Théorème des deux carrés

n∈N
n=a2+b2a, b ∈N
Σ = {n∈N/∃a, b ∈Nn=a2+b2}.
Z[i] = {a+ib ∈C/ a, b ∈Z}
N(a+ib) = |z|2=a2+b2
p∈N
p∈Σ⇐⇒ p= 2 p≡1 [4].
p∈Σ⇐⇒ pZ[i].
p∈Σp=a2+b2= (a+ib)(a−ib)p
a6= 0 b6= 0 a+ib a −ib Z[i]∗=
{1,−1, i, −i}pZ[i]
p=zz0z, z06∈ Z[i]∗={1,−1, i, −i}N(p) =
N(z)N(z0) = p2N(z), N(z0)6= 1 p=N(z) = N(z0)
p=N(z) = N(a+ib) = a2+b2∈Σ
Z[i]
pZ[i]⇐⇒ (p)
⇐⇒ Z[i]/(p).
Z[i]∼
=Z[X]/(X2+ 1) ϕ:Z[X]→Z[i]
ϕ(X) = i ϕ P ∈Z[X]
ϕ(P) = 0 P X2+1 P= (X2+1)Q+R
ϕ(P)=0 R(i)=0 R61 (1, i)
R= 0 Ker(ϕ)=(X2+ 1) Z[X]/Ker(ϕ)∼
=Z[i]
Z[i]/(p)∼
=Z[X]/(X2+ 1)/(p)
∼
=Z[X]/(X2+ 1, p)
∼
=Z[X]/(p)/(X2+ 1)
∼
=(Z/pZ)[X]/(X2+ 1) = Fp[X]/(X2+ 1).

Z[i]/(p)⇐⇒ Fp[X]/(X2+ 1)
⇐⇒ (X2+ 1)
⇐⇒ X2+ 1 Fp,
Fp
Fp[X]
X2+ 1 Fp⇐⇒ X2+ 1 Fp
⇐⇒ −1F∗
p
⇐⇒ p= 2 p≡1 [4].
p6= 2 q=p
q=prr>1
ϕ:F∗
q−→ F∗2 q
x7−→ x2
F∗2
q={x∈F∗ q/∃y∈F∗ q, x =y2} {−1,1}
X2−1−1 1
Im(ϕ) = F∗2
q∼
=F∗
q/Ker(ϕ) Card(F∗2 q) = q−1
2
F∗2
q=nx∈Fq/ xq−1
2= 1o.
X6q−1
2
Xq−1
2−1x∈F∗2
qx=y2
xq−1
2=yq−1= 1 F∗2
q⊂X
−1∈F∗2
p⇔(−1)p−1
2= 1 ⇔p−1
2⇔p≡1 [4].
n∈N∗n6= 1 n
n=Qp∈P pvp(n)
n∈Σ⇐⇒ vp(n)p≡3 [4].
n∈Σ
n∈Σ⇐⇒ ∃z∈Z[i]n=N(z).

N n, n0∈Σn=N(z)n0=N(z0)
nn0=N(zz0)∈Σ
n
pipi∈Σpi= 2 pi≡1 [4] pi≡3 [4] p2
i∈Σ
Σp2
i=p2 i+ 0
p≡3 [4] vp(n)
vp(n)vp(n)=0 p n =a2+b2= (a+ib)(a−ib)
p6∈ Σp≡3 [4] p
Z[i]p(a+ib) (a−ib)p p
a b a =pa0b=pb0n
p2=a02+b02∈Σvpn p2=vp(n)−2
vp(n)
f:A→B
I= Ker(f)J⊂I A p :A→A/J
f:A/J →B f =f◦p
f J =I
f f
Im(f)∼
=A/Ker(f)
FpFpFp
1Fp{0}1
x6= 0 x−1x= 1 FpFp{0}= (0)
Fp= (1) Fp
Z[i]∗={−1,1,−i, i}
z∈Z[i]∗z0∈Z[i]zz0= 1 N(zz0) =
N(z)N(z0) = N(1) = 1 N(z), N(z0)∈NN(z) = N(z0)=1
z=a+ib a2+b2= 1
Z[i]N
z, t ∈Z[i]\{0}z t
z
t∈Cz tq
z
t=x+iy q =a+ib a b x y
z
t−q=p(x−a)2+ (y−b)26s1 22
+1
22
=√2
2<1.

r=z−qt ∈Z[i]|r|=|t|z
t−q<|t|
N(r)< N(t)z=qt +r N(r)< N(t)
Z[i]/(p)∼
=(Z/pZ)[X]/(X2+ 1).
ϕ:Z[i]−→ (Z/pZ)[X]/(X2+ 1)
a+ib 7−→ a+bΠ(X)
Π : (Z/pZ)[X]→(Z/pZ)[X]/(X2+ 1)
ϕ(p) = pZ[i]
A
(p)p
A p
(p)
(p)p=ab ab ∈(p)
(p)a∈(p)b∈(p)
a∈(p)p a a =pc c ∈A p =ab =pbc A
bc = 1 i.e. b ∈A∗p
p ab ∈(p)k∈A ab =kp
A a, b k
ab kp p a b a ∈(p)b∈(p)
(p)