Groupes, anneaux et arithmétique

A S A
1∈S06∈ S x, y ∈S xy ∈S
RA×S
(a, s)R(a0, s0)⇐⇒ as0−a0s= 0.
S−1ARa
s
(a, s)A×S
S−1A
π A S−1A A
π(A)
B
φ∈Hom(A;B)φ(S)⊂B∗φ∈Hom(S−1A;B)
φ◦π=φ
S=A\{0}S−1A
A F rac(A)
A=ZS={10k/k ∈N}S−1A
Z[X]/(3, X)Z/3Z
x n > 0xn= 0
N
u x
u= 1 + x
Q
x∈Nexp(x) = 1 + x
1+... +xn n!+...
uln(u)
exp ln N
U

A
A
a
<a>
<a>
a, b, c ∈A a a bc a
b c
a A u
α1, α2, ..., αrr≥
a=uα1...αr
r α1, α2, ..., αr
nZ/nZn
KφZ K
φ(1) = 1; φ(n) = n.1; φ(−n) = n.(−1) ∀n∈N∗.
φ
K
Kpnp
n
x2+ 2 = y3
Z[i√2]
Z[i√2] = {a+ib√2/(a, b)∈Z2}z N(z) = z×z
Z[i√2] Cz∈Z[i√2] N(z)
(z1, z2)∈Z[i√2]2N(z1.z2) = N(z1)N(z2)
zZ[i√2] N(z) = 1
Z[i√2]
Z[i√2]
z=x+i√2y x, y ∈Qγ
Z[i√2] N(z−γ)<1
α β Z[i√2] γ δ
Z[i√2]
α=β×γ+δ N(δ)< N(β)
Z[i√2]
i√2
x2+ 2 = y3
(x, y)

x+i√2x−i√2Z[i√2]
Z[i√2]
x+i√2Z[i√2]
n= 3 n= 4
Z[i√5] = {a+bi√5/(a, b)∈Z2}z N(z) = z.z
Z[i√5] C
xZ[i√5]
N(x)=1 Z[i√5]
2,3,1−i√5,1 + i√5Z[i√5]
Z
p a p ap−1≡1
(mod p)
a560 ≡1 (mod 561).
an−1a= 2 n
Mp= 2p−1
p
q Mp
2Z/qZ
q≡1 (mod 2p)
M17, M19 M23
1 + 2 + 3 = 6
n > 0
p n = 2p−1MpMp
pFp=Z/pZ.
F∗
p
F∗
pF∗ p
a∈F∗
pap−1
2≡1 (mod p)
a∈F∗
pap−1
2≡ −1 (mod p)
p≡1 (mod 4)
F∗
17
n n p
a n ≡a2(mod p)
n
p=
0,
1,
−1,
1
p−3 17 

n, n0nn0
p=n pn0 p
Z/n Zn≥(Z/n Z)∗
ϕ
n ϕ(n)q
n n ϕ(1) = 1
p α ϕ(pα) = pα1−1
p
∀n∈N∗, ϕ(n) = # (Z/nZ)∗
m, n
ϕ(m×n) = ϕ(m)×ϕ(n).
ϕ(n)n
G n ≥1G
G ω(G)
ω(G)n
x, y G p q
xy pq
x, y G p q
ppcm(p, q)
ω(G)
p s (Z/p Z)∗
(Z/p Z)∗p−1 (Z/p Z)∗
Xs−1
(Z/p Z)∗
(Z/13Z)∗
(Z/17Z)∗
p α
k
(p+ 1)pk≡1 + pk+1 (mod pk+2).
p+ 1 pα−1(Z/pαZ)∗.
(Z/pαZ)∗
(Z/2Z)∗(Z/4Z)∗
α
k
52k≡1+2k+2 (mod 2k+3)
5 (Z/2αZ)∗−1
5
(Z/2αZ)∗Z/2α−2Z×Z/2Z