四変数基本対称式の解放
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1 The second-thought of the Galois-style way to solve a quartic equation Oomori, Yasuhiro in Himeji City, Japan Jan.6, 013 Abstract v ρ (v) Step1.5 l l 3 7. Step - V v Step groupe groupe groupe
2 1. (1 ) f(x) = x 4 + ox 3 + px + qq + r = 0 o, p, q, r C = (x α)(x β)(x γ)(x δ) α, β, γ, δ {ρ C f(ρ) = 0} (1) α + β + γ + δ = o αβ + αγ + αδ + βγ + βδ + γδ = p (1 ) αβγ + αβδ + αγδ + βγδq = q αβγδ = r {o, p, q, r} {α, β, γ, δ} V () V(α, β, γ, δ; A, B, C, D) = Aα + Bβ + Cγ + Dδ () A, B, C, D l 3 (3) l 3 = α + β γ δ (3) l 3 (3) V (4) l 3 = V(α, β, γ, δ; 1,1, 1, 1) (4) V(α, β, γ, δ; 1,1, 1, 1) ψ(α, β, γ, δ) (5) V(α, β, γ, δ; 1,1, 1, 1) = ψ(α, β, γ, δ) (5) l 3 Step1 (6) : l 3 (o, p, q, r) l 3 = 1 Ø 3 + ω Ø 3 + Ø + ω C 3 1 Ø Ø = α + β γ δ (6) Ø 1 = o Ø = 4f o (6.) Ø 3 = 3Ø C D Ø 4 = Ø 3 4C 3 C = 4f D = o f o 4 3 f 3 o f o (6.3) 4
3 dd(x) f (x) = dd = 4x3 + 3ox + pp + q f (x) = d f(x) = 1x + 6oo + p dd (6.4) Step Step3 l 3 (6 ) α α β, γ, δ l 3
4 . Step - Step - l 3 = α + β γ δ ( 1.1.1) o = α + β + γ + δ p = αβ + αγ + αδ + βγ + βδ + γδ ( 1.1.) q = αβγ + αβδ + αγδ + βγδq r = αβγδ Step.1 ( 1.1) l 3 α ( 1.) {l 3 (α + β γ δ)} = 0 ( 1.) groupe (G) α τ n n {1,,3,4,5} l 3 ( 1.3) g = {l 3 τ 1 ψ(α, β, γ, δ)}{l 3 τ ψ(α, β, γ, δ)}{l 3 τ 3 ψ(α, β, γ, δ)}{l 3 τ 4 ψ(α, β, γ, δ)}{l 3 τ 5 ψ(α, β, γ, δ)} ( 1.3) τ 1 = (γδ) τ = (βδ) τ 3 = (βγ) τ 4 = (βγδ) τ 5 = (βδγ) ( ) F (l 3, α, β, γ, δ) = {l 3 ψ(α, β, γ, δ)}g = 0 {l 3 (α + β γ δ)}{l 3 (α + γ δ β)}{l 3 (α + δ β γ)} ( ) {l 3 (α + β δ γ)}{l 3 (α + δ γ β)}{l 3 (α + γ β δ)} = 0 (l 3 l 3 )(l 3 l )(l 3 l 4 )(l 3 l 3 )(l 3 l ){l 3 l 3 } = 0 l 1 l l 3 l 4 = α + β + γ + δ = α β + γ δ = α + β γ δ = α β γ + δ ( 3)
5 ( ) ( 1.1.) ( 4) F (l 3, α, o, p, q) = F (l 3, α) = l 3 3 l 3 (4α + o) + 1 l 3{(4α + o) o } d = 0 ( 4.1) c = 4f o 4 43 f o = l 4 l 3 + l l 4 + l 3 l 4 o = 4f o = l 4 + l 3 + l ( 4.) d = 8f o = l 4 l 3 l 4 F (l 3, α, o, p, q) ( 4) ( ) groupe (G) σ G ( 5) σf (l 3, α, o, p, q) = 0 σ G ( 5) σ 1 σ 7 ( 6) σ 1 = (αβ)(γδ) = βαδγ σ 9 = (αβ) = ( 6) βαγδ ( 7) σ 1F (l 3, α, o, p, q) = F (σ 1 l 3, σ 1 α, σ 1 o, σ 1 p, σ 1 q) = F (l 3, β, o, p, q) = 0 ( 7) σ 9 F (l 3, α, o, p, q) = F (σ 9 l 3, σ 9 α, σ 1 o, σ 1 p, σ 1 q) = F (l 3, β, o, p, q) = 0 α β ξ ( 8) ξ {α, β} ( 8) ( 4.1) ( 1.1.) α β ( 9) f(ξ) = ξ 4 + oξ 3 + pξ + qq + r = 0 F (l 3, ξ) F (l 3, ξ, o, p, q) = l 3 3 l 3 (4ξ + o) + 1 l 3{(4ξ + o) o } d ( 9) = 0
