Feuille de TD 1 - Université de Rennes 1

X2+X+1
2 2[X]/(X2+X+1)
(cl(1), cl(X))
d
p(p, d) = (2,3) (3,2) P
p[X]/(P)pd
k k
p k p
p k p
ϕp:k→k x 7→ xp
ϕpk k
p n >1k
pXpn−X
Xpn−X k k pn

p
Xpn−X pn
pn
pn
pXpn−X
pn
p k
pn>1k
pnk ϕn
ppn
pd⊂pnd n d
npn
pdn/d ϕd
p
pnn>1pk
k q n >1I(k, n)
n k
Xqn−X=Y
d|nY P∈I(k,d)
P
qn=X
d|n
d#I(k, d).
17
1722
k q P n
k P k
P Xqd−X1d
n
P k
P k
P
n>1 Φn∈[X]
n
∗n
∀n>1, Xn−1 = Y
d|n
Φd.

pΦpΦ15
p n >1
Φpn= ΦpXpn−1.
[X]A B
[X]B
(Q, R) [X]
R= 0 deg(R)<deg(B)
A=B Q +R
A B [X]B
B A [X]B A
[X]
nΦn∈[X]
n k
Φnk
kΦn
→k
p k p
n>1k[X] Φp n = Φp
n
n p Φn
k[X]
kΦn
K k n
K
k q K
k n k
kΦnK
ΦnK
Φnk[X]
q(/n )∗
2
k n
Φpn−1k[X]

X4+
X+ 1 2X16 −X2
42
p x ∈∗
pAx
{x, −x, x−1,−x1}R∗
pxRy
y∈Ax
∗
p
−1pp≡1 [4]
p
p≡1 [3]
⇔∗
p3
⇔X2+X+ 1 p
⇔ − 3p.
G n
α G d αd
G d n
α∈G ` n
αn
`6= 1 α G
` n
α`(α`)
n
`6= 1 n`n `
α=Q`αn`
`G
p
p= 2,3,5,7,11,31,43,71,139,313.

X4+ 1 pp
P n >2
P P
n[X]/(n, P )
(/n )[X]/(P)
[X] 2 X2+ 3 X−5
p[i]/p [i]
p[X]/X2+ 1 [j]/p [j]
p[j] [j]
X2+3 X−5x X
[X]/(X2+ 3 X−5) [X]/(X2+
3X−5) a+bx (a, b)∈2
[X]/(X2+ 3 X−5)
(a, b)∈2(a, b)6= (0,0)
(a+bx) (a−b(3 + x))
[X]/(X2+ 3 X−5)