Corrigé de l`exercice 3.4, questions 3 et 4

Q(√2 + √3) = Q(√2,√3) [Q(√2 + √3) : Q]
Q(√2 + √3) = Q(√2,√3)
Q⊂Q(√2)
[Q(√2 + √3) : Q]=[Q(√2,√3) : Q] = [Q(√2,√3) : Q(√2)] [Q(√2) : Q] = 2 [Q(√2,√3) : Q(√2)].
[Q(√2,√3) : Q(√2)] β=√3
X2−3Q Q(√2)
Q(√2)[X]
2X2−3
Q(√2) ∆ = 12 = 3.22Q(√2)
3Q(√2) 3 a+b√2a, b ∈Q
a2+ 2b2+ 2ab√2=3 a2+ 2b2= 3 ab = 0
2Q
X2−3Q(√2)[X] [Q(√2,√3) : Q(√2)] = 2
[Q(√2 + √3) : Q]=4
√2 + √3Q(Z/pZ)[X]
p∈N
α=√2 + √3α−√2 = √3
α2−2√2α+ 2 = 3.
α2−1 = 2√2α α4−2α2+ 1 = 8α2α
P=X4−10X2+ 1 Q⊂Q(√2 + √3)
4fα,QαQ4P fα,Q
P p p p = 2 P=X4+ 1 =
(X+ 1)4p6= 2
F×
p2 3 6
2 2 = a2a∈Fp
P
α
α2−1 = 2√2α X2−2√2X−1αQ(α)
Q
K02 3
α0=√2 + √3K0Fp
2 = a2FpX2−2aX −1
P
X4−10X2+ 1 = (X2−2aX −1)(X2+ 2aX −1).

3 3 = b2a∈Fp
P√3
√2α−√3 = √2α2−2√3α+ 3 = 2
X2−2√3X+ 1 αFp3 = b2
X2−2bX + 1
X4−10X2+ 1 = (X2−2bX + 1)(X2+ 2bX + 1).
Fp6 = c2c∈Fp
α=√2 + √3√6
α2= 2 + 3 + 2√6.
Fp6 = c2X2−5−2c
X4−10X2+ 1 = (X2−5−2c)(X2−5+2c).
X4−10X2+ 1 p