Énoncé Partie I. Le théorème de Liouville Partie II. Le théorème de l

α∈Cα P ∈Q[X]
P(α) = 0
Q
α
P∈Z[X]P(α)=0
P d
Mα= sup
[α−1,α+1] |P0|
(ak)k≥1ak∈ {0,1,··· ,9}
m n m > n
m
X
k=n+1
10−k!<10
910−(n+1)!
(sn)n≥1sn=
n
X
k=1
ak10−k!s
s
r
|r−α| ≥ Cα
qd
Cα= min(1,1
Mα)r=p qp∈Zq∈N∗qdPp q
s
sm−snn
α p
(1, α, ··· , αp)
α d p
(1, α, ··· , αp)d α
Q[α] = Vect(1, α, ···αd−1)
(1, α, ··· , αd−1)Q[α]
αn∈Q[α]n
α−1∈Q[α]
P∈Q[X]P(α) = 0
α p β q
Q[α, β]αmβnm
p−1n q −1
∀l∈N: (α+β)l∈Q[α, β],(αβ)l∈Q[α, β]
α+β αβ
C
P Q s
QsQ s −X X Q
Qs=b Q(s−X) = Q◦(s−X)
P=X2+X+ 1 Q=X2−2
s P Qs
P Qs
C(s) = 0 C C
Q[X]j+√2
Q[X]√3+√2

H
H
n(z1, z2,··· , zn)
A=
n
Y
i=1
(X−zi)SA=Y
(i,j)∈{1,··· ,n}2
(X−zi−zj)
SAA SA
H
R[X]
R[X]
A SAA
A A SA
α P1∈Q[X]˜ P1(α) = 0
X−r r
P1P2Q
α
P2P
P2Z
1
10
m
X
k=n+1
10−k!<10−(n+1)! + 10−(n+1)!−1+ 10−(n+1)!−2+··· + 10−m!
<10−(n+1)! 1−(1
10 )m!−(n+1)!
1−1
10
<10−(n+1)! 1
1−1
10
=10
910−(n+1)!
ak
720 = 6!
P α r P
d
qd˜ P(p
q)∈Z
r=p
q∈[α−1, α + 1]
|P(r)|=|P(r)−P(α)|≤|r−α|Mα
|r−α| ≥ 1
Mα|P(r)| ≥ 1 Mαqd|qdP(p q)| ≥ 1 Mαqd

qdP(p
q)P
|r−α| ≥ 1
|r−α| ≥ 1≥1
qd
|r−α| ≥ 1≥Cα
qd
Cα= min(1,1
Mα)
s
0< sm−sn<10
910−(n+1)!
m n
0< s −sn≤10
910−(n+1)!
s
Cα
10−d n!≤10 910−(n+1)! ⇒10(n+1)!−d n!≤10 9Cα
(n+ 1)! −d n!→+∞
p(1, α, α2,··· , αp)Q
Ra0,··· , ap
a01 + a1α+a2α2+··· +apαp= 0
P=a0+a1X+··· +apXp
α α
(1, α, α2,··· , αd−1)Q[α]
d p (1, α, α2,··· , αp)
Q[α]
(1, α, α2,··· , αd) (1, α, α2,··· , αd−1)
(a0,··· , ap)6= (0,··· ,0)
a01 + a1α+a2α2+··· +adαd= 0
ad
(a0,··· , ad−1)6= (0,··· ,0) a01 + a1α+a2α2+··· +ad−1αd−1= 0
(1, α, α2,··· , αd−1)
αdα
α
αd=−a0
ad− ··· − ad−1 ad
αd−1
⇒αd+1 =−a0
ad
α− ··· − ad−2
ad
αd−1−ad−1
ad
αd∈Vect(1, α, α2,··· , αd−1)
α αn∈Q[α]
n
α−1∈Q[α] (a0,··· , ad)6=
(0,··· ,0)
a0+a1α+··· +adαd= 0
a06= 0 α
a0= 0 ⇒αa1+··· +adαd−1= 0 ⇒a1+··· +adαd−1= 0
(a1,··· , ad)6= (0,··· ,0) d
(1, α, α2,··· , αp)
a06= 0 α−1
1 + αa1
a0
+··· +ad
a0
αd−1= 0 ⇒α−1=−a1
a0− ··· − ad a0
αd−1∈Q[X]
P d α
d
Q R
d P =QR α
d p (1, α, α2,··· , αp)

P d α
Q α
P−Q d
d
αmβnm n
(αβ)l(α+β)n
Q
αm∈Q[α]) αii0p−1
βm∈Q[β]) βjj0q−1
(α+β)l(αβ)lQ[α, β]
Q Q[α, β]
p
(1,(α+β),··· ,(α+β)p)
Q[α, β]p+ 1 >dim Q[α, β]
α+β αβ
α+β
α β
0
P Qs
(1 + 2s)X+ 3 −s21+2s6= 0
7−2s−s2+ 2s3−s4= 0
1 + 2s6= 0 −1
2s
C(s)=0 P Qs
Cu∈CP
QsQs(u) = Q(s−u)s−u v Q
Q(s) = 0 u P v Q
s=u+v s u v
P Q s −v=u P P (u) = 0 Qs
Qs(s−v) = Q(s−(s−v)) = Q(s) = 0
C P Q
C
j P √2
Q j +√2
R= 7 −2X−X2+ 2X3−X4

P Q R
R
j+√2j−√2j2+√2j2−√2
RQ[X]
R
Q[X]
R T ∈Q[X]
2X T R
√3 + √2
√3 + √2
X4−10X2+ 1
SAn2A
SA
HA=P Q
P Q P Q
A
z P P (z)>0
P
R[X]
1 2
P
P1z∈RP=X−z P z ∈ H
z P
P2z∈C
P= (X−z)(X−z) = X2−2 Re(z)X+|z|2
|z| ≥ 0 Re(z)≤0z∈ H
A SAz A
z+z= 2 Re z SA2 Re z < 0z∈ H
A
A
A
SA
A
A∈R[X]SAA
A SA
A A
A B =AA A B
B