TD12 : Algèbre et géométrie, Nullstellensatz, lemme de Nakayama

P∈C[X, Y ]C[X, Y ]/(P)
C[T]C[T, T −1]P
C
K(x, y)∈K2
5x2+ 3y2= 2
(x2+y2)2+ 3x2y−y3= 0
K f1, ...fr∈K[X1, ..., Xn]g1, ..., gs∈K[X1, ..., Xm]
B=K[X1, ..., Xn]/(f1, ..., fr)C=K[X1, ..., Xm]/(g1, ..., gs)VB
f1=... =fr= 0 VCg1=... =
gs= 0 f:B→C
f∗
X:VC(X)→VB(X)K X
X→Y K
VC(X)// VB(X)
VC(Y)//VB(Y).
f∗:VC→VB

C[X, Y ]/(Y3−X5)
C[T]
V y3=x5A2
C
f∗:A1
C→V
V
f∗
C:C→V(C)
g∗:V→A1
Cg∗ C
f∗
Cf∗
VA1
C
C[X, Y ]/(Y3−X5)[X−1]C[T, T −1]
V x 6= 0
A1
Ct6= 0 VA1 C
C[X, Y ]/(Y2−X3−X2)
C[X, Y ]/(Y2−X4−X5)
C[X, Y ]/(Y2+ 2iXY −X3+X−1)
P Q R C[X1, ..., Xn]P
x∈CnP(x) = 0 Q(x)6= 0 R(x) = 0
P|Q P |R
k A B k
A B A ⊗kB
V →Am
QV(C) = ∅
V(Q) = ∅
f1, ..., fn∈Z[X1, ...., Xm]
f1(x) = ... =fn(x) = 0 Cm
Fpp
A=Z[X, Y ]/(X2+Y2−1) Fp∈Q[X]

ζ(A, s) = QpFp(p−s).
ζ(A, s) = ζ(s−1) ∞
X
n=1
(−1)n−1
(2n−1)s!.
A=Z[X, Y ]/(XY −1), B =Z[X1, X2, X3, X4, Y ]/((X1X4−X2X3)Y−1),
C=Z[X]/(X21 −1), D =Z[X, Y ]/(X2−Y3),
E=Z[X, Y ]/(Y2−X2(X+ 1)).
A I A
A M A
IM =M M
AmN A M
M=N+mM N M
A M A
M
Z
A M A
X=A
d:X→N
p7→ dimk(p)Mp/pMp
.
p∈X n =d(p)α:k(p)n→Mp/pMp
k(p)
f∈Arp
A[f−1]n//
γ
An
p
//
β
k(p)n
α
M[f−1]//Mp
//Mp/pMp

α β :An
p→MpAp
f /∈pβ γ :
A[f−1]n→M[f−1]A[f−1]
d
{p∈X|d(p)≥n}n≥0
A A
M→N→P→0. 
A Q 
M⊗AQ→N⊗AQ→P⊗AQ→0.
M→N
M⊗AQ→N⊗AQ
A Q
A M →N M ⊗AQ→N⊗AQ
A
A A
A
Q A Q
I A Q ⊗AI→Q
A A
S A A
S−1A
Q A Q
mA AmQm
0→M→N→P→0A
P A Q
0→M⊗AQ→N⊗AQ→P⊗AQ→0
A Q A
Q