Satz: Sei
D
≡
0
,
1
(
m
o
d
4
)
,
{\displaystyle D\equiv 0,1\,(\mathrm {mod} \,4),}
D
{\displaystyle D}
Fundamentaldiskriminante . Dann gibt es eine bijektive Korrespondenz zwischen den echten Äquivalenzklassen bijektiv quadratischer Formen mit Diskriminante
D
{\displaystyle D}
O
Q
(
d
)
{\displaystyle {\rm {\mathcal {O}}}_{{\mathbb {Q} }({\sqrt {d}})}}
h
Q
(
d
)
+
{\displaystyle h_{{\mathbb {Q} }({\sqrt {d}})}^{+}}
h
(
D
)
{\displaystyle h(D)}
.
Beweis: Sei
a
{\displaystyle {\rm {\mathfrak {a}}}}
(
0
)
{\displaystyle (0)}
Ideal von
O
Q
(
d
)
{\displaystyle {\rm {\mathcal {O}}}_{{\mathbb {Q} }({\sqrt {d}})}^{}}
λ
∈
a
{\displaystyle \lambda \in {\rm {\mathfrak {a}}}}
(
λ
)
⊆
a
{\displaystyle (\lambda )\subseteq {\rm {\mathfrak {a}}}}
a
|
(
λ
)
{\displaystyle {\rm {\mathfrak {a}}}|(\lambda )}
N
(
a
)
|
N
(
λ
)
{\displaystyle N({\rm {\mathfrak {a}}})|N(\lambda )}
Abbildung
ϕ
:
a
→
Q
,
ϕ
(
a
)
=
N
(
λ
)
N
(
a
)
{\displaystyle {\phi :{\rm {\mathfrak {a}}}\to {\mathbb {Q} }{\rm {\;}},{\rm {\;\;\;\;\;\;\;}}\phi ({\rm {\mathfrak {a}}})={\frac {N(\lambda )}{N({\rm {\mathfrak {a}}})}}}}
Werte in
Z
{\displaystyle {\mathbb {Z} }}
a
{\displaystyle {\rm {\mathfrak {a}}}}
Basis
{
α
,
β
}
{\displaystyle \left\{\alpha ,\beta \right\}}
a
=
Z
α
⊕
Z
β
≃
Z
2
{\displaystyle {\rm {\mathfrak {a}}}={\mathbb {Z} }\alpha \oplus {\mathbb {Z} }\beta \simeq {\mathbb {Z} }^{2}}
ϕ
{\displaystyle \phi }
f
{\displaystyle f}
Z
2
{\displaystyle {\mathbb {Z} }^{2}}
f
(
x
,
y
)
=
ϕ
(
x
α
+
y
β
)
=
(
x
α
+
y
β
)
(
x
α
′
+
y
β
′
)
N
(
a
)
=
N
(
α
)
N
(
a
)
x
2
+
α
β
′
+
α
′
β
N
(
a
)
x
y
+
N
(
β
)
N
(
a
)
y
2
{\displaystyle f(x,y)=\phi (x\alpha +y\beta )={\frac {(x\alpha +y\beta )(x\alpha '+y\beta ')}{N({\rm {\mathfrak {a}}})}}={\frac {N(\alpha )}{N({\rm {\mathfrak {a}}})}}x^{2}+{\frac {\alpha \beta '+\alpha '\beta }{N({\rm {\mathfrak {a}}})}}xy+{\frac {N(\beta )}{N({\rm {\mathfrak {a}}})}}y^{2}}
Wir erhalten also eine binäre quadratische Form:
f
(
x
,
y
)
=
a
x
2
+
b
x
y
+
c
y
2
w
o
b
e
i
a
:=
N
(
α
)
N
(
a
)
,
b
:=
α
β
′
+
α
′
β
N
(
a
)
,
c
:=
N
(
β
)
N
(
a
)
(
2
)
