Examen - Université Paris-Sud

3
a R n >1
1A=a N =n r = 1
2N > 0
•N r =r×A
•A=A×A
•N=N2
3r
(a, n)
C Z
n>1Z/nZn(Z/nZ)×
Z/nZϕ
n>1 (Z/nZ)×p
Z/pZ Fp
p n n
p
n
p=
0p|n ,
1n p ,
−1n p .
n
p≡np−1
2(mod p)
p
pZ
AZ[i] = {a+ib;a, b ∈Z} ⊂ CN:A→N
i j j 6=i

p=a2+b2a b Zp≡1(mod 4)
N(zz0) = N(z)N(z0)A1−1i−i
A N
(a, b)∈A2(q, r)∈A2a=bq +r06N(r)< N(b)
z t A a
z/t b q =a+ib
|z/t −q|<1
r=z−qt q r
z t A N
p1 4
pZ
(p) := {pz;z∈A}p A (X2+ 1)
X2+1 Fp[X]π:Z→Fpp
a+ib 7→ π(a) + π(b)X A/(p)
Fp[X]/(X2+ 1)
p≡1(mod4) Fp[X]/(X2+ 1)
x y z A pz =xy p x y A
p A A
λ λ p A
µ∈A p =λµ
N(λ) = p p Z
N p =λµ
p
x A N(x)Zx
A
a b p =a2+b2p= (a+ib)(a−ib)
p A
p
n>3n=
pα1
1· · · pαr
ra∈Za n
a
n:= a p1α1
· · · a
prαr
.
mm0
n=m nm0 n,m nn0=m nm n0,
(m, m0)∈Z2(n, n0)
3

n>3
a∈Za
n6≡ an−1
2(mod n)
a
n≡an−1
2(mod n)a∈Z
n=p
n>3k>1
1i:= 1
2i6k a ∈ {1, . . . , n −1}
•a
n0n
•a
na(n−1)/2n
n
•i:= i+ 1
3n
n>3a∈ {1, . . . , n −
1}a∈ {1, . . . , n−1}
n2
n
G=na∈(Z/nZ)×;a
n≡an−1
2(mod n)o.
G(Z/nZ)×e G (Z/nZ)×
e ϕ(n)G
e G 6= (Z/nZ)×
a∈(Z/nZ)×n
G6= (Z/nZ)×n
G= (Z/nZ)×
an−1
2≡ ±1(mod n)a n
a n b n
r>3s>3
n=rs , b ≡1(mod r), b ≡a(mod s).
an−1
2≡1(mod n)a n
4 4
4
p q p
a0
n=pq , a0
pq =−1.
α p n
α
p=−1
n>3
1−2−k