I. Soient F5 un corps à 5 éléments et le polynôme (1) Montrer que P

F55
P(X) = X5−X+ 1 ∈F5[X].
P(X)F5
Falg
5F5α∈Falg 5
α2= 2
F5(α) 2 F5
P(X)F5(α)
P(X)F5
F55K=F5(T)
Q(X) = X5−X+T∈K[X].
Q(X)K
Kalg K θ ∈Kalg
Q(X)
θ0Q(x)Kalg
θ−θ0∈F5K(θ)/K
Gal(K(θ)/K)→F5, σ 7→ σ(θ)−θ

P(X) = X4+ 2X2−2∈Q[X].
α=p√3−1β=ip√3+1
P(X)Q
C±α±β
Q(α, β) = Q(α, i√2)
αβ =i√2
Q(α)/Q(√3) Q(α, β)/Q(α)
2
Q(α, β)/Q8
HomQ(Q(√3),C)u1u2
u1(√3) = √3u2(√3) = −√3,
u1u2Q
Q(α)Cu1,1u1,2u2,1u2,2
u1,1(α) = α u1,2(α) = −α,
u2,1(α) = β u2,2(α) = −β,
HomQ(Q(α),C) = {ui,j / i, j = 1,2}
ua,b a, b = 1,2Q
Q(α, β) = Q(α, i√2) ua,b,c c= 1,2
ua,b,1(i√2) = i√2ua,b,2(i√2) = −i√2,
G:= Gal(Q(α, β)/Q) = {ua,b,c / a, b, c = 1,2}
Q(α, β) = Q(α+i√2)
σ=u2,2,2τ=u1,1,2
G= Gal(Q(α, β)/Q)
◦(σ) = 4 ◦(τ) = 2 τσ =σ3τ
G=< σ, τ >=Id, σ, σ2, σ3, τ, στ, σ2τ, σ3τ,
Q(α, β)<τ > =Q(α)Q(α, β)<σ> =Q(i√6)
Q(α, β)<στ > =Q(α−β)

pFpp K =Fp(T)
Kalg K
P(X) = Xp2−T Xp−T∈K[X],
α∈Kalg P(X)β, γ ∈Kalg
βp=T , γp(p−1) =T.
γp−1=β β =αp/(α+ 1)
P(X)K
P(X)Kalg
{α+λγ / λ ∈Fp}.
K(β)/K
K(γ)/K(β)p−1
K(αp)/K p
K(α)/K(αp)p
K(α, γ)K
p2(p−1) p(p−1)
G= Gal(K(α, γ)/K)
(K(α, γ))G=K(β).
Fp×F×
p
(λ, µ)·(λ0, µ0) = (λ+µλ0, µµ0),
Fp×F×
p
G→Fp×F×
p, σ 7→ γ−1(σ(α)−α), γ−1σ(γ)