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Proof.
If
V
is reducible, then
V
=
W
1
»
V
1
, where
W
1
and
V
1
are two proper sub-
varities of
V
. If one of these, say
V
1
is reducible, then
V
1
=
W
2
»
V
2
, where
W
2
and
V
2
are two proper subvarities of
V
1
. If the theorem is false for some
V
, then we could
continue on like this to get a strictly decreasing chain
V
…
V
1
…
V
2
… . . . of subvari-
eties
V
i
. This would give a strictly increasing chain of ideals I(
V
) Ã I(
V
1
) Ã I(
V
2
) Ã
...,contradicting the Hilbert Basis Theorem.
Definition.
A union
VV V
=»»»
1
...
V
2
n
is called an
irredundant union
if no
V
i
is contained in any
V
j
for i π j.
10.8.16. Theorem.
Let
VV V
= » »» = » »»
...
V W W
...
W
,
1
2
s
1
2
t
where the
V
i
and
W
j
are irreducible and we have irredundant unions. Then {
V
i
} = {
W
j
}.
Proof.
We have
=«=«
Ê
ˆ
˜
=
UU
(
)
VV VV
W
V W
«
.
Á
i
i
i
j
i
j
j
j
Since
V
i
is irreducible, we must have
V
i
=
V
i
«
W
j
for some j. In other words,
V
i
Õ
W
j
. A similar argument shows that
W
j
Õ
V
s
for some s. The inclusions
V
i
Õ
W
j
Õ
V
s
and the irredundancy of the unions imply that
V
i
=
W
j
.
Theorems 10.8.15 and 10.8.16 say that every variety can be expressed as a finite
irredundant union of irreducible varieties in only essentially one way. This justifies
the next definition that generalizes one in Section 10.5.
Definition.
If a variety
V
is written as a finite irredundant union of irreducible vari-
eties
V
i
, then the
V
i
are called the
irreducible components
of
V
.
Do the analogs of the last two results about varieties hold for ideals? The relevant
facts are:
(1) The Hilbert Basis Theