Graphics Reference
In-Depth Information
10.5.13. Theorem.
(1) A hypersurface in
C
n
with minimal equation f = 0 is irreducible if and only if
f is irreducible.
(2) Every hypersurface in
C
n
is the union of irreducible hypersurfaces in essen-
tially a unique way.
Proof.
To prove (1), assume that V(f) is irreducible. Assume that f has minimal
degree and
fff
=
12
...
k
,
where the f
i
are irreducible and k > 1. If g = f
2
f
3
...f
k
, then f = f
1
g and
()
=
()
»
()
Vf
Vf
Vg
.
1
If V(g) Õ V(f
1
), then f
1
would vanish on V(g). This would imply that f
1
divides g, which
is a contradiction. Therefore, V(g) À V(f
1
). A similar argument shows that V(f
1
) À V(g).
But we now have a contradiction to the fact that V(f) is irreducible.
Next, assume that f is irreducible and that V(f) =
V
1
»
V
2
. Assume that
V
1
π
V(f). We shall prove that
V
2
= V(f), which will show that V(f) is not reducible.
Let
V
1
= V({g
i
}) and
V
2
= V({h
i
}). Now one of the g
i
, say g
1
, does not vanish on
V(f); otherwise, V(f) Õ
V
1
and it would follow that
V
1
= V(f). But g
1
h
j
vanishes on
V(f) for all j. By Corollary 10.5.7, f divides g
1
h
j
. Since f is irreducible and does
not divide g
1
, it must divide h
j
. It follows that V(f) Õ
V
2
. Clearly,
V
2
Õ V(f) and so
V
2
= V(f).
To prove (2), assume that the hypersurface
S
is defined by the minimal equation
f = 0, where
fff
=
12
...
k
,
the f
i
are irreducible, and k > 1. If
S
i
= V(f
i
), then the
S
i
are irreducible and distinct.
Furthermore,
=»»
»
.
..
S
.
SS S
1
2
k
To prove that the
S
i
are unique, assume that
=»»
»=»»»
.
..
STT
...
T
,
SS S
1
2
k
1
2
m
where the
T
j
are irreducible hypersurfaces and distinct. If
T
j
= V(g
j
), where g
j
are irre-
ducible polynomials, then each g
j
must be an associate of some f
i
by Theorem 10.5.8
and so
T
j
=
S
i
and (2) is proved.
Definition.
Let V(f) be a hypersurface defined by a polynomial f that has a factor-
ization as shown in equation (10.36). The sets V(f
i
) are called the
irreducible compo-
nents
, or simply
components
, of V(f).
The next result is useful when we want to localize a problem to affine space.
