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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 3 1 6. l 3 7. Step - V v - 3 8. Step1.3 - - groupe groupe groupe

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 3 3 1 Ø 3 + Ø + ω C 3 1 Ø 3 3 + 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 + 3 43 f o 4 3 f 3 o 4 33 4 3 f o (6.3) 4

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

. 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)

( ) ( 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

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

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)

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 3 1 3 Ø 3 + Ø + ω C 3 3 1 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 + 3 43 f o 4 3 f 3 o 4 33 4 3 f o (4 1.3.3) 4 dd(x) f (x) = dd = 4x3 + 3ox + pp + q (4 1.4) f (x) = d f(x) = 1x + 6oo + p dd

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 4 3 + 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 3 4 + 4f 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

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) 4 4 + 1 (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)

(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

(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 (7 10..1) 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 (7 10.3.1) μ 3 (x) = d μ 4 (x) = 4o d x c = 4f o 4 43 f o 4 o = 4f o (7 10.3.) 4 d = 8f o

8. Step1.3 - - 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 = 4 11 5 Δ (8.1) Δ = {(α β)(α γ)(α δ)(β γ)(γ δ)(δ β)} (8.) τ 1 = (γδ)

Reference Évariste Galois Mémoire sur les conditions de résolubilité des équations par radicaux Internet Archive http://archive.org/details/uvresmathmatiqu00frangoog 00 pp33 50

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