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27
Thus, our rational map is given by
T
1
7→
T
2

1
T
2
+ 1
, T
2
7→

2
T
T
2
+ 1
.
Next note that the obtained map is birational. The inverse map is given by
T
7→
T
2
T
1

1
.
In particular, we see that
R
(
V
(
T
2
1
+
T
2
2

1))
∼
=
k
(
T
1
)
.
The next theorem, although sounding as a deep result, is rather useless for
concrete applications.
Theorem 4.10.
Assume
k
is of characteristic
0
. Then any irreducible aﬃne
algebraic
k
set is birationally isomorphic to an irreducible hypersurface.
Proof.
Since
R
(
V
)
is a ﬁnitely generated ﬁeld over
k
, it can be obtained as an
algebraic extension of a purely transcendental extension
L
=
k
(
t
1
, . . . , t
n
)
of
k
.
Since char
(
k
) = 0
, R
(
V
)
is a separable extension of
L
, and the theorem on a
primitive element applies (M. Artin, ”Algebra”, Chapter 14, Theorem 4.1): an
algebraic extension
K/L
of characteristic zero is generated by one element
x
∈
K
. Let
k
[
T
1
, . . . , T
n
+1
]
→
R
(
V
)
be deﬁned by sending
T
i
to
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 Spring '09
 wei
 Algebra, Geometry

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