Corrigé de l`examen d`Algèbre 2

1e
b=Y2+rR[Y]r > 0
B[X]B=R[Y]
X2+b b B
X2+Y2+rR[X, Y ]R[X, Y ]
X2+Y2+r A
P∈R[X, Y ]Y2+X2+r Y
R[X, Y ]P P = (Y2+X2+r)U+V
V Y
P A V (x, y) = f(x) + yg(x)f g R[X]
f(x)+yg(x) = 0
A f g 2
f(X) + Y g(X)Y2+X2+r
R[X, Y ]Y f =g= 0
mA6= (0) π−1(m)R[X, Y ]
(X2+Y2+r)
A I A/I A
I A π−1(m) (a)a
X2+Y2+r
a a 2
π−1(m) = (X2+Y2+r)R[X, Y ]
π−1(m)
R[X, Y ] (X2+Y2+r)
A
mA
mAm=π(π−1(m))
π−1(m)mπ(X2+
Y2+r) = x2+y2+r(= 0) m
m= (x−α, y −β)x2+y2+r≡α2+β2+r(mod m)
x≡α(mod m)y≡β(mod m)x2+y2+r= 0 A α2+β2+r∈R∗
m= (x−α, P (y)) Y2+α2+r∈π−1(m)
y2+x2+r−(x−α)(x+α)my2+x2+r= 0
(x−α)∈mP Y 2
π−1(m)Y2+α2+r
R[Y]
UP +V(Y2+α2+r) = 1 1 π−1(m)
m= (x−α, y2+α2+r)y2+α2+r=y2+x2+r−(x−α)(x+α)∈
(x−α)y2+x2+r= 0 m= (x−α)

A
(P(x), y −αx −β)P X 2α
β
P X2+ (αX +
β)2+r2
π−1(m)
y2+x2+r−(y−αx −β)(y+αx +β) = −(y−αx −β)(y+αx +β)∈m
P
R[X]π−1(m) 1
m=π(π−1(m)) (x2+ (αx +β)2+r, (y−αx −β)
x2+ (αx +β)2+r∈(y−αx −β)m= (y−αx −β)
a∈E a u(a)u
GL(E)H
u∈H a ∈E
u(a)
a∈E a u(a)
u
e1λ1e2λ16=λ2
e1+e2
E≥3
H1H
a u(a)a u(a)
e1, . . . , en−2(e1, . . . , en−2, a, u(a))
H1=< e1, . . . , en−2, a > a ∈H1
τ H1ka
k
φ=τuτ−1u−1H u ∈H H τuτ−1
u−1∈H φ
1φ
φ(u(a)) = τuτ−1(a) = τ(u(a)) τ−1
H1A∈H1τ−1(a) = a τ u(a)
τ(u(a)) 6=u(a)τ(u(a)) = u(a) + λa
λ6= 0 u(a)a u(a)
H∩SL(E)φ
H∩SL(E) SL(E)H
G H ∩K K K G
PSL(E)n≥3H∩SL(E) = H
H⊃SL(E)

d´et : G−→ k∗SL(E)
HSL(E)H= d´et−1(d´et(H))
d´et−1(d´et(H)) ⊂H
u∈d´et−1(d´et(H)) v H d´et(u) = d´et(v)
v−1u
1H u =v(v−1u)∈H K = d´et(H)
K k∗H= d´et−1(K)
k∗
S G G S
G/S
G−→ G/S
A(H) GL(E)u
v∈A(H)u−1uv A(H)
τ
H τ 6= id τ|H= idHd´et(τ) = 1
T(H)
T(H)H
A(H)
λ∈k∗
en/∈H u u(x) = x x ∈H u(en) = λen
n−1
H en
λ u
1λ
T(H)A(H)
k∗
f E k H en/∈H
e1, . . . , en−1H f(ei) = 0 i < n
f(en)=1 τ H
a∈H τ(x) = x+f(x)a a
f τ(en)−enf
φ:T(H)−→ H τ a φ
τ7→ τ(en)−enf
τ= id 0 τ
∀x∈E τ(x) = x+f(x)φ(τ)φ
u v ∈T(H)φ(u) = a φ(v) = b uv(en) = u(en+b)
v(x) = x+f(x)φ(v)u(v(en)) = en+a+u(b) = en+a+b b ∈H
u(b) = b φ(uv) = φ(u) + φ(v)φ
a= 0 a∈H
τ(f, a)τ(x) = x+f(x)a a

φ A(H)T(H)
H
B(H)
A(H)τ∈T(H)τ6= id u∈B(H)τ
H uτu−1
u(H)u(H) = H uτu−1∈T(H)
T(H)B(H)v∈A(H)
x∈H uvu−1(x) = u(v(u−1(x)))
u∈B(H)u−1(x)∈H u−1B(H)
v∈A(H)v(u−1(x)) = u−1(x)
uvu−1(x) = x x ∈H uvu−1∈A(H)A(H)
B(H)
φ:B(H)−→ k∗×GL(H)u∈B(H)
u(H) = H u|HH
Ker(φ) = {u∈B(H) d´et(u) = 1 u|H= idH}T(H)
T(H)
(λ, v)∈k∗×GL(H)en/∈H E =H⊕ken
u u(en) = λ/d´et(v).enu(x) =
v(x)x∈H u ∈B(H)x∈H u(x) = v(x)∈
Hd´et(u) = d´et(v).λ/d´et(v) = λ6= 0
φ(u) = (λ, v)φ
(e1, . . . , en−1)H en/∈HB= (e1, . . . , en)
E D(H) = {u∈GL(E)u(H) = H
enu}u(en) = λen
λ∈k∗0
D(H)B(H)enu
v uv u−1(λ, v)u
D(H)φ D(H)
u∈D(H)
φ(u) = (d´et(u|H)λ, u|H)λ en
uB(1,idH)
u|H= idH1λ= 1
u|H= idHu(en) = enu= id φ D(H)