Cryptography Reference
In-Depth Information
2
2
. Then
Example 5.5.3
Let
X
=
V
(
x
−
y
)
⊆ A
and
Y
=
V
(
x
−
z
)
⊆ P
φ
(
x,y
)
=
(
x
:
xy
:
y
)
is a rational map from
X
to
Y
. Note that this formula for
φ
is not defined at (0
,
0). However,
φ
is regular at (0
,
0) since taking
g
=
x
−
1
=
(
x
−
1
x
:
gives the equivalent form
φ
(
x,y
)
x
−
1
xy
:
x
−
1
y
)
=
(1 :
y
:
y/x
) and
y/x
≡
1in
k
(
X
). Also note that the image of
φ
is not
equal to
Y
(
) as it misses the point (0 : 1 : 0).
Similarly,
ψ
(
x
:
y
:
z
)
k
(
x/y,z/y
) is a rational map from
Y
to
X
. This map is not
regular at (1 : 0 : 1), but it is surjective to
X
. The composition
ψ
=
◦
φ
maps (
x,y
)to
(1
/y,
1
/x
).
V
(
y
2
z
(
x
3
Axz
2
))
2
1
. Consider the rational
Example 5.5.4
Let
X
=
−
+
⊆ P
and
Y
= P
map
φ
(
x
:
y
:
z
)
=
(
x/z
:1)
.
Note that this formula for
φ
is defined at all points of
X
except
P
0
=
(0:1:0).Let
g
(
x
:
y
:
z
)
=
(
x
2
+
Az
2
)
/y
2
∈ k
(
X
). Then the map (
x
:
y
:
z
)
→
(
gx/z
:
g
) can be written
as (
x
:
y
:
z
)
→
(1 :
g
) and this is defined at (0 : 1 : 0). It follows that
φ
is regular at
P
0
and that
φ
(
P
0
)
=
(1 : 0).
Lemma 5.5.5
Let X and Y be varieties over
k
and let φ
:
X
→
Y be a rational map. Then
there is an open set U
⊆
X such that φ is regular on U.
It immediately follows that Theorem
5.4.8
generalises to rational maps.
Theorem 5.5.6
Let X and Y be varieties. Suppose φ
1
,φ
2
:
X
→
Y are rational maps that
are regular on non-empty open sets U
1
,U
2
⊆
X. Suppose further that φ
1
|
U
1
∩
U
2
=
φ
2
|
U
1
∩
U
2
.
Then φ
1
=
φ
2
.
Exercise 5.5.7
Prove Theorem
5.5.6
.
Definition 5.5.8
Let
X
and
Y
be algebraic varieties over
k
. A rational map
φ
:
X
→
Y
defined over
k
is a
birational equivalence
over
k
if there exists a rational map
ψ
:
Y
→
X
over
k
such that:
1.
ψ
◦
φ
(
P
)
=
P
for all points
P
∈
X
(
k
) such that
ψ
◦
φ
(
P
) is defined;
2.
φ
◦
ψ
(
Q
)
=
Q
for all points
Q
∈
Y
(
k
) such that
φ
◦
ψ
(
Q
)
=
Q
is defined.
Varieties
X
and
Y
are
birationally equivalent
if there is a birational equivalence
φ
:
X
→
Y
between them.
2
and
Y
2
are birationally
Exercise 5.5.9
Show that
X
=
V
(
xy
−
1)
⊆ A
=
V
(
x
1
−
x
2
)
⊆ P
equivalent.
Exercise 5.5.10
Verify that birational equivalence is an equivalence relation.
