Corrigé du contrôle 4 - Jean

n∈N∗P=anXn+an−1Xn−1+· · · +a1X+a0n
p
p an
∀k∈ {0,1, . . . , n −1}p ak
p2a0
PQ[X]
P1= 2X4−3X3+ 12X2+ 9X−6.
p= 3
32a0= 6
P1Q[X]
P2=X10 −5.
p= 5
Q[X]
P3=X2−4X+ 4.
p= 2 22a0= 4
P3= (X−2)2Q[X]
pQ[X]
Xp−1X−1
P=Xp−1
X−1Xp−1X−1
PQ[X]
Xp−1=(X−1)(Xp−1+Xp−2+· · · +X+ 1) P
p−1
Xp−1p
P
p P
P=
p−1
Y
k=1
(X−e
2ikπ
p).
QQ[X]P X + 1 Q(X) = P(X+ 1)
Q=P(X+ 1) = (X+1)p−1
(X+1)−1
(X+ 1)p=Pp
k=0 p
kXkQ=Pp
k=1 p
kXk−1=Xp−1+p p−1Xp−2+· · · p 2X+
p
1

p
Q ap−1= 1 p2a0=p
1=p
Q
PQ[X]
R S Q[X]P=RS Q =P(X+ 1) = R(X+ 1)S(X+ 1)
R(X+ 1) S(X+ 1) Q
Q P
Q[X]
M z
z
Q[X] 2kk∈N
p
e
2iπ
p
p
p2π
p
p
r r(e
2iπ
p−1)
e
2iπ
p
e
2iπ
pP P
p−1Q[X]
e
2iπ
p
P2kk
p−1 2kp2k+ 1
n n ≥3n
n
n= 2jp1p2· · · pm,
j m pi
pi= 2ki+ 1 ki∈N
n
n3,4,5,6,8,10,12,15,16,17

u v R2u v
C={nu +mv ∈R2|n∈Z, m ∈Z}.
θRRθO θ R
Rθ(R,◦)
Id
G={Rθ∈ R | ∀w∈C, Rθ(w)∈C}
θ θ0RRθRθ0G Rθ◦Rθ0
G w ∈C Rθ0(w)∈C Rθ∈G Rθ(Rθ0(w)) ∈C
Rθ◦Rθ0(w)∈C C
Rθ◦Rθ0∈G G
w∈C Id(w) = w∈C Id ∈G
G
(u, v)C
u= (1,0)
v= (0,1) u= (2,0) v= (0,1) u= (1,0)
v= (1
2,√3 2)
G={Id, Rπ
2, Rπ, R3π
2}
G2={Id, Rπ}
G3={Id, Rπ
3, R2π
3, Rπ, R4π 3, R5π 3}
G
u v C θ ∈R
Rθ∈G
Rθ
Mθ=cos(θ)−sin(θ)
sin(θ) cos(θ)∈ M2(R).
R2
w=w1
w2R2Rθ(w) = Mθw

u=u1
u2v=v1 v2
MθC u v
MθC n, m, p q
Mθu=nu +mv Mθv=pu +qv
P=u1v1
u2v2
MθP Mθu Mθv
MθP=nu1+mv1pu1+qv1
nu2+mv2pu2+qv2
u v
u1v2−u2v16= 0 Q=1
u1v2−u2v1v2−v1
−u2u1P Q
I2P Q
N=P−1MθP N =n p
m q
A=a b
c d∈ M2(R)ATr(A) = a+d
A=a b
c dB=a0b0 c0d0M2(R)AB =aa0+bc0ab0+bd0 ca0+dc0cb0+dd0
Tr(AB) = aa0+bc0+cb0+dd0BA =a0a+b0c a0b+b0d
c0a+d0c c0b+d0dTr(BA) =
a0a+b0c+c0b+d0dTr(AB) = Tr(BA)
A=P−1B=MθPTr(P−1MθP) =
Tr(MθP P −1) Tr(N) = Tr(Mθ)
Mθ2 cos(θ)N n +q
2 cos(θ)
cos(θ) 2 cos(θ)
cos(θ)±1
2±1θ2π
±π
2±π 3±2π 3π θ
Rθ
θ
Rπ
2Rπ 3
Rπ
2◦Rπ 3=R5π
6
{Id},{Id, Rπ},{Id, R2π
3, R4π 3},{Id, Rπ
2, Rπ, R3π
2},{Id, Rπ
3, R2π
3, Rπ, R4π 3, R5π 3}.