avec correction

a b A ha, bia b
A δ :Ar{0} → N
a∈A b ∈Ar{0}q, r ∈A a =bq +r r = 0 δ(r)< δ(b)
I A a ∈I I =aA =hai
A A
A I A I I ={0}I= 0A I
I6={0}
{δ(a) : a∈I, a 6= 0} ⊂ N
b∈I δ(b)≤δ(a)a∈I
I=bA b ∈I bA ⊂I I a ∈I a =bq +r
q, r ∈A r
r= 0 δ(r)< δ(b).
r=a−bq ∈I r 6= 0 δ(r)≥δ(b)b
a∈I a =bq q ∈A I ⊂bA I =bA
A a ∈A
a=bc ⇒b∈A×c∈A×.
a∈A a a =ηp1···pkη
p1,...,pkk≥0 1
a=ηp1···pk=θq1···q`k=`
q1,...,qkp1∼q1,...,pk∼qk∼x∼y x |y y |x
a∈A a
A A
Z[i] = {x+yi :x, y ∈Z} ⊂ C

Z[i]
z, w ∈Z[i]z±w, zw ∈Z[i]Z[i]
C C Z[i]
Q(i)t2+ 1 Q Q(i)
Q(i)Z[i]
KZ[i]
K={z/w :z, w ∈Z[i], w 6= 0}.
Z=z/w ∈K z =x+yi ∈Z[i] 0 6=w=u+vi ∈Z[i]Z=z¯w
w¯w=xu+yv
u2+v2+i−xv+yu u2+v2∈Q[i]
Z∈Q[i]x+yi
ux, y, u ∈Zu6= 0 Z=z/u z =x+yi ∈Z[i]
u∈Z[i]
Z[i]
z=x+yi ∈CN(z) = z¯z=|z|2=x2+y2
z, w ∈CN(zw) = N(z)N(w)
N(zw) = zwzw =z¯zw ¯w=N(z)N(w)
z, w ∈Z[i]z|wZ[i]N(z)| N (w)Z
z|w w =zz0z0∈Z[i]N(w) = N(z)N(z0)N(w),N(z),N(z0)∈Z
N(z)| N (w) =⇒z|w
N(2 + i) = N(2 −i)=5 N(2 + i)| N(2 −i) 2 + i-2−i2−i
2+i=3 5−i4 5/∈Z[i]
z∈Z[i]Z[i]N(z)=1
z∈Z[i]×z|1z|1N(z)| N(1) = 1 N(z)≥0
N(z)=1
N(z)=1 z¯z= 1 z|1
Z[i]×
z=x+yi ∈Z[i]×x2+y2= 1 x, y ∈Z
x=±1, y = 0 x= 0, y =±1Z[i]×={±1,±i}
z∈Z[i]N(z)zZ[i]
z=w1w2w1, w2∈Z[i]w1w2
N(z) = N(w1)N(w2).
N(z)N(w1)=1 N(w2)=1
z= 3
N(3) = 9 3 Z[i] 3 = zw
z w ∈Z[i]N(z) = N(w)=3
z=x+yi x2+y2= 3 x2≤3|x| ≤ √3x= 0 ±1x= 0 y2= 3
x=±1y2= 2
Z[i]
z∈Cq∈Z[i]N(z−q)≤1/2
z=x+yi u v x y
u, v ∈Z,|u−x| ≤ 1/2,|v−y| ≤ 1/2.
q=u+vi ∈Z[i]N(z−q)=(x−u)2+ (y−v)2≤(1/2)2+ (1/2)2= 1/2
Z[i]
a, b ∈Z[i]b6= 0 z=a/b q ∈Z[i]
N(a/b −q)≤1/2r=a−bq N(r) = N(a/b −q)N(b)≤(1/2)N(b)<N(b)N(b)>0
Z[i]N
Z[i]
Z[i]

