Devoir à la maison

C C E
n>2σ:E→E σn= Id ζ
nexp(2iπ/n)
i, j 0n−1
n−1
P
k=0
ζikζ−jk
i6=j
n−1
P
k=0
ζikζ−jk =
n−1
P
k=0
(ζi−j)k=(ζi−j)n−1
ζi−j−1= 0 i=j
n−1
P
k=0
ζikζ−jk =
n−1
P
k=0
1 = n
nδi,j
A n ai,j ζ−ij 06i6n−1
06j6n−1A
1n0n−1
B
bi,j 1
nζij (i, j)BA (1/n)Pn−1
k=0 ζikζ−kj
BA = Id A B
A
n x1, . . . , xn
(i, j) (xi)jQj>i (xj−xi)
1, ζ−1, ζ−2, . . . , ζ−(n−1) A
ζ−j−ζ−i
j > i
k∈ {0, . . . , n−1}pk=1
n
n−1
P
i=0
ζ−ikσi
E E

E E
Ekσ ζkpk
Ek
x∈E
σ(pk(x)) = 1
n
n−1
X
i=0
ζ−ikσi+1 σ,
=ζk1
n
n
X
j=1
ζ−jkσjj=i+ 1,
=ζk1
n
n−1
X
j=0
ζ−jkσjσn= IdE,
=ζkpk(x).
pk(x)∈Ek
k, l 0n−1pk◦pl
x∈E pl(x)∈Elσ(pl(x)) = ζlpl(x)
σi(pl(x)) = ζilpl(x)i>1
(pk◦pl)(x) = 1
n
n−1
X
i=0
ζ−ikζilpl(x) = 1
n
n−1
X
i=0
ζ−ikζil!pl(x).
0k6=l pl(x)k=l pk◦pl= 0 k6=l
(pk)2=pk
p0+p1+· · ·+pn−1= Id E
EkpkEkE
x∈E
p0(x) + p1(x) + · · · +pn−1(x) = 1
n
n−1
X
k=0
n−1
X
i=0
ζ−ikσi(x) =
n−1
X
i=0 1
n
n−1
X
k=0
ζ−ik!σi(x)
=
n−1
X
i=0
δi,0σi(x)j= 0
=σ0(x) = x.

p0+p1+· · · +pn−1= Id Ek
x∈E x =p0(x) + p1(x) + · · · +pn−1(x)∈E0⊕E1⊕. . . En−1
EkE pk
EkE
x=p0(x) + p1(x) + · · · +pn−1(x)
k ek,i Eki∈ {1,...,dim(Ek)}
B={ek,i}k i E
σ∗:E∗→E∗B∗={e∗
k,i}e∗ k,i
σ∗ζk
σ∗σ∗(ϕ) = ϕ◦σ ϕ ∈E∗
B∗={e∗
k,i}B={ek,i}E∗e∗ k,i(el,j ) = 1 (k, i)=(l, j)
e∗
k,i(el,j ) = 0 σ∗(e∗ k,i) = ζke∗ k,i
E x ∈EB
x=Pλl,j el,j λl,j ∈C
σ∗(e∗
k,i)(x) = e∗ k,i(σ(x)) σ∗,
=e∗
k,i Xλl,j σ(el,j )
=e∗
k,i Xλl,j ζlel,j el,j σ ζl,
=λk,iζke∗
k,i,
=ζke∗
k,i(x)e∗
k,i .
dimC(E) = n σ Xn−1
Ek
⇒EkEdimC(E) =
PdimC(Ek) = n Ek6= 0 ζkX−ζk
µσσ k
Xn−1µσµσXn−1
σn= Id µσ=Xn−1
⇒Ek6= 0 Ek= 0 pkEk
Pn−1
i=0 ζ−ikσi= 0 P(X) = Pn−1 i=0 ζ−ikXi
n−1σ
σ Xn−1Ek6= 0 dim(Ek)>1
n E n
1

EC
6n−1σ P (X)
P(ζX)σn= Id σ Xn−1
σ
i>1σiP(X)P(ζiX)i=n
σn= Id 0 6k6n−1ek=Xk
ζkσ(ek) = (ζX)k=ζkXk=ζkek
n Ek
ek⇒
σ Xn−1