{\displaystyle f(x,y)=ax^{2}+bxy+cy^{2}{\rm {\;\;wobei\;\;}}a:={\frac {N(\alpha )}{N({\rm {\mathfrak {a}}})}},{\rm {\;}}b:={\frac {\alpha \beta '+\alpha '\beta }{N({\rm {\mathfrak {a}}})}},{\rm {\;}}c:={\frac {N(\beta )}{N({\rm {\mathfrak {a}}})}}\qquad \qquad (2)}
Für die Diskriminante findet man leicht
b
2
−
4
a
c
=
(
α
β
′
+
α
′
β
)
2
−
4
N
(
α
)
N
(
β
)
N
(
a
)
2
=
(
α
β
′
−
α
′
β
)
2
N
(
a
)
2
=
D
(
a
)
N
(
a
)
2
=
D
,
(
3
)
{\displaystyle b^{2}-4ac={\frac {(\alpha \beta '+\alpha '\beta )^{2}-4N(\alpha )N(\beta )}{N({\rm {\mathfrak {a}}})^{2}}}={\frac {(\alpha \beta '-\alpha '\beta )^{2}}{N({\rm {\mathfrak {a}}})^{2}}}={\frac {D({\rm {\mathfrak {a}}})}{N({\rm {\mathfrak {a}}})^{2}}}=D,\qquad \qquad \quad (3)}
wobei die Diskriminante
D
(
a
)
{\displaystyle D({\rm {\mathfrak {a}}})}
Determinante von
(
α
β
α
′
β
′
)
{\displaystyle \left({\begin{array}{cc}{\alpha }&{\beta }\\{\alpha '}&{\beta '}\end{array}}\right)}
a
=
N
(
α
)
N
(
a
)
,
{\displaystyle a={\frac {N(\alpha )}{N({\rm {\mathfrak {a}}})}},}
c
=
N
(
β
)
N
(
a
)
∈
Z
{\displaystyle c={\frac {N(\beta )}{N({\rm {\mathfrak {a}}})}}\in {\mathbb {Z} }}
D
=
b
2
−
4
a
c
∈
Z
{\displaystyle D=b^{2}-4ac\in {\mathbb {Z} }}
b
∈
Z
{\displaystyle b\in {\mathbb {Z} }}
f
{\displaystyle f}
Diskriminante
D
{\displaystyle D}
{
α
1
,
β
1
}
{\displaystyle \left\{\alpha _{1},\beta _{1}\right\}}
a
{\displaystyle {\rm {\mathfrak {a}}}}
{
α
,
β
}
{\displaystyle \left\{\alpha ,\beta \right\}}
{
α
1
,
β
1
}
{\displaystyle \left\{\alpha _{1},\beta _{1}\right\}}
T
=
(
p
q
r
s
)
∈
Z
2
×
2
{\displaystyle T=\left({\begin{array}{cc}{p}&{q}\\{r}&{s}\end{array}}\right)\in {\mathbb {Z} }^{2\times 2}}
p
s
−
q
r
=
±
1
{\displaystyle ps-qr=\pm 1}
ϕ
{\displaystyle \phi }
{
α
1
,
β
1
}
{\displaystyle \left\{\alpha _{1},\beta _{1}\right\}}
f
1
{\displaystyle f_{1}}
(
x
,
y
)
{\displaystyle (x,y)}
(
p
x
+
q
y
,
r
x
+
s
y
)
{\displaystyle (px+qy,rx+sy)}
{
α
,
β
}
{\displaystyle \left\{\alpha ,\beta \right\}}
a
{\displaystyle {\rm {\mathfrak {a}}}}
positiv orientiert , wenn für die rationale Zahl
α
′
β
−
α
β
′
D
>
0
{\displaystyle {\frac {\alpha '\beta -\alpha \beta '}{\sqrt {D}}}>0}
(
α
′
β
−
α
β
′
