S ) ⊂ OP ∩ S and if P ∈Π is a polar prime divisor

linear operator D1 (the Lie derivative) characterized─of u ∈ S /K then so is the restriction to S of P, P /s.

byHence we may take the set of prime divisors of S/KD1(adb) = D(a)db + adDb(a, b ∈ R),{P /s /some element of S has a pole at P ∈Π }which in particular satisﬁes D1d=dD1 on R. According

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to [R], D1(adb) = d(aDb) holds provided a, b ∈ R areProof

algebraically dependent over C.

of solutions,Φ, of the system of linear differential equ-R/K is generated by a fundamental system

×ations DΦ = AΦ over K. Since the quotient ﬁeld of R ○,The following is an analogue of the Picard s

K R is generated with constants, R/K has nomovable

theorem on analytic representation of a plane curve.

algebraic singularities. So, it remains to prove inthe case of n=3. As seen above S=K(u, v)/K has noLet R/K be a Fuchsain differential ﬁeldmovable singularities in the sense of Matusda withextension with Π, a set of prime divisors of R/K. Suppose

genus 1. Hence it is an elliptic function ﬁeld over K

that there exists a subﬁeld S of R which is a one-([N] or [M]), whence an abelian extension. Noting

dimensional algebraic function ﬁeld over C withthe following Remark, S must agree with K, which

genus greater than one. Then CR ≠ C.

shows our assertion.

Proof Let udv ∈ΩS/C ⊂ ΩKS/K be regular. We then haveRemarkTheorem

1Suppose R/K is a Picard-Vessiot extension

D (udv)=d(uDv). Assume that uDv is transcendental overwith K being algebraically closed in R and S/K anK. Then there exists a prime divisor P ∈Π at which uDv

abelian extension with C RS = C. Then R and S are

has a pole. Let t ∈ KS be a prime element associatedlinearly disjoint over K. In fact, the differential Galois

with the restriction of P to KS. By assumption theregroup G(R ∩ S/K) is the C-homomorphic image of1

is an element w∈ OP with udv=wdt, whence D (udv)=G(R/K) as well as G(S/K) (Theorem 4 in [K, p.401]).

D(w)dt + wdDt. Since Dt, Dw ∈ OP, it follows d(uDv)

Hence it is afﬁne as well as complete, consequently

1= D (udv)∈OPd OP and a contradiction. Thus uDv ∈ K.

R ∩ S=K by [K, pp.359 and 377]. Since by Theorem

Since S and K are linearly disjoint over C and KS ⊂5 in [K, p.403] G(RS/S) G(R/K) it follows trans.deg

R, KS/K has the same genus as S/C and is Fuchsian,

RS/S = trans.deg R/K, which implies R and S are free

therefore having no movable singularities in theover K, hence they are linearly disjoint over K sincesense of Matsuda as mentioned in the introduction.

R/K is regular.

By the theorem of Poincaré in [N] or [M], CKS ≠ C.

We shall end this note by explaining our framework applies to another proof of Sperber's theorem

Let n be an integer greater than 3.

[S]: Let each of nonzero y1, y2,..., yn(n>1)satisfy some

Suppose that R/K has no movable algebraic singu-linear ordinary differential equation over K, andlarities with CR = C and there are elements u, v in R

2yn,n where the mi are positive
suppose they fulﬁll y1=y2 ...

with un +vn = 1 . Then they in fact are algebraic over K.

integers. Assume N, the order of the linear ordinary

Corollarymm

differential equation over K satisﬁed by y1, does not
Assume that u is transcendental over K, andexceed min{m2,...,mn} then all the Dyi/yi are algebraic

let S =C(u, v). Then S is a one-dimensional algebraic

over K. In fact suppose the converse, namely, somefunction ﬁeld over C with genus greater than 1. By

Dyj /yj is transcendental over K. Since the differential

the theorem we have CR ≠ C, a contradiction.

extension R=K〈y1, y2,...,yn〉/K is Fuchsian, there is a

Proofpolar prime divisor P of Dyj/yj with the valuation ring
Let n be an integer greater than 2.

OP being stable under D. Let v and t be the valuation

Suppose that R/K is a Picard-Vessiot extension andand a prime element for P. If v(Dt)>0 then v(Dyj/yj) >Corollarynn

there are elements u, v in R with u +v = 1. Then they

0 because describing yj = t rz(v(z)=0) we have v(Dyj/in fact are algebraic over K .

yj)=v(rDt/t+Dz/z)>0. This is absurd. Hence v(Dt)=0.

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Linear Ordinary Differential Equations and Fermat Equations

This time v(Dz)=v(z)-1 holds for z ∈ R with v(z) ≠ 0.

References

Since inequality v(yi)<0 would derive that yi satisﬁes[BB] Bialynicki-Birula, A., “On the Galois theory of fields with operators”, In Amer. J. Math., Vol. 84, No. 1, 1962, pp.89-109.
[B] Buium, A., “Differential function fields and moduli of algebraic
varieties”, In LMN, 1226, Berlin, Springer, 1986.
[K] Kolchin, E.R., Differential Algebra and Algebraic Groups, Academic Press, 1973.
[M] Matsuda, M., “First order algebraic differential equations”, In
LNM, 804, Berlin, Springer, 1980.
[N] Nishioka, K., “Algebraic differential equations of Clairaut typefrom the differential-algebraic standpoint”, In J. Math. Soc. Japan, Vol. 31, 1979, pp.553-560.
[P] Picard, E., “Sur une propriété des fonctions uniformes d'unevariable, liées par une relation algébrique”, In Œble de Émile
Picard, CNRS, Vol. 1, 1978, pp.65-67.
[R] Rosenlicht, M., “The nonminimality of the differential closure”,
In Pacific J. Math., Vol. 52, No. 2, 1974, pp.529-537.
[S] Sperber, S., “On solutions of differential equations which satisfycertain algebraic relations”, In Pacific J. Math., Vol. 124, No.1,1986, pp.249-256.

no linear ordinary differential equation over K, it
follows v(yi)> 0 for any i and clearly v(yj)> 0 . ThenN>v(y1)=m2v(y2)+ ... +mnv(yn) > mjv(yj)> mj ,and so that N>min{m2,..., mn}, a contradiction, which
completes the proof.

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