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Proof.
See [Shaf94].
Note that Theorem 10.13.15 can be rephrased as saying that every irreducible
variety “projects” to a hyperplane.
Let us again look at the special case of plane curves more closely.
The variety V(y
5
- x
2
) described in Example 10.13.7 is bira-
10.13.16. Example.
tionally equivalent to
R
. See [CoLO97].
Example 10.13.16 shows that isomorphism is stronger than birational equiva-
lence.
10.13.17. Theorem.
Every rational transform of an affine rational curve is rational.
Proof.
The theorem follows easily from Theorem B.11.16.
10.13.18. Theorem.
Let
C
be an irreducible plane curve defined by an equation
f(X,Y) = 0. The following two conditions are equivalent:
(1) There exist two rational functions p(t), q(t) Πk(t) so that
(a) f(p(t),q(t)) = 0 for all but a finite number of t, and
(b) for all but a finite number of points (x,y) on
C
there is a unique t Πk sat-
isfying x = p(t) and y = q(t).
(2) The curve
C
is rational.
Proof.
Because of Theorem 10.13.14, condition (2) is equivalent to
(3) The function field k(
C
) is isomorphic to the field of rational functions k(t) for
a transcendental variable t.
We follow the proof in [Walk50] and show that (1) and (3) are equivalent.
(1) fi (3): Condition (1b) implies that the functions p and q are not both con-
stants. Assume that p is not constant. It will then be transcendental over k. Since
k(p,q) Õ k(t), Theorem B.11.5 implies that k(p,q) = k(l) for some lŒk(t). But k(
C
) ª
k(p,q), and so we have proved (3).
(3) fi (1): If the function field of
C
is isomorphic to k(l), where l is transcendental
over k, then
C
is birationally equivalent to the line s = 0 in (s,t)-space, that is,
()
()
xpt andyqt
=
=
for rational functions p and q. These functions satisfy (1a) and (1b) because, except
for a finite number of points on
C
and s = 0, the birational equivalence is a bijection
between the points of these two curves.
Theorem 10.13.18 is proved.
