純粋・応用数学・数学隣接分野(含むガロア理論)12at MATH
純粋・応用数学・数学隣接分野(含むガロア理論)12 - 暇つぶし2ch238:現代数学の系譜 雑談
23/01/01 09:36:32.87 x1AjdVpC.net
>>237
つづき
4. Ambiguity by units
P8
Let qo generate p, the ideal lying under P in Z[ω], where P defines the Kummer (-Teichm¨uller) character.
Identify (Z/m)× with the Galois group of Q(ω) over Q, which we know acts transitively on primes over p in Z[ω].
6. Numerical examples
P13
[6.7] p = 11 and order m = 5 Since ω = ω5 satisfies ω^4 + ω^3 + . . . + ω + 1 = 0,
0 =((ω + 2) - 2)^4+((ω + 2) - 2)^3+ . . . +((ω + 2) - 2)+ 1 = (ω + 2)^4 + . . . + 11
The constant term 11 = (2^5 + 1)/(2 + 1) is the norm of qo = ω + 2, so
11 = (ω + 2)(ω^2 + 2)(ω^3 + 2)(ω^4 + 2)
The fifth power of the quintic Gauss sum is
γ(χ^-2_P )^5 = η ・ (ω + 2) (ω^2 + 2)^3(ω^3 + 2)^2(ω^4 + 2)^4
and the congruence for η is
-η (ω^2 + 2)^2(ω^3 + 2) (ω^4 + 2)^3 = (-1/((11-1)/5)!)5 mod (ω + 2)
Using ω = -2 mod ω + 2, this is
η ((-2)^2 + 2)^2((-2)^3 + 2) ((-2)^4 + 2)^3 =1/2^5 mod (ω + 2)
or
η ・ 6^2・ (5) ・ (7)^3 = -1 mod (ω + 2)
which simplifies to η ・ 3 ・ 5 ・ 2 = -1 mod (ω + 2) and then 3η = 1 mod (ω + 2), so η = 4 mod (ω + 2). Since
ω = -2 mod (ω + 2), this gives η = ω^2. Thus,
γ(χ^-2_P )^5 = ω^2・ (ω + 2) (ω^2 + 2)^3(ω^3 + 2)^2(ω^4 + 2)^4
and the quintic subfield of Q(ω5, ζ11) is generated over Q(ω5) by the fifth root of this.
つづく


次ページ
続きを表示
1を表示
最新レス表示
レスジャンプ
類似スレ一覧
スレッドの検索
話題のニュース
おまかせリスト
オプション
しおりを挟む
スレッドに書込
スレッドの一覧
暇つぶし2ch