Cryptography Reference
In-Depth Information
be decomposed in this way. This concept is most useful when working over an algebraically
closed field, but we give some of the theory in greater generality.
n
defined over
Definition 5.3.1
An affine algebraic set
X
⊆ A
k
is
k
-
reducible
if
X
=
X
1
∪
X
2
with
X
1
and
X
2
algebraic sets defined over
k
and
X
i
=
X
for
i
=
1
,
2. An affine
algebraic set is
-
irreducible
if t
he
re is no such decomposition. An affine algebraic set is
geometrically irreducible
if
X
is
k
k
-irreducible. An
affine variety
over
k
is a geometrically
irreducible algebraic set defined over
k
.
n
defined over
A projective algebraic set
X
⊆ P
k
is
k
-
irreducible
(respectively,
geomet-
n
rically irreducible
)if
X
is n
o
t the union
X
1
∪
X
2
of projective algebraic sets
X
1
,X
2
⊆ P
defined over
k
(respectively,
k
) such that
X
i
=
X
for
i
=
1
,
2. A
projective variety
over
k
is a geometrically irreducible projective algebraic set defined over
.
Let
X
be a variety (affine or projective). A
subvariety
of
X
over
k
k
is a subset
Y
⊆
X
that is a variety (affine or projective) defined over
k
.
This definition matches the usual topological definition of a set being irreducible if it is
not a union of proper closed subsets.
V
(
x
2
y
2
)
2
over
Example 5.3.2
The algebraic set
X
=
+
⊆ A
R
is
R
-irreducible. However,
over
C
we have
X
=
V
(
x
+
iy
)
∪
V
(
x
−
iy
) and so
X
is
C
-reducible.
yz,x
2
3
Exercise 5.3.3
Show that
X
=
V
(
wx
−
−
yz
)
⊆ P
is not irreducible.
It is often easy to determine that a reducible algebraic set is reducible, just by exhibiting
the non-trivial union. However, it is not necessarily easy to show that an irreducible algebraic
set is irreducible. We now give an algebraic criterion for irreducibility and some applications
of this result.
Theorem 5.3.4
Let X be an algebraic set (affine or projective). Then X is
k
-irreducible if
and only if I
k
(
X
)
is a prime ideal.
Proof
(
⇒
): Suppose
X
is irreducible and that there are elements
f,g
∈ k
[
x
] such that
fg
∈
I
k
(
X
). Then
X
⊆
V
(
fg
)
=
V
(
f
)
∪
V
(
g
), so
X
=
(
X
∩
V
(
f
))
∪
(
X
∩
V
(
g
)). Since
X
∩
V
(
f
) and
X
∩
V
(
g
) are algebraic sets it follows that either
X
=
X
∩
V
(
f
)or
X
=
X
∩
V
(
g
), and so
f
∈
I
k
(
X
)or
g
∈
I
k
(
X
).
(
⇐
): Suppose
I
=
I
k
(
X
) is a prime ideal and that
X
=
X
1
∪
X
2
where
X
1
and
X
2
are
k
I
k
(
X
2
). By parts 3 and 4 of Proposition
5.1.13
or parts 4 and 5 of Proposition
5.2.12
we have
I
-algebraic sets. Let
I
1
=
I
k
(
X
1
) and
I
2
=
⊆
I
1
,
I
⊆
I
2
and
I
=
I
1
∩
I
2
. Since
I
1
I
2
⊆
I
1
∩
I
2
=
I
and
I
is a prime ideal, then either
I
1
⊆
I
or
I
2
⊆
I
. Hence, either
I
=
I
1
or
I
=
I
2
and so, by part 6 of Proposition
5.1.13
or part 7 of Proposition
5.2.12
,
X
=
X
1
or
X
=
X
2
.
x
2
) is irreducible in
2
(
Exercise 5.3.5
Show that
V
(
y
−
A
k
).
