◦
4
√5, j +√2,√3−3
√5,2
3 + √13,−2 + √2 + i√2
2, e2iπ
n.
Z[X]
1
2
k K k n K
k n b ∈K k
µb:K−→ K
x→bx.
norme b trace b
µbNK/k(b)T rK/k(b)PcarK/k (b)
NK/k(b)T rK/k(b)k PcarK/k (b)
k
P b ∈K µP(b)=P(µb)
µbb
b→T r(b)k T r(a) = na a ∈k
b→N(b)N(a) = ana∈k
L K m b ∈K NL/k(b) =
NK/k(b)mT rL/k(b) = mT rK/k(b)PcarL/k (b) = PcarK/k (b)m
k=QK=Q[√d]NK/k
x∈CfQ C f
f=Qn
i=1(X−xi)
xif
Q[X]
Z
x K =Q(x)T rK/Q(x)∈Z
NK/Q(x)∈ZPcarK/Q (x) = f(x)∈Z[X]
x=3
√2K=Q(3
√2) A=Z[x]B K
A=B
A⊂BZA
z=a+bx +cx2∈K a, b, c ∈Qz
T rK/Q(z), T rK/Q(xz)T rK/Q(x2z) 6B⊂A
z B a =a0/6, b =b0/6c=c0/6
a0, b0, c0∈ZPcar(z)a0, b0, c0∈6Z
x x
µxQ
(1, x, x2, . . . , xn−1)n x
x y x +y xy
i+j ij
A
f=Xn+an−1Xn−1+··· +a0∈A[X].
A C
C A n!
λ1, . . . , λn∈C f =
n
Q
i=1
(x−λi)
n
B=A[X]/(f)z∈B
NB/A(z)
z−h(x) = h(X)f x X B h ∈A[X]
S=
n
Q
i=1
h(Li)∈A[L1, . . . , Ln]Li
S=T(σ1, . . . , σn)σi
LiT A
−NB/A(z) = T(−an−1, an−2,...,(−1)na0)
A K f ∈A[X]
f=gh g h K[X]g h
A[X]f K
A R A P ∈R[X]
P A[X]A
P Q(Y) = Ym+F1Ym−1+···+FmFiA[X]
P1=P−Xrr P1Q(Y+Xr) = Ym+G1Ym−1+···+Gm
(−P1)(Pm−1
1+···+Gm−1) =GmR[X]r Gm=Q(Xr)
−P1Pm−1
1+···+Gm−1
P A
A R A n A0
A R A[X1+··· +Xn]
R[X1+···+Xn]A0[X1+···+Xn]A A[X1+···+Xn]
A B A
B A ⇔B A −
x x
x=a0+1
a1+1
a2+1
,
a0∈Zaix= [a0, a1, a2, . . .]
x
x= [a0, a1, . . . , an]
e= [2,1,2,1,1,4,1,1,6,1,1,8,1,1,10, . . .] = [2,1,2k, 1]k≥1
α[a0, a1, a2,···]
α0=α n ≥0an= [αn], αn+1 =1
αn−an
,
[y]y
n≥0pn
qn= [a0, . . . , an]pn∧qn= 1 pnqn
pn=anpn−1+pn−2pnqn−1−pn−1qn= (−1)n+1
qn=anqn−1+qn−2pnqn−2−pn−2qn= (−1)nan.
n≥0αn= [an, an+1, . . .]α= [a0, . . . , an−1, αn]
n
α=αn+1pn+pn−1
αn+1qn+qn−1
.
|α−pn
qn|<1
qnqn+1
<1
q2
n
.
pn
qnn α
α
α=1
2(√5−1) α
1
αα
3 + √2 2 + √6 2 + √15
5√15 ≈3,8
a, b ∈Na6= 0
√a2+b=a+b
2a+b
2a+b
.
√5√10 √17 √26 √37
α
α= [a0, a1, . . . , aN, aN+1, . . . , aN+T, aN+1,··· , aN+T, aN+1, . . .]=[a0, . . . , aN, aN+1, . . . , aN+T].
α
β= [aN+1, . . . , aN+T]β= [aN+1, . . . , aN+T, β].
β
α∈Q(β)α
α=√d d
α[a0, a1, . . . , an]n
an= 2a0
d x2−dy2= 1
(x1, y1)Q(√d)M
x1+y1√d(1,√d)
n(x2
1−dy2
1)n= 1 ( xn
yn) = Mn(x1
y1)
n
(x1, y1)
(x1, y1)
(p, q)|√d−p
q|<1
q2
p
q√d
p
q√d
T√d n n + 1
Tpn
qn√d
(pn, qn)
n
α=√d
1
αn+2 =1
α1=α−a0
(qn+1 −a0qn)√d+dqn=pn√d+ (pn+1 −a0pn)
p2
n−dq2
n= (−1)n+1
√d√d= [a0, a1, . . . , aT]
TpT−1
qT−1= [a0, . . . , aT−1]T
p2T−1
q2T−1= [a0, . . . , a2T−1] (pk, qk)
M
d= 3,5,6,13,17,19,34,37,53
√34 = [5,1,4,1,10] √53 = [7,3,1,1,3,14]
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