A A
Z[i][t]
Z[i]Z[i][t]
A[t]AZ[i][t]
Z[i][t]I=h3, ti
I=hfif∈I f |3f f ∼1f∼3
3Z[i] 3 -t f ∼1 1 ∈I
u(t), v(t)∈Z[i][t] 1 = 3u(t) + tv(t) 1 = 3u(0) 3 |1
A I J A I +J=A
I J IJ =I∩J
IJ ⊂I IJ ⊂J IJ ⊂I∩J
I+J=A a ∈I b ∈J a +b= 1 x∈I∩J xa, xb ∈IJ
x=x(a+b)∈IJ I ∩J⊂IJ
I J A/IJ ∼
=A/I ×A/J
ϕIϕJA→A/I A →A/J
ϕ:A→A/I ×A/J ϕ(x)=(ϕI(x), ϕJ(x)) ker ϕ= ker ϕI∩ker ϕJ=I∩J=IJ
ϕ ϕI
ϕJ(¯u, ¯v)∈A/I ×A/J x ∈A ϕ(x) = (¯u, ¯v)ϕIϕJ
u, v ∈A ϕI(u) = ¯u ϕJ(v) = ¯v x =ub +va a ∈I
b∈J a +b= 1 ϕI(a)=0 ϕI(b) = ϕI(1) −ϕI(a)=1−0=1 ϕI(x) = ϕI(u) = ¯u
ϕJ(x) = ¯v ϕ(x) = (¯u, ¯v)
ϕ
ϕ A/I ×A/J ∼
=A/ ker ϕ=A/IJ
A a, b ∈Ahai hbi
pgcd(a, b)=1
Aha, bi=hdid= pgcd(a, b)ha, bi=hai+hbi hai
hbipgcd(a, b)=1
A A
a∈A a A →A x 7→ xa
+x∈A xa = 0 A
{0}x∈A xa = 1
A
z∈Z[i]Z[i]/hzi
C > 0Z[i]
C
Z[i]Nz
N(z)
zZ[i]Z[i]/hzi
zZ[i]hziZ[i]/hzi
Z[i]/hzi
1 + i, 2 + i, 2,3,5∈Z[i]Z[i]
3 1 + i2 + i
2 5 2 = (1 + i)(1 −i) 5 = (2 + i)(2 −i)

Z[i]/ < 2 + i > Z[i]/ < 3>Z[i]/ < 5>
ϕ:Z[i]→Z[i]/ < 2 + i > ϕ0=ϕ|ZϕZϕ(i) = ϕ0(−2)
z=x+yi ϕ(z) = ϕ0(x−2y)∈ϕ0(Z)ϕ0
Z[i]/ < 2 + i >∼
=Z/ker ϕ05 = (2 + i)(2 −i)∈ker ϕ01/∈ker ϕ0ϕ0
Z5 1 ker ϕ0= 5Z Z[i]/ < 2 + i >∼
=F5
Z[i]/ < 2−i >∼
=F5
Z[i]∼
=Z×Z Z[i]/ < 3>∼ =Z/3Z×Z/3Z|Z[i]/ < 3>|= 9 3
Z[i]Z[i]/ < 3>
Z[i]/ < 3>∼
=F9
Z[i]/ < 5>5 = (2 + i)(2 −i) 2 + i2−i
pgcd(2 + i, 2−i)=1
Z[i]/h5i∼
=Z[i]/ < 2 + i > ×Z[i]/ < 2−i >∼ =F5×F5.
K L K 2f(t)∈K[t] 3
f(t)K[t]f(t)L[t]
f(t)L[t]K[t]
f(t)L[t] deg f= 3
L K[t]f(t)
K3 2
2 4 5
2t2+ 1 R[t]C[t]
4t4+ 1 Q[t]Q(i)[t]
t4+ 1 = (t2−i)(t2+i).
5f(t)∈K[t] 5 L[t]f(t) = g(t)h(t)
g(t), h(t)∈L[t] deg g≤2g(t) 4 K f(t)
K[t]f(t) 5 K
t3−t+ 1 F3[t]F9[t]
F3F3[t]
F9[t]
2014t3+ 2013t2+ 2015t+ 2014 Z[t]Z[i][t]Q[t]
Q(i)[t]
ϕ:Z[t]→F3[t] 3
ϕ(2014t3+ 2013t2+ 2015t+ 2014) = t3−t+ 1.
F3[t] 2014t3+ 2013t2+ 2015t+ 2014
Z[t]Q[t]
Q(i)[t]Z[i][t]
f(t) = t3−t+ 1 ∈F3[t]
¯
F3F3f(t)F3[t]¯ F3
F33F33¯ F3
F27 f(t)F27 F27
F27 f(t)
α, β, γ f(t) = t3−t+ 1 ∈F3[t]
α13 =β13 =γ13 ∈ {1,−1}
F27 26 α26 =β26 =γ26 = 1
α13, β13, γ13 ∈ {1,−1}α13 =β13 =γ13 θ=α13 α t13 −θ∈F3[t]
{g(t)∈F3[t] : g(α)=0}F3[t]f(t)
t13 −θ t13 −θ=f(t)h(t)h(t)∈F3[t]
β13 −θ=f(β)h(β)=0 β13 =θ γ13 =θ
α4+β4+γ4
α3−α+ 1 = 0 α4=α2−α β γ
α4+β4+γ4=α2+β2+γ2−(α+β+γ)=(α+β+γ)2−2(αβ +αγ +βγ)−(α+β+γ).
f(t)
α+β+γ= 0, αβ +αγ +βγ =−1.
α4+β4+γ4= 02−2·(−1) −0=2=−1