D
)
2
=
D
(
a
)
D
=
N
(
a
)
2
{\displaystyle \left({\frac {\alpha '\beta -\alpha \beta '}{\sqrt {D}}}\right)^{2}={\frac {D({\rm {\mathfrak {a}}})}{D}}=N({\rm {\mathfrak {a}}})^{2}}
Matrix eines Basiswechsels zwischen positiv orientierten Basen immer die Determinante
+
1
{\displaystyle +1}
f
(
x
,
y
)
=
a
x
2
+
b
x
y
+
c
y
2
{\displaystyle f(x,y)=ax^{2}+bxy+cy^{2}}
a
{\displaystyle {\rm {\mathfrak {a}}}}
a
{\displaystyle {\rm {\mathfrak {a}}}}
(
τ
)
a
{\displaystyle (\tau ){\rm {\mathfrak {a}}}}
τ
∈
Q
(
d
)
{\displaystyle \tau \in {\mathbb {Q} }({\sqrt {d}})}
N
(
τ
)
>
0
{\displaystyle N(\tau )>0}
(
τ
α
,
τ
β
)
{\displaystyle (\tau \alpha ,\tau \beta )}
(
τ
)
a
{\displaystyle (\tau ){\rm {\mathfrak {a}}}}
N
(
(
τ
)
a
)
=
|
N
(
τ
)
|
N
(
a
)
=
N
(
τ
)
N
(
a
)
{\displaystyle N\left((\tau ){\rm {\mathfrak {a}}}\right)=\left|N(\tau )\right|N({\rm {\mathfrak {a}}})=N(\tau )N({\rm {\mathfrak {a}}})}
(
τ
)
a
{\displaystyle (\tau ){\rm {\mathfrak {a}}}}
(
x
,
y
)
↦
N
(
x
τ
α
+
y
τ
β
)
N
(
(
τ
)
a
)
=
N
(
τ
)
N
(
x
α
+
y
β
)
N
(
τ
)
N
(
a
)
=
N
(
x
α
+
y
β
)
N
(
a
)
{\displaystyle (x,y)\mapsto {\frac {N(x\tau \alpha +y\tau \beta )}{N\left((\tau ){\rm {\mathfrak {a}}}\right)}}={\frac {N(\tau )N(x\alpha +y\beta )}{N(\tau )N({\rm {\mathfrak {a}}})}}={\frac {N(x\alpha +y\beta )}{N({\rm {\mathfrak {a}}})}}}
mit der Form
f
{\displaystyle f}
D
{\displaystyle D}
f
(
x
,
y
)
=
a
x
2
+
b
x
y
+
c
y
2
,
a
,
b
,
c
,
∈
Z
,
b
2
−
4
a
c
=
D
{\displaystyle f(x,y)=ax^{2}+bxy+cy^{2},{\rm {\;\;\;}}a,b,c,\in {\mathbb {Z} }{\rm {\;}},{\rm {\;\;}}b^{2}-4ac=D}
eine quadratische Form mit Diskriminante
D
{\displaystyle D}
D
{\displaystyle D}
(
a
,
b
,
c
)
=
1
{\displaystyle (a,b,c)=1}
f
{\displaystyle f}
a
>
0
{\displaystyle a>0}
a
z
2
−
b
z
+
c
=
0
{\displaystyle az^{2}-bz+c=0}
z
1
=
b
+
D
2
a
,
z
2
=
b
−
D
2
a
{\displaystyle z_{1}={\frac {b+{\sqrt {D}}}{2a}},{\rm {\;}}z_{2}={\frac {b-{\sqrt {D}}}{2a}}}
Nun ist
a
=
Z
⊕
Z
z
1
{\displaystyle {\rm {\mathfrak {a}}}={\mathbb {Z} }\oplus {\mathbb {Z} }z_{1}}
a
{\displaystyle {\rm {\mathfrak {a}}}}
τ
=
u
+
v
d
2
∈
O
Q
(
d
)
