ガロア第一論文と乗数イデアル他関連資料スレ18

ガロア第一論文と乗数イデアル他関連資料スレ18 (541ﾚｽ)
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27: 現代数学の系譜 雑談 ◆yH25M02vWFhP  [] 2025/05/28(水) 14:53:16.49 ID:vzADU7Bh

つづき

Let  ai=bi pi, i=1,…s,
where the pi  are s
different primes and the bi
positive integers not divisible by any of them. The author proves by an inductive argument that, if xj
are positive real roots of xnj−aj=0, j=1,...,s,
and P(x1,...,xs)
is a polynomial with rational coefficients and of degree not greater than nj−1
with respect to xj,
then P(x1,...,xs)
can vanish only if all its coefficients vanish.  Reviewed by W. Feller.

15,404e 10.0X
Mordell, L. J.
On the linear independence of algebraic numbers.
Pacific J. Math. 3 (1953). 625-630.

Let K
be an algebraic number field and x1,…,xs
roots of the equations  xnii=ai (i=1,2,...,s)
and suppose that (1) K
and all xi
are real, or (2) K
includes all the ni
th roots of unity, i.e. K(xi)
is a Kummer field. The following theorem is proved. A polynomial P(x1,...,xs)
with coefficients in K
and of degrees in xi
, less than ni
for i=1,2,…s
, can vanish only if all its coefficients vanish, provided that the algebraic number field K
is such that there exists no relation of the form  xm11 xm22⋯xmss=a
, where a
is a number in K
unless  mi≡0modni (i=1,2,...,s)
. When K
is of the second type, the theorem was proved earlier by Hasse [Klassenkorpertheorie, Marburg, 1933, pp. 187--195] by help of Galois groups. When K
is of the first type and K
also the rational number field and the ai
integers, the theorem was proved by Besicovitch in an elementary way. The author here uses a proof analogous to that used by Besicovitch [J. London Math. Soc. 15b, 3--6 (1940) these Rev. 2, 33].  Reviewed by H. Bergstrom.

https://arxiv.org/abs/2006.07951
Journal reference: Can. Math. Bull. 64 (2021) 877-885 [Submitted on 14 Jun 2020 (v1), last revised 27 May 2021 (this version, v2)]
https://arxiv.org/pdf/2006.07951
ON THE DEGREE OF REPEATED RADICAL EXTENSIONS
FERNANDO SZECHTMAN
Abstract. We answer a question posed by Mordell in 1953, in the case repeated radical extensions, and find necessary and sufficient conditions for [F[ m1 √N1,..., mℓ √Nℓ] : F] = m1 ···mℓ, where F is an arbitrary field of characteristic not dividing any mi.
(引用終り)
以上

http://rio2016.5ch.net/test/read.cgi/math/1748354585/27

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