81
number of connected components of the corresponding
K
set equals the number of
distinct roots of
F
(
Z
)
in
K
.
Corollary 9.6.
Assume
k
is a perfect ﬁeld. Let
V
be a projective algebraic
k
set,
n
= #
π
0
(
V
)
. Then there is an isomorphism of
k
algebras
O
(
V
)
∼
=
k
1
⊕
. . .
⊕
k
n
where each
k
i
is a ﬁnite ﬁeld extension of
k
. Moreover
n
X
i
[
k
i
:
k
] = #¯
π
0
(
V
)
.
In particular, if
V
is connected as an algebraic
K
set,
O
(
V
) =
K
.
Proof.
Let
V
1
, . . . , V
n
be connected components of
V
. It is clear that
O
(
V
)
∼
=
O
(
V
1
)
⊕
. . .
⊕O
(
V
n
)
so we may assume that
V
is connected. Let
f
∈ O
(
V
)
. It deﬁnes a regular
map
f
:
V
→
A
1
(
K
)
. Composing it with the inclusion
A
1
(
K
)
±
→
P
1
k
(
K
)
, we obtain
a regular map
f
0
:
V
→
P
1
k
(
K
)
. By Theorem
9.1
,
f
(
V
) =
f
0
(
V
)
is closed in
P
1
k
(
K
)
.
Since
f
(
V
)
⊂
A
1
k
(
K
)
, it is a proper closed subset, hence ﬁnite. Since
V
is connected,
f
(
V
)
must be connected (otherwise the preimage of a connected component of
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This note was uploaded on 01/08/2012 for the course MATH 299 taught by Professor Wei during the Spring '09 term at SUNY Stony Brook.
 Spring '09
 wei
 Algebra, Geometry

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