Cryptography Reference
In-Depth Information
n
be an algebraic set over
Exercise 5.3.6
Let
X
⊂ A
k
. Suppo
s
e there exist polynomials
f
1
,...,f
n
∈ k
[
t
] such that
X
={
(
f
1
(
t
)
,f
2
(
t
)
,...,f
n
(
t
)) :
t
∈ k}
. Prove that
X
is geomet-
rically irreducible.
Remark 5.3.7
A
k
-algebraic set
X
is the vanishing of polynomials in
k
[
x
1
,...,x
n
].
However, we say
X
is defined over
k
if
I
(
X
) is generated by polynomials in
k
[
x
1
,...,x
n
].
k
Hence, it is clear that an algebraic set defined over
-algebraic set. The converse
does not hold in general. However, if
X
is absolutely irreducible and
k
is a
k
is a perfect field
then these notions are equivalent (see Corollary 10.2.2 of Fried and Jarden [
197
] and use
the fact that when
X
is absolutely irreducible then the algebraic closure of
k
k
in
k
(
X
)is
k
).
Note that Corollary
5.1.20
proves a special case of this result.
2
or X
2
. Then X is geomet-
Theorem 5.3.8
Let X
=
V
(
f
(
x,y
))
⊂ A
=
V
(
f
(
x,y,z
))
⊆ P
rically irreducible if and only if f is irreducible over
k
.
Proof
If
f
=
gh
is a non-trivial factorisation of
f
then
X
=
V
(
f
)
=
V
(
g
)
∪
V
(
h
)is
reducible. Hence, if
X
is geometrically irreducible then
f
is
-irreducible.
Conversely, by Corollary
5.1.20
(respectively, Theorem
5.2.26
)wehave
I
k
(
V
(
f
))
k
=
(
f
).
Since
f
is irreducible it follows that (
f
) is a prime ideal and so
X
is irreducible.
Example 5.3.9
It is necessary to work over
k
for Theorem
5.3.8
. For example, let
f
(
x,y
)
=
y
2
x
2
(
x
1)
2
. Then
V
(
f
(
x,y
))
2
(
+
−
⊆ A
R
) consists of two points and so is reducible, even
though
f
(
x,y
)is
R
-irreducible.
Lemma 5.3.10
Let X be a variety and U a non-empty open subset of X. Then I
k
(
U
)
=
I
k
(
X
)
.
Proof
Since
U
⊆
X
we have
I
k
(
X
)
⊆
I
k
(
U
). Now let
f
∈
I
k
(
U
). Then
U
⊆
V
(
f
)
∩
X
.
Write
X
1
=
V
(
f
)
∩
X
, which is an algebraic set, and
X
2
=
X
−
U
, which is also an
=
X
1
∪
X
2
and, since
X
is irreducible and
X
2
=
=
algebraic set. Then
X
X
,
X
X
1
.In
∈
other words,
f
I
k
(
X
).
Exercise 5.3.11
Let
X
be an irreducible variety. Prove that if
U
1
,U
2
⊆
X
are non-empty
open sets then
U
1
∩
U
2
= ∅
.
5.4 Function fields
If
X
is a variety defined over
then
I
k
(
X
) is a prime ideal and so the affine or homogeneous
coordinate ring is an integral domain. One can therefore consider its field of fractions. If
X
is affine then the field of fractions has a natural interpretation as a set of maps
X
k
.
When
X
is projective then a ratio
f/g
of polynomials does not give a well-defined function
on
X
unless
f
and
g
are homogeneous of the same degree.
→ k
Definition 5.4.1
Let
X
be an affine variety defined over
k
.The
function field
k
(
X
)isthe
set
k
(
X
)
={
f
1
/f
2
:
f
1
,f
2
∈ k
[
X
]
,f
2
∈
I
k
(
X
)
}
