DS 2012/13 (avec correction) - Institut de Mathématiques de Bordeaux

a b A ha, bi
a b
A a, b ∈Apgcd(a, b)
d∈Apgcd(a, b)d a b d |a d |b
a b d pgcd(a, b)
Apgcd(a, b)a, b ∈A
a, b 6= 0 {p1,...,ps}ab
a b a =pα1
1···pαs
sb=pβ1
1···pβs
sαi, βi∈Z≥0pmin{α1,β1}
1···pmin{αs,βs}
s= pgcd(a, b)
a= 0 pgcd(a, b) = b
A P (t)∈A[t]
pgcd 1 d
P P =de Pe P
P(t)Q(t)P(t)Q(t)
cont(P Q) = cont(P) cont(Q)P Q
P Q p
P Q p
P(t) = a0+a1t+..., Q(t) = b0+b1t+..., P(t)Q(t) = c0+c1t+...,
P i p -aik i ` j
p-bj
p-ak, p |ai(i= 0,...,k−1); p-b`, p |bj(j= 0,...,`−1).
ck+`=
k−1
X
i=0
aibk+`−i+akb`+
k+`
X
i=k+1
aibk+`−i=
k−1
X
i=0
aibj+akb`+
`−1
X
j=0
ak+`−jbj
p akb`p-ck+`
P=de P Q =ee Q d = cont(P)e= cont(Q)e P , e Q
P Q =de e Pe Qe Pe Qcont(P Q) = de
P(t)=(t+ 2)(2t+ 3)(3t+ 4) · · · (2011t+ 2012) ∈Z[t].
nt +n+ 1 n+ 1 −n= 1 P(t)
F(t, u)=(t+ 2)3−(u+ 3)2R[t, u]
F(t, u)G(t, u) = F(t−2, u −3) = t3−u2G
R[t]G=H1H2deguH1= deguH2= 1
G u R(t)R(t)
S I P (t, u)∈R[t, u]
P(a2−2, a3−3) = 0 a∈S.
IR[t, u]F(t, u)∈I
P1(a2−2, a3−3) = P2(a2−2, a3−3) = 0 (P1+P2)(a2−2, a3−3) = 0 P(a2−2, a3−3) = 0
Q∈R[t, u] (QP )(a2−2, a3−3) = 0 I
a∈RF(a2−2, a3−3) = a6−a6= 0 F∈I

f:R[t, u]→R[x]
P(t, u)7→ P(x2−2, x3−3)
IhF(t, u)i
f
P∈ker f P (a2−2, a3−3) = 0 a∈RP∈I P ∈I
P(x2−2, x3−3) ∈R[x]S P (x2−2, x3−3)
P∈ker fker f=I
˜
IR(t)[u]I
F F R(t)[u]
˜
I=hFiP∈I F P R(t)[u]F
uR[t]PR[t][u]I=hFi
t3−t−30 Q[t]Z[t]t3+t+ 30
at3+bt2+ct +d∈Z[t] 3 Z[t]
αt +β∈Z[t]α|a β |d a = 1 f d
30 t3+t+ 30 −3t3−t−30
30
t2012 + 21t49 + 49t21 + 70
p= 7
2F2[t]F22
t2+bt +c∈F2[t]c= 1 2 t2+ 1 = (t+ 1)2
t2+t+ 1 F2
t5+t2+ 1 F2[t]
F22
t2+t+ 1
t2+t+ 1 t5+t2+ 1
2007t5+ 2008t4+ 2010t3+ 2011t2+ 2012t+ 2013
Z[t]
Z[t]→F2[t] 2 t5+t2+ 1
F2[t]Z[t]
t4−tZ[t]
t4−t=t(t−1)(t2+t+ 1)
pFppZ[t]/ht4−t+p, t2+t+ 1i
Fp[t]/ht2+t+ 1i
A I J A ¯ J J A/I
A/(I+J)∼
=(A/I)/¯
J.
A→A/I →(A/I)/¯ J
I+J
A=Z[t] (t2+t+ 1) |(t4−t)
ht4−t+p, t2+t+ 1i=hp, t2+t+ 1i=I+J, I =hpi, J =ht2+t+ 1i,
A/I =Fp[t]¯ J=ht2+t+ 1i
R[t, u]/ht4−t+u, t2+t+ 1iC
A=R[t, u]
ht4−t+u, t2+t+ 1i=hu, t2+t+ 1i=I+J, I =hui, J =ht2+t+ 1i,
A/I =R[t]¯ J=ht2+t+ 1i
R[t, u]/ht4−t+u, t2+t+ 1i∼
=R[t]/ht2+t+ 1i.
R[t]→Ct7→ −1+√−3
2ht2+t+ 1i
R[t]/ht2+t+ 1i∼
=C