6 Step. - - ( 9) ξ ξ l 3 ( 9) ξ ( 10) l 3 (4ξ + o) l 3 (4ξ + o) + l 3 3 o l 3 d = 0 ( 10) ξ l 3 ρ (l 3 ) ( 11) ξ = ρ (l 3 ) = 1 o + l 3 ± o l 3 + d l 3 ( 11.1) c = 4f o 4 43 f o 4 o = 4f o ( 11.) 4 d = 8f o
7 4. Step3 - - Step ( 7) σ 1F(l 3, α) = F(σ 1 l 3, σ 1 α) = F(l 3, β) = 0 ( 7) σ 9 F(l 3, α) = F(σ 9 l 3, σ 7 α) = F(l 3, β) = 0 σ 6 (3 1) σ 6 = (αγ)(βδ) = (3 1) γδαβ ( 7) (3 ) σ 6F(l 3, α) = F(σ 6 l 3, γ) = 0 (3 ) σ 6 F(l 3, β) = F(σ 6 l 3, δ) = 0 γ δ η ( 10) η {γ, δ} (3 3) γ δ (3 4) f(η) = 0 (3 4) F (σ 6 u, η) = 0 γ δ (3 5) η = ρ (σ 8 l 3 ) (3 5) σ 6 u (3 6) σ 6 l 3 = σ 6 ψ(α, β, γ, δ) = ψ(γ, δ, α, β) = α β + γ + δ = l 3 (3 6) η l 3 ρ (x) (3 7) η = ρ ( l 3 ) (3 7)
8 5. Step Step3 Step1 3 (4 1) { ρ f(ρ) = ρ 4 + oρ 3 + pρ + qρ + r } = { ρ 1 (l 3 ), ρ 1 ( l 3 ), ρ l 3, ρ ( l 3 ) } (4 1) ρ1 (x) = 1 o + x o 4 x + d x (4 1.) ρ (x) = 1 o + x + o 4 x + d x c = 4f o 4 43 f o 4 o = 4f o (4 1..) 4 d = 8f o l 3 = 1 Ø 3 + ω Ø 3 + Ø + ω C Ø 3 + Ø ω: ω + ω + 1 = 0 (4 1.3) Ø 1 = o Ø = 4f o (4 1.3.) Ø 3 = 3Ø C D Ø 4 = Ø 3 4C 3 C = 4f D = o f o 4 3 f 3 o f o ( ) 4 dd(x) f (x) = dd = 4x3 + 3ox + pp + q (4 1.4) f (x) = d f(x) = 1x + 6oo + p dd
9 6. l 3 l 3 groupe (G) l 3 = α + β γ δ (6 1) (α + β γ δ), (α β γ + δ), (α β + γ δ), Gl 3 = (6 1) ( α β + γ + δ), ( α + β + γ δ), ( α + β γ + δ) (6 1) (6 ) {l 3 (α + β γ δ)}{l 3 (α β γ + δ)}{l 3 (α β + γ δ)} {l 3 ( α β + γ + δ)}{l 3 ( α + β + γ δ)}{l 3 ( α + β γ + δ)} = 0 (6 ) (6 ) (6 6) {l 3 l 3 }{l 3 l 4 }{l 3 l }{l 3 ( l 3 )}{l 3 ( l 4 )}{l 3 ( l )} = 0 (6 3.1) l 1 = α + β + γ + δ l = α β + γ δ (6 3.) l 3 = α + β γ δ = α β γ + δ l 4 l 3 l 3 l 3 l l 3 l = 0 (6 4) l 6 3 l + l 3 + l l l l 3 + l l 4 + l 3 l l 3 (l l 3 l 4 ) = 0 (6 5) l 3 6 4f o 43 f o l f o l 3 8f o = 0 (6 6) l 3 (6 7) l 6 3 o l 4 3 +c l 3 d 3 = 0 (6 7.1) c = 4f o 4 43 f o = l 4 l 3 + l l 4 + l 3 l 4 o = 4f o = l 4 + l 3 + l (6 7.) d = 8f o = l 4 l 3 l 4
10 7. Step - V v - V v V (7 1) l 1, l, l 3, l (7 ) v (7 3) V = Aα + Bβ + Cγ + Dδ (7 1) l 1 = α + β + γ + δ = o l = α β + γ δ (7 ) l 3 = α + β γ δ l 4 = α β γ + δ v = α + iβ γ iδ (7 3) (7 4) V = Aα + Bβ + Cγ + Dδ = 1 (A + B + C + D)(α + β + γ + δ) + 1 (A B + C D)(α β + γ δ) (A ib C + id)(α + iβ + γ iδ) + 1 (A + ib C id)(α iβ + γ + iδ) 4 4 = 1 (a 4 1l 1 + a l + a 4 v + a 3 v ) (7 4) a 1 = A + B + C + D a = A B + C D (7 