独学ノート（土筆の子）コミュのf[x]=x^4 - 2*x^2 - 2

http://userdisk.webry.biglobe.ne.jp/020/691/47/N000/000/000/129499861005616220389.gif
　　の　紫の囲み　部分　は　もう　辟易でしょうか？
次数を　ひとつ上げ　模倣したい；

f[x]=x^4 - 2*x^2 - 2(のSqrt[判別式]=48*I*Sqrt[2]の確認を）

もかいを全て　遺漏なきよう　漏らし　ます；
σ1[α]=α
σ2[α]=-α,
σ3[α]= 2*I*Sqrt[2/3]*α - (I*α^3)/Sqrt[6],
σ4[α]=-2*I*Sqrt[2/3]*α + (I*α^3)/Sqrt[6]}
せめてσ3[α]も　ホンマに　かい　即　f[σ3[α]]=Zero か　真を問うてください(信は問わないでください)；
例えば
Eliminate[{α^4-2*α^2-2=0,β=2*I*Sqrt[2/3]*α - (I*α^3)/Sqrt[6]},α]を
例の処に　挿入し　真偽を　問うて、下さい。
　　　★商空間　K[x]/f[x]*K[x]の考察をします★；

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[1] mixiユーザー 01月16日 11:02

f[x]=x^4 - 2*x^2 - 2
=(x-(Sqrt[1 + Sqrt[3]]) (x+Sqrt[1 + Sqrt[3]]) (x^2-1 + Sqrt[3])

f[x]=0

x = -sqrt(1+sqrt(3))
x = sqrt(1+sqrt(3))
x = -i*sqrt(sqrt(3)-1)
x = i*sqrt(sqrt(3)-1)

これが、以下に関係するのでしょうが、見えにくいですね。

σ1[α]=α
σ2[α]=-α,
σ3[α]= 2*I*Sqrt[2/3]*α - (i*α^3)/Sqrt[6],
σ4[α]=-2*I*Sqrt[2/3]*α + (i*α^3)/Sqrt[6]}

Delta = -4608
=-2^9×3^2 (11 prime factors, 2 distinct)

[2] mixiユーザー 01月16日 11:08

f(x)=x^3-x-1

Delta = -23

すごい複雑になる。

-(i (-108 i x+118 i 2^(1/3) (3 (9+sqrt(69)))^(2/3)+15 2^(2/3) sqrt(23) (3 (9+sqrt(69)))^(1/3)-540 i 2^(2/3) (3 (9+sqrt(69)))^(1/3)-30 2^(1/3) 3^(1/6) (9+sqrt(69))^(2/3)+59 i 2^(2/3) 3^(5/6) sqrt(23) (9+sqrt(69))^(1/3)-36 2^(2/3) 3^(5/6) (9+sqrt(69))^(1/3)+118 i 2^(1/3) (3 (9-sqrt(69)))^(2/3)+15 2^(2/3) sqrt(23) (3 (9-sqrt(69)))^(1/3)-540 i 2^(2/3) (3 (9-sqrt(69)))^(1/3)+30 2^(1/3) 3^(1/6) (9-sqrt(69))^(2/3)-59 i 2^(2/3) 3^(5/6) sqrt(23) (9-sqrt(69))^(1/3)+36 2^(2/3) 3^(5/6) (9-sqrt(69))^(1/3))

(216 i x+118 i 2^(1/3) (3 (9+sqrt(69)))^(2/3)+15 2^(2/3) sqrt(23) (3 (9+sqrt(69)))^(1/3)-513 i 2^(2/3) (3 (9+sqrt(69)))^(1/3)-30 2^(1/3) 3^(1/6) (9+sqrt(69))^(2/3)+59 i 2^(2/3) 3^(5/6) sqrt(23) (9+sqrt(69))^(1/3)-27 2^(2/3) 3^(5/6) (9+sqrt(69))^(1/3)+118 i 2^(1/3) (3 (9-sqrt(69)))^(2/3)+15 2^(2/3) sqrt(23) (3 (9-sqrt(69)))^(1/3)-513 i 2^(2/3) (3 (9-sqrt(69)))^(1/3)+30 2^(1/3) 3^(1/6) (9-sqrt(69))^(2/3)-59 i 2^(2/3) 3^(5/6) sqrt(23) (9-sqrt(69))^(1/3)+27 2^(2/3) 3^(5/6) (9-sqrt(69))^(1/3))

(216 i x+118 i 2^(1/3) (3 (9+sqrt(69)))^(2/3)+15 2^(2/3) sqrt(23) (3 (9+sqrt(69)))^(1/3)-567 i 2^(2/3) (3 (9+sqrt(69)))^(1/3)-30 2^(1/3) 3^(1/6) (9+sqrt(69))^(2/3)+59 i 2^(2/3) 3^(5/6) sqrt(23) (9+sqrt(69))^(1/3)-45 2^(2/3) 3^(5/6) (9+sqrt(69))^(1/3)+118 i 2^(1/3) (3 (9-sqrt(69)))^(2/3)+15 2^(2/3) sqrt(23) (3 (9-sqrt(69)))^(1/3)-567 i 2^(2/3) (3 (9-sqrt(69)))^(1/3)+30 2^(1/3) 3^(1/6) (9-sqrt(69))^(2/3)-59 i 2^(2/3) 3^(5/6) sqrt(23) (9-sqrt(69))^(1/3)+45 2^(2/3) 3^(5/6) (9-sqrt(69))^(1/3)))/5038848

[3] mixiユーザー 01月16日 11:22

f(x)=x^3-x-1

の一つの根をαとすると、

α^3-α-1=0

他の根、β、γ
β^3-β-1=0

α+β+γ=0
αβ+βγ+γα=-1
αβγ=1

γ=1/αβ=-(α+β)

αβ+1/α+1/β=-1=αβ+(α+β)/αβ=αβ-1/(αβ)^2

(αβ)^3+(αβ)^2-1=0

(α+1)(β+1)+(αβ)^2-1=0

2(αβ)^2+(α+β)=0

2(αβ)^2+(α+β)=2(αβ)^2-1/αβ=0

2(αβ)^3=1

αβ=[3]√(1/2)

[4] mixiユーザー 01月16日 11:27