Sommes de deux carrés et entiers de Gauss. Corrigé.

Q⊂N
Q={a2+b2|a, b ∈N}
n∈Q n 6≡ 3 mod 4
a∈Za2≡0 1 mod 4 a≡0,1,2−1 mod 4 02= 0
(±1)2= 1 22= 4 ≡0 mod 4 n=a2+b2n≡0 + 0 0 + 1 1 + 0 1 + 1 mod 4
Z[i]Q
m=a2+b2n=c2+d2z=a+ib w =c+id zw =e+if e =ac −bd
f=ad +bc m =z¯z n =w¯w mn = (z¯z)(w¯w)=(zw)(zw) = e2+f2=
(ac −bd)2+ (ad +bc)2
p p 6≡ 3 mod 4
p= 2 p≡1 mod 4
Z[i]d(a+bi) = a2+b2Z[i]
Z[i]
p∈Z Z 5Z[i] 5 = (1+2i)(1−2i)
p∈Z Z[i]pZ
pZp=nm m > 1n > 1
m=m+ 0i∈Z[i]n=n+ 0i∈Z[i]mZ[i]
n x ∈Z[i]mx = 1 |x|= 1/|m|<1
pZpZ[i]p
p=ππ π Z[i]π p Z[i]
ππ p
p
pZ[i]
pZ[i]π p
Z[i]|π|>1z∈Z[i]p=zπ = ¯z¯π¯π p ¯π
π=xy ¯π= ¯x¯y π¯π p π¯π=p p
π¯π∈Zπ¯π=|π|2>1π¯π p
⇐⇒
=⇒p=a2+b2p=π¯π π =a+ib π |π|= 1
|¯π|= 1 p=|π¯π|= 1
=⇒pZ[i]p=π¯π π ∈Z[i]π=a+ib
p=a2+b2

Z[i]/(p)'Fp[X]/(X2+ 1) Fp=Z/pZ
pZ[i]X2+ 1 Fp[X]
p
pZ[i]
−1Fp
ϕ:Z[i]/(p)→Fp[X]/(X2+ 1) z
Z[i]/(p)a+ib ϕ(z)a+bX a b
Fp
ϕ a0+ib0z(a−a0)+ i(b−b0)∈(p)
a0+ib0=a+ib + (c+id)p= (a+cp) + i(b+dp)a0+b0X= (a+cp) + (b+dp)X=a+bX
a=a+cp b =b+dp Fp
ϕ z =a+ib w =c+id
ϕ(z+w) = ϕ(z) + ϕ(w)ϕ(z)ϕ(w) = (a+bX)(c+dX) = (ac +bdX2) + (ad +bc)X=
(ac −bd)+(ad +bc)X=ϕ(ac −bd +i(ad +bc)) = ϕ(zw)
ker ϕ= 0 zker ϕ a +ib ϕ(z) = a+bX = 0
Fp[X]/(X2+ 1) a=b= 0 Fpa≡b≡0 mod p a +ib ∈(p)
ϕ=Fp[X]/(X2+ 1) Fp[X]/(X2+ 1) a+bX
0≤a, b < p ϕ(a+ib)
¡pZ[i]¢⇐⇒ ¡Z[i]/(p)¢⇐⇒ ¡Fp[X]/(X2+ 1) ¢⇐⇒
¡X2+ 1 Fp[X]¢
⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒
G n m G
gm= 1 g∈G
G(Z/m1Z)× · · · × (Z/msZ)
m1|m2|. . . |ms
m n m =n G
F∗
p
Xm−1G n
G= (Z/m1Z)×· · ·×(Z/msZ)
n=d1. . . dsm=dsm=n s = 1 d1=m=n G =Z/nZ
mF∗
pF∗ p
Xm−1|F∗
p| ≤ m
m≤ |F∗
p|m=|F∗ p|F∗ p
−1
p= 3,5,7

p−1FpF∗
p
4−1 2 F∗
p
F∗
pp−1−1
p p = 2 p≡1 mod 4
p
pZ[i]
−1Fp
p= 2 p≡1 mod 4
−1 = 22mod 5 −1
02= 0 (±1)2= 1 02= 0 (±1)2= 1 (±2)2= 4 (±3)2= 9 ≡2
−1F∗
pX2−1
deg(X2−1) = 2 ±1
x4x2x2=−1
x x2=−1x
p≡1 mod 4 k= (p−1)/4aF∗
p(a2k)2=
a4k=ap−1= 1 a2k6= 1 a2k=−1
b=akb2=a2k=−1
p6≡ 1 mod 4 −1p
⇐⇒ ⇐⇒ ⇐⇒
n∈Nn∈Q
Zn=pα1
1. . . pαk
k
3 mod 4
4410 = 2 ×9×5×49 924 =
4×3×7×11
P±p p ≡ ±1 mod 4
n∈Q n =z¯z z ∈Z[i]z=pα1
1. . . pαk
kπβ1
1. . . πβl
l
nZ[i]p1, . . . , pk∈Zπ1, . . . , πl6∈ Zqj=πj¯πj
p1, . . . , pk∈P−q1, . . . , ql∈P+∪ {2} ⇔
nZn=z¯z=p2α1
1. . . p2αk
kqβ1
1. . . qβl
l
P−
n=p2α1
1. . . p2αk
kqβ1
1. . . qβl
lqjqj=πj¯πjπj∈Z[i]
n=z¯z z =pα1
1. . . pαk
kπβ1
1. . . πβl
ln∈Q
4410 = 2 ×9×5×49 = (1 + i)(1 −i)×32×(2 + i)(2 −i)×72=¡3×7×(1 + i)(2 + i)¢¡3×
7×(1 −i)(2 −i)¢=¡21 ×(1 + 3i)¢¡21 ×(1 −3i)¢= (21 + 63i)(21 −63i) = 212+ 632.
924 = 4 ×3×7×11 6∈ Q