Factorisation des polynômes sur les corps finis 1 L`algorithme de

K=FqP∈K[X]P
P P
P
P0= 0 Q P (X) = Q(Xp)R
Q F
K P (X) = (R(X))pR
P∧P06= 1 P P ∧P0
P A =K[X]/P K
n K F −Id r
N= ker(F−Id) r≥1K⊂N r = 1 P
r≥2N\K Q ∈K[X]
α∈K P Q −α
α
P=X9+X6−X+ 1 K=F3
K K
3
P0=−1
P∧P0= 1 P0=−1
A=K[X]/(X9+X6−X+ 1) .
K9{1, X, X2, . . . , X8}
F−Id X3X24
X10 X11
X9=−X6+X−1
X10 =−X7+X2−X
X11 =−X8+X3−X2
X12 =−(−X6+X−1) + X4−X3=X6+X4−X3−X+ 1
X15 = (−X6+X−1) + X7−X6−X4+X3=X7+X6−X4+X3+X−1
X18 = (−X7+X2−X)+(−X6+X−1) −X7+X6+X4−X3=X7+X4−X3+X2−1
X21 = (−X7+X2−X) + X7−X6+X5−X3=−X6+X5−X3+X2−X
X24 =−(−X6+X−1) + X8−X6+X5−X4=X8+X5−X4−X+ 1

F−Id
000−1 1 −1−101
0−101−110−1−1
0 0 −1000110
010−1−1 1 −1−1 0
00000−110−1
00000−1011
001−111−1−1 0
0000011−1 0
000000000
1
X
X2
X3
X4
X5
X6
X7
X8
(0,1,−1,−1,1,1,−1,0,1)
Q(X) = X8−X6+X5+X4−X3−X2+X
P∧(Q−α)α α = 0 1 2
X9+X6−X+ 1 = (X8−X6+X5+X4−X3−X2+X−α)X+ (X7−X5+X4+X3−X2+ (α−1)X+ 1)
X8−X6+X5+X4−X3−X2+X−α= (X7−X5+X4+X3−X2+ (α−1)X+ 1)X−α(X2+ 1)
α= 0 X7−X5+X4+X3−X2−X+ 1 P
P(X) = (X7−X5+X4+X3−X2−X+ 1)(X2+ 1)
α6= 0
X2+ 1 P
X2+ 1 F3
P1=X7−X5+X4+X3−X2−X+ 1
X7=X5−X4−X3+X2+X−1
X9=−X6+X−1
X12 =X6+X4−X3−X+ 1
X15 =X6+X5+X4+X2−X+ 1
X18 =X5+X3−X2+X+ 1
F−Id
000−1111
0−101−1−1 1
0 0 −1001−1
010−1−101
0000010
0000001
001−111−1
1
X
X2
X3
X4
X5
X6
1P
X9+X6−X+ 1 (X7−X5+X4+X3−X2−X+ 1)(X2+ 1)

2 3
K P ∈K[X]n P
≤n/2
≤3
F3
2X2+ 1 X2+X−1X2−X−1
2
2X X2+aX =
(X+1
2a)2+. . . P P (X+a)
2X X2+ 1
X+a a ∈F3X X2+X−1X2−X−1
3X3−X
F3P3
P−(X3−X)≤2
aQ a ∈F×
3Q1
2Q= 1 Q2
P=X3−X+aQ 3
X3−X+ 1 ,(X3−X)+(X2+X−1) = X3+X2−1,
X3−X−1,(X3−X)−(X2+X−1) = X3−X2+X+ 1 ,
(X3−X)+(X2+ 1) ,(X3−X)+(X2−X−1) = X3+X2+X−1,
(X3−X)−(X2+ 1) ,(X3−X)−(X2−X−1) = X3−X2+ 1 .
card I(n, q) = 1
nX
d|n
µ(d)qn/d
card I(2,3) = 1×32+(−1)×3
2= 3 card I(3,3) = 1×33+(−1)×3 3= 8