TD10 : Modules de type fini sur un anneau principal

Z
Z/4Z⊕Z/8Z⊕Z/6Z⊕Z/18Z⊕Z/7Z.
n≥1u1unZn
QnMZn
u1, ..., unMZn
u1, ..., un
MZ3u1= (2,1,1) u2= (1,2,1)
u3= (1,1,2) Z3/M
(ui,j)1≤i≤n,1≤j≤nui,j = 2
i=j ui,j = 1 u1un
MZnu1, ..., un
Zn/M
Z/(n+ 1)Z
ZMZ4
(2,−1,0,0) (−1,2,−1,−1) (0,−1,2,0) (0,−1,0,2)
Z4/M
(Z/2Z)2
ZMZ3(4,8,16)
(1,5,10) (6,2,4) (5,8,6)
Z/40Z

Z35x+7y+35z= 0
Z
Z3x+ 2y+ 3z≡0
mod 4
Z/4Z
C[[X]] C[[X]]2
((1 −X)−1,(1 −X2)−1) ((1 + X)−1,(1 + X2)−1)
C[[X]]/(X2)
ZM N Z2
(e1, e2)Z2
a, b, c, d (ae1, be2)M(ce1, de2)
N
Z[i]
Z[i] 3
5 9
Z[i]Z[i]M
M∼
=
n
M
r=1
Z[i]/(zr)
zr∈Z[i]r z1|z2|...|zn
|M|=|z1...zr|2Z[i]
Z[i]Z[i]/(2 + i)
Z[i]/(2 −i)Z[i]Z[i]/(3)
Z[i] 53·64
Z[i]M
M∼
=
n
M
r=1
m
M
s=1
(Z[i]/(zs
r))ar,s ,
z1, ..., znar,s ≥0
|M|=Qr|zr|2sar,s Z[i]
•π2|π2|2= 2
•p≡1 mod 4 πpπ0
p
p
•p≡3 mod 4 πpp2
53·64= 24×34×53
5×2×(4 + 4 + 2) = 100
p
Z/pZ Z[i]a

b p =a2+b2
A A
A
A I A A
a1, ..., an∈A I = (a1, ..., ar)I A
A
x, y ∈I A
yx −xy = 0 I
A M ∈ Mm,n(A)
P∈ Mm(A)Q∈ Mn(A)
P MQ
d10 0 ... 0... 0
0d20... 0... 0
0 0 d3... 0... 0
...
0 0 0 ... dr... 0
...
0 0 0 ... 0... 0
d1, ..., drA d1|d2|d3|...|dr
M∈GLn(A)P∈ Mn(A)
P M =
1 0 ... 0 0
0 1 ... 0 0
0 0 ... 1 0
0 0 ... 0 (M)
.
GLn(A)
SLn(A)

A n ∈N∗S(A, n)
A n
X∈S(A, n)nA ⊆X
S(A, n)S(A/nA, n)
m∈N∗N∈N∗m∧n= 1
S((Z/mnZ)N, mn)S((Z/mZ)N, m)×S((Z/nZ)N, n)
S(Z2,2)
S(Z2, n)
X∈S(Z2, n)X∩(Z⊕0) = aZ⊕0⊆Z2
a n |S(Z2, n)|=Pa|na
Pr≥0|S(Z2, pr)|TrPn≥1|S(Z2, n)|n−s
S(Z3, n)|S(Z3, n)|=
Pab|na2bPr≥0|S(Z3, pr)|Tr
Pn≥1|S(Z3, n)|n−s
p V0Z Z2=Z⊕ZV1
Zp V0
V1p+ 1
V1V1V2p
V2p+1
V0
V1Z2
Zn≥0
V1∼
=ZnV1Z2V1∼ =Z2
V2p+ 1 V2∈ V2V0
m∈NV2=mV0[V0:V2] = m2= [V0:V1][V1:V2] = p2
V2=pV0V2V0
Tp={Z Z2p}/( )
v v0
V V 0v v0V
p V 0
v→v0
v0→v
v→v0V V 0
v v0V p V 0
pV 0⊆V[V:pV 0] = [V0:pV 0]
[V0:V]=p v0→v
Tpv v0Tp

V(0) V(n)v v0
V(i)1≤i≤n−1V(0) ⊇V(1) ⊇
· · · ⊇ V(n)V(i+1) p V(i)i
V(0) W v v0m≥0
V(0) pmV0V0=pmW
V0⊆V(0) V0=V(0) V(0) 6=V0
V(0)/V 0
V(0)/V 0∼
=
k
M
i=1
Z/priZ.
p V(0)
V(1) p V 0V(1) 6=V0
V(1) V(2) p V 0
V(i)1≤i≤n
V(0) ⊇V(1) ⊇ · · · ⊇ V(n)V(n)=V0V(i+1) p V(i)
i
(e1, e2)V(0) a, b > 0V0=
paZe1⊕pbZe2
V(0) ⊇pZe1⊕Ze2⊇p2Ze1⊕Ze2⊇... ⊇paZe1⊕Ze2⊇paZe1⊕pZe2⊇... ⊇V0.
v0, v1, . . . , vn−1, vn=v0vivi+1
n= 0 n≥2 1 ≤i≤n−1
vi+1 =vi−1
n > 0v w d(v, w)
v w r n d(v0, vr) =
maxid(v0, vi)
V0Vrv0vr(e1, e2)V0Vr=Ze1⊕paZe2
a > 0d(v0, vr) = a
VrZe1⊕pa−1Ze2Vk,a = (e1+
bpae2)Z⊕pa+1Ze2b∈ {0, ..., p −1}vk,a
Vk,a d(v0, vk,a) = a+ 1 vr−1vr+1
Ze1⊕pa−1Ze2vr−1=vr+1
A B |A[n]|=|B[n]|
n > 0A B