Exercices sur les extensions de corps, chapitre 3 indications de

g=Y4+Y3+Y2+Y+ 1
F2[Y]Y2+Y+ 1 F2
2
1Y2+aY + 1 a= 0 Y2+ 1 = (Y+ 1)2
a= 1 Y2+Y+ 1 ≤3
g=ij i, j i j (1,3)
(2,2) gF2
(2,2) i, j i =j=Y2+Y+ 1 g6= (Y2+Y+ 1)2
gF2[Y]/(g)K24= 16
α=Y α5−1=(α−1)(α4+α3+α2+α+ 1) = 0 α5= 1 α5
K∗K∗15
K∗
α15 α
α5= 1 α−1=α4=−(α3+α2+α+ 1) = α3+α2+α+ 1
F2−1 = +1
α−1α5α−1
K∗
F2(α+ 1)3=α3+ 3α2+ 3α+ 1 = α3+α2+α+ 1
α+ 1 α−115 K∗
f=X4+X+ 1 K K
α d ≤3f
fF2d= 0
r+sα r, s ∈F2d= 1
r+sα +tα2
r4+s4α4+t4α8+r+sα +tα2+ 1 = 0.
α8=α3r4=r s4=s t4=t x ∈F2
xnn≥1r=s=t= 1 z= 1 + α+α2
f F :K→K, x 7→ x2K
F(z) = z2F2(z) = z4F3(z) = z8f
f K
{1 + α+α2,1 + α2+α4,1 + α3+α4,1 + α+α3}.
F2[X]→K X 1+α+α2
u:F2[X]/(f)→K f F2(X2+X+1)2
gF2[X]/(f)
uF24

K9K∗8α2= 1 + α α4= 2 α8= 1
α8x8= 1 x∈K∗
8
r∈F∗
3T−r K∗
F∗
3r+sα r, s ∈F3s6= 0
r+sα f(s−1(T−r)) r+sα α
T7→ r+sT F3[T]
r+sα α T 7→ s−1(T−r)
K∗Z/8Z8
1,3,5,7K∗K∗α α3α5α7
{1, α}
{α, 2α+ 1,2α, α + 2}
X9−X=X(X8−1) = X(X4−1)(X4+ 1) = X(X−1)(X+ 1)(X2+ 1)(X4+ 1).
X2+ 1 X4+ 1
X2+X+ 2 X2+ 2X+ 2
X9−X=X(X−1)(X+ 1)(X2+ 1)(X2+X+ 2)(X2+ 2X+ 2).
F3[X] 2
P∈K[X]K⊂L
P K[X]
L[X]L[X]
P L[X]K[X]
F16
F4
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F2
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B
B
B
B
B
B
B
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Fq0Fq
q0q
X16 −XF2[X]
F4F8X16 −X
F2[X] 4 1,2 4

4
(X2+X+ 1)2
X16 −X=X(X−1)(X2+X+ 1)(X4+X+ 1)(X4+X3+ 1)(X4+X3+X2+X+ 1).
F4[X]F
F4[X]
F4[X] = F2(β)β X2+X+ 1 F4={0,1, β, β + 1}β β + 1
F β2β+ 1
≤1,2F4[X]
4F4[X]
1 2 X2+rX +s r, s
F4
s0,1 + r, 1 + β(1 + r), r +β(1 + r)
(r, s)∈(1, β),(1, β2),(β, 1) ,(β, β),(β2,1) ,(β2, β2).
F2[X] 4
F4P P F2
β β2P(X+β)(X+β2)
X2+X+1 PF4[X]PF2[X]
F4[X]F2[X]F2[X]
PF4[X] 2
F2⊂L α ∈L
P∈F2[X]F(α)
Q∈L[X]P∈F2[X]L[X]F(Q)P L[X]F(Q)
Q
F P =QR
F(P) = F(Q)F(R)F(P) = P
P=X2+X+β
F2[X] 4 F(P) = X2+X+β2F4[X]
2
(X2+X+β)(X2+X+β2) = X4+X+ 1,
(X2+βX + 1)(X2+β2X+ 1) = X4+X3+X2+X+ 1,
(X2+βX +β)(X2+β2X+β2) = X4+X3+ 1.
4F4[X]
2X16 −XF4[X]
X16 −X=X(X−1)(X+β)(X+β2)(X2+X+β)(X2+X+β2)
×(X2+βX + 1)(X2+β2X+ 1)(X2+βX +β)(X2+β2X+β2).
F8[X]
X16 −X=X(X−1)(X2+X+ 1)(X4+X+ 1)(X4+X3+ 1)(X4+X3+X2+X+ 1),
F2[X]F8
X2+X+ 1 F8α∈F8F2[X]→F8

X α F4'F2[X]/(X2+X+ 1) →F8
X2+X+ 1 F8P
4F2PF8
αF16 'F2[X]/(P)→F8X α
PF8
2Q∈F8[X]
2P γ
2F8F64 γ
P4F2(γ) 4 F2F16
F8=F2(α)α P =X3+X+ 1 P
F8α2α4α4=α2+αF8[X]
P= (X+α)(X+α2)(X+α+α2).
xF8x=r+sα +tα2{1, α, α2}
x∈F2s=t= 0 X−r
x∈F2st 6= 0 1
3 3
P(X+r) = (X+r+α)(X+r+α2)(X+r+α+α2).
α α2α+α2P(X) 1 + α
1 + α21 + α+α2P(X+ 1)
K β f(X−1)
X2+X+ 2 X2+ 2 F5
P=X2+X+ 2 P(β6)6= 0
f1:k1→k2f1(α) = β6
P(2β+ 2) = 0 f2:k1→k2
f2(α) = 2β+ 2
f2