Basis (Modul) Sei
R
{\displaystyle R}
Hauptidealring und
M
{\displaystyle M}
R
{\displaystyle R}
R
{\displaystyle R}
N
<
M
{\displaystyle N<M}
Sei
{
m
1
,
…
,
m
n
}
{\displaystyle \{m_{1},\dotsc ,m_{n}\}}
M
{\displaystyle M}
N
i
=
N
∩
<
m
1
,
…
,
m
i
>
{\displaystyle N_{i}=N\cap <m_{1},\dotsc ,m_{i}>}
Das Ideal
{
r
∈
R
:
∃
m
∈
N
i
+
1
mit
m
=
m
′
+
r
⋅
m
i
+
1
und
m
′
∈<
m
1
,
…
,
m
i
>
}
{\displaystyle \{r\in R:\exists m\in N_{i+1}{\text{ mit }}m=m'+r\cdot m_{i+1}{\text{ und }}m'\in <m_{1},\dotsc ,m_{i}>\}}
a
i
+
1
{\displaystyle a_{i+1}}
n
i
+
1
=
m
′
+
a
i
+
1
⋅
m
i
+
1
∈
N
i
+
1
mit
m
′
∈<
m
1
,
…
,
m
i
>
{\displaystyle n_{i+1}=m'+a_{i+1}\cdot m_{i+1}\in N_{i+1}{\text{ mit }}m'\in <m_{1},\dotsc ,m_{i}>}
dann gilt
N
i
+
1
=
N
i
⊕
R
⋅
n
i
+
1
{\displaystyle N_{i+1}=N_{i}\oplus R\cdot n_{i+1}}
Sei
M
=
Z
3
=<
(
1
,
0
,
0
)
,
(
0
,
1
,
0
)
,
(
0
,
0
,
1
)
>
{\displaystyle M=\mathbb {Z} ^{3}=<(1,0,0),(0,1,0),(0,0,1)>}
Z
{\displaystyle \mathbb {Z} }
N
:=
{
z
∈
Z
3
:
2
z
1
+
3
z
2
+
4
z
3
=
0
∧
5
teilt
z
2
}
{\displaystyle N:=\{z\in \mathbb {Z} ^{3}:2z_{1}+3z_{2}+4z_{3}=0\land 5{\text{ teilt }}z_{2}\}}
Eine Basis von
N
{\displaystyle N}
N
1
=
N
∩
<
(
1
,
0
,
0
)
>=
{
z
∈
Z
3
:
2
z
1
=
0
}
=
{
(
0
,
0
,
0
)
}
{\displaystyle N_{1}=N\cap <(1,0,0)>=\{z\in \mathbb {Z} ^{3}:2z_{1}=0\}=\{(0,0,0)\}}
N
2
=
N
∩
<
(
1
,
0
,
0
)
,
(
0
,
1
,
0
)
>=
{
z
∈
Z
3
:
2
z
1
+
3
z
2
=
0
∧
5
|
z
2
}
{\displaystyle N_{2}=N\cap <(1,0,0),(0,1,0)>=\{z\in \mathbb {Z} ^{3}:2z_{1}+3z_{2}=0\land 5\vert z_{2}\}}
Wir suchen nun das kleinste positive
z
2
{\displaystyle z_{2}}
⇒
N
2
=<
(
−
15
,
10
,
0
)
>
{\displaystyle \Rightarrow N_{2}=<(-15,10,0)>}
N
3
=
N
∩
<
(
1
,
0
,
0
)
,
(
0
,
1
,
0
)
,
(
0
,
0
,
1
)
>=
N
{\displaystyle N_{3}=N\cap <(1,0,0),(0,1,0),(0,0,1)>=N}
Wir suchen das kleinste positive
z
3
{\displaystyle z_{3}}
⇒
N
3
=
N
2
⊕
<
(
−
2
,
0
,
1
)
>
{\displaystyle \Rightarrow N_{3}=N_{2}\oplus <(-2,0,1)>}
Wir haben eine Basis
N
=<
(
−
15
,
10
,
0
)
,
(
−
2
,
0
,
1
)
>
{\displaystyle N=<(-15,10,0),(-2,0,1)>}