4.1) a 3 = A + ib C id a 4 = A ib C + id v = (1 i)l 3 + i v = α iβ γ + iδ (7 4.) v, v l 3, l (7 5) v = 1 (l 3 + l 4 ) + i (l 3 l 4 ) v = 1 (l 3 + l 4 ) i (l (7 5.1) 3 l 4 ) l 3 = 1 (v + v ) i (v v ) l 4 = 1 (v + v ) + i (7 5.) (v v ) ( 4.) (7 5.) (7 6) c = l l 3 + l l 4 + l 3 l 4 = l (v + v ) (v v ) o = l + l 3 + l 4 = l + (v + v ) d = l l 3 l 4 = l (v + v ) (v v ) + (v v ) + (v + v ) + (v v ) (7 6)
11 (7 6) (7 4) V, l 1, l, v (7 7) 4V = a 1l 1 + a l + a 4 v + a 3 v o = l 1 c = vv l + (v + v ) = 4f o 4 43 f o o = l + vv = 4f o d = l (v + v ) = 8f o (7 7) (7 7) V, l, v (7 8) 4V ( a 1o + a l + a 4 v + a 3 v ) = 0 l 5 o l 3 + (o + 4v 4 )l 4d v = 0 l 6 o l 4 + c l d = 0 (7 8) v = o l v (7 8) V, l (7 9) o l 4V a 1 o + a l + a 4 v + a 3 = 0 v l 4 + 4v4 + c o l o + 4d v l o d (7 9) = 0 o ( 4v 4 + c + o )l 3 + ( 4o v 4 + 4d v o 3 )l d l + 4o d v = 0
12 (7 9) l 1 v V (7 10) V = 1 a o l (v) 4 1o + a l (v) + a 4 v + a 3 v (7 10.1) = l (v) l μ 1 (x)λ (x)λ 3 (x) μ (x)λ 1 (x)λ 3 (x) + μ 4 (x)λ 1 (x) l (x) = μ 1 (x)λ 1 (x)λ 3 (x) μ 1 (x)λ (x) + μ (x)λ 1 (x)λ (x) μ 3 λ ( ) 1 (x) λ 1 (x) = 4x4 + c o μ o 1 (x) + μ (x) μ 1 (x)μ 3 (x) λ (x) = 4d x μ o 1 (x) + μ (x)μ 3 (x) μ 1 (x)μ 4 (x) (7 10..) λ 3 (x) = d μ o 1 (x) + μ (x)μ 4 (x) μ 1(x) = 4x 4 + c + o μ (x) = 4o x 4 + 4d x o 3 ( ) μ 3 (x) = d μ 4 (x) = 4o d x c = 4f o 4 43 f o 4 o = 4f o ( ) 4 d = 8f o
13 8. Step groupe (G) αβγδ, βαδγ, γδαβ, δγβα, αγδβ, γαβδ, (G) δβαγ, βδγα, αδβγ, δαγβ, βγαδ, γβδα, αβδγ, βαγδ, δγαβ, γδβα, αδγβ, αβγδ, αγδβ, αδβγ, δαβγ, βαδγ, γαβδ, δαγβ, G γβαδ, γδαβ, δβαγ, βγαδ, βγδα, δγβα, βδγα, γβδα; αγβδ, γαδβ, βδαγ, δβγα; Fig. 1 αβγδ = σ 1, αγδβ = σ 5, αδβγ = σ 9, βαδγ = σ, γαβδ = σ 6, δαγβ = σ 10, γδαβ = σ 3, δβαγ = σ 7, βγαδ = σ 11, δγβα = σ 4, βδγα = σ 8, G γβδα = σ 1 G Fig. groupe G v = φ(α, β, γ, δ) 1 θ (8 1) = σ i v σ j v σ k vσ l v σ m v σ n v (8 1) θ σ i, σ j, σ k σ l, σ m, σ n G i<j<k<l<m<n Θ (8 ) Θ = θ τ 1 θ = Δ (8.1) Δ = {(α β)(α γ)(α δ)(β γ)(γ δ)(δ β)} (8.) τ 1 = (γδ)
14 Reference Évariste Galois Mémoire sur les conditions de résolubilité des équations par radicaux Internet Archive 00 pp33 50
四変数基本対称式の解放
Solving the simultaneous equation of the symmetric tetravariate polynomials and The roots of a quartic equation Oomori, Yasuhiro in Himeji City, Japan Dec.1, 2011 Abstract 1. S 4 2. 1. {α, β, γ, δ} (1)
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