{\displaystyle \tau ={\frac {u+v{\sqrt {d}}}{2}}\in {\rm {\mathcal {O}}}_{{\mathbb {Q} }({\sqrt {d}})}}
u
,
v
∈
Z
,
u
≡
v
D
{\displaystyle u,v\in {\mathbb {Z} },{\rm {\;\;}}u\equiv vD{\rm {\;}}}
α
=
x
+
y
z
∈
a
{\displaystyle \alpha =x+yz\in {\rm {\mathfrak {a}}}}
τ
α
=
(
u
+
v
d
2
)
(
x
+
y
b
+
y
d
2
a
)
=
x
u
2
+
y
b
u
4
a
+
y
v
D
4
a
+
(
x
v
2
+
y
b
v
4
a
+
u
y
4
a
)
D
=
x
u
2
+
y
b
u
4
a
+
y
v
(
b
2
−
4
a
c
)
4
a
+
(
x
v
2
+
y
b
v
4
a
+
u
y
4
a
)
(
2
a
z
−
b
)
=
(
x
u
−
v
b
2
−
y
v
c
)
+
(
x
v
a
+
y
u
+
v
b
2
)
z
,
{\displaystyle {\begin{array}{l}{\tau \alpha =\left({\frac {u+v{\sqrt {d}}}{2}}\right)\left(x+{\frac {yb+y{\sqrt {d}}}{2a}}\right)}\\{{\rm {\;\;\;\;}}={\frac {xu}{2}}+{\frac {ybu}{4a}}+{\frac {yvD}{4a}}+\left({\frac {xv}{2}}+{\frac {ybv}{4a}}+{\frac {uy}{4a}}\right){\sqrt {D}}}\\{{\rm {\;\;\;\;}}={\frac {xu}{2}}+{\frac {ybu}{4a}}+{\frac {yv(b^{2}-4ac)}{4a}}+\left({\frac {xv}{2}}+{\frac {ybv}{4a}}+{\frac {uy}{4a}}\right)\left(2az-b\right)}\\{{\rm {\;\;\;\;}}=\left(x{\frac {u-vb}{2}}-yvc\right)+\left(xva+y{\frac {u+vb}{2}}\right)z,}\end{array}}}
und damit
Z
⊕
Z
z
1
{\displaystyle {\mathbb {Z} }\oplus {\mathbb {Z} }z_{1}}
b
2
≡
D
(
m
o
d
4
a
)
,
{\displaystyle b^{2}\equiv D\,(\mathrm {mod} \,4a),}
⇒
{\displaystyle \Rightarrow }
b
≡
D
(
m
o
d
2
)
,
{\displaystyle b\equiv D\,(\mathrm {mod} \,2),}
⇒
{\displaystyle \Rightarrow }
u
≡
v
D
≡
v
b
(
m
o
d
2
)
,
{\displaystyle u\equiv vD\equiv vb\,(\mathrm {mod} \,2),}
z
1
−
z
2
2
a
>
0
{\displaystyle {\frac {z_{1}-z_{2}}{2a}}>0}
{
1
,
z
1
}
{\displaystyle \left\{1,z_{1}\right\}}
a
{\displaystyle {\rm {\mathfrak {a}}}}
D
(
a
)
=
det
(
1
z
1
1
z
2
)
2
=
(
z
1
−
z
2
)
2
=
D
a
2
,
{\displaystyle D({\rm {\mathfrak {a}}})=\det \left({\begin{array}{cc}{1}&{z_{1}}\\{1}&{z_{2}}\end{array}}\right)^{2}=(z_{1}-z_{2})^{2}={\frac {D}{a^{2}}},}
und da wie in
(
3
)
{\displaystyle (3)}
D
(
a
)
N
(
a
)
2
=
D
{\displaystyle {\frac {D({\rm {\mathfrak {a}}})}{N({\rm {\mathfrak {a}}})^{2}}}=D}
N
(
a
)
=
1
a
{\displaystyle N({\rm {\mathfrak {a}}})={\frac {1}{a}}}
a
{\displaystyle {\rm {\mathfrak {a}}}}
(
x
,
y
)
↦
N
(
x
+
y
z
)
N
(
a
)
=
x
2
+
b
a
x
y
+
c
a
y
2
1
a
=
f
(
x
,
y
)
.
{\displaystyle (x,y)\mapsto {\frac {N(x+yz)}{N({\rm {\mathfrak {a}}})}}={\frac {x^{2}+{\frac {b}{a}}xy+{\frac {c}{a}}y^{2}}{\frac {1}{a}}}=f(x,y).}
Ist hingegen
a
{\displaystyle {\rm {\mathfrak {a}}}}
{
α
,
β
}
{\displaystyle \left\{\alpha ,\beta \right\}}
N
(
α
)
>
0
{\displaystyle N(\alpha )>0}
a
,
b
,
c
{\displaystyle a,b,c}
(
2
)
{\displaystyle (2)}
b
+
D
2
a
=
α
β
′
+
α
′
β
+
N
(
a
)
D
2
N
(
α
)
=
α
β
′
+
α
′
β
+
(
α
′
β
−
α
β
′
)
2
N
(
α
)
=
β
α
.
{\displaystyle {\begin{array}{l}{{\frac {b+{\sqrt {D}}}{2a}}={\frac {\alpha \beta '+\alpha '\beta +N({\rm {\mathfrak {a}}}){\sqrt {D}}}{2N(\alpha )}}}\\{{\rm {\;\;\;\;\;\;\;\;\;\;\;\;}}={\frac {\alpha \beta '+\alpha '\beta +(\alpha '\beta -\alpha \beta ')}{2N(\alpha )}}={\frac {\beta }{\alpha }}}.\end{array}}}
Damit ist also
Z
⊕
Z
b
+
D
2
a
=
Z
⊕
Z
β
α
=
(
α
−
1
)
a
{\displaystyle {\mathbb {Z} }\oplus {\mathbb {Z} }{\frac {b+{\sqrt {D}}}{2a}}={\mathbb {Z} }\oplus {\mathbb {Z} }{\frac {\beta }{\alpha }}=(\alpha ^{-1}){\rm {\mathfrak {a}}}}
a
{\displaystyle {\rm {\mathfrak {a}}}}
Sei nun
a
<
0
{\displaystyle a<0}
D
>
0
{\displaystyle D>0}
Z
τ
⊕
Z
τ
z
1
{\displaystyle {\mathbb {Z} }\tau \oplus {\mathbb {Z} }\tau z_{1}}
τ
∈
Q
(
D
)
{\displaystyle \tau \in {\mathbb {Q} }({\sqrt {D}})}
N
(
λ
)
<
0
{\displaystyle N(\lambda )<0}
{
λ
,
λ
z
}
{\displaystyle \left\{\lambda ,\lambda z\right\}}
a
{\displaystyle {\rm {\mathfrak {a}}}}
f
{\displaystyle f}
a
{\displaystyle {\rm {\mathfrak {a}}}}
{
α
,
β
}
{\displaystyle \left\{\alpha ,\beta \right\}}
N
(
α
)
<
0
{\displaystyle N(\alpha )<0}
f
{\displaystyle f}
a
<
0
{\displaystyle a<0}
Z
τ
⊕
Z
τ
z
1
{\displaystyle {\mathbb {Z} }\tau \oplus {\mathbb {Z} }\tau z_{1}}
a
{\displaystyle {\rm {\mathfrak {a}}}}
h
Q
(
d
)
+
=
h
(
D
)
{\displaystyle h_{{\mathbb {Q} }({\sqrt {d}})}^{+}=h(D)}
◻
{\displaystyle \Box }
