TY - JOUR

T1 - Transcendental constants over the coefficient fields in differential elliptic function fields

AU - Nishioka, Keiji

N1 - Copyright:
Copyright 2017 Elsevier B.V., All rights reserved.

PY - 1978/1

Y1 - 1978/1

N2 - Let k be a differential field of characteristic 0, and Ω be a universal extension of k. Suppose that the field of constants k0 of k is algebraically closed. Consider the following differential polynomial of the first order over k in a single indeterminate y: T(y) = (y′)2 - λS (y; k) λ ∈ k; λ ≠ 0; here S(y; k) = y(1 - y)(1 - κ2y); κ ∈ k; k2 ≠ 0, 1; k′ = 0. Take a generic point z of the general solution of T. Then, z is transcendental over k, and k(z, z′) is called a differential elliptic function field. We prove the following: THEOREM. Let k(z, z′) be a differential elliptic function field over k. Then, there exists a finitely generated differential extension field k* of k such that the following three conditions are satisfied: (i) z is transcendental over k*; (ii) the field of constants of k* is the same as k0; (iii) there exists an element ζ of Ω such that k*(z, z′) = k*(ζ, ζ′) and (ζ′)2 = 4S(ζ; κ) with the same modulus as κ.

AB - Let k be a differential field of characteristic 0, and Ω be a universal extension of k. Suppose that the field of constants k0 of k is algebraically closed. Consider the following differential polynomial of the first order over k in a single indeterminate y: T(y) = (y′)2 - λS (y; k) λ ∈ k; λ ≠ 0; here S(y; k) = y(1 - y)(1 - κ2y); κ ∈ k; k2 ≠ 0, 1; k′ = 0. Take a generic point z of the general solution of T. Then, z is transcendental over k, and k(z, z′) is called a differential elliptic function field. We prove the following: THEOREM. Let k(z, z′) be a differential elliptic function field over k. Then, there exists a finitely generated differential extension field k* of k such that the following three conditions are satisfied: (i) z is transcendental over k*; (ii) the field of constants of k* is the same as k0; (iii) there exists an element ζ of Ω such that k*(z, z′) = k*(ζ, ζ′) and (ζ′)2 = 4S(ζ; κ) with the same modulus as κ.

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U2 - 10.2140/pjm.1978.74.191

DO - 10.2140/pjm.1978.74.191

M3 - Article

AN - SCOPUS:84972494000

VL - 74

SP - 191

EP - 197

JO - Pacific Journal of Mathematics

JF - Pacific Journal of Mathematics

SN - 0030-8730

IS - 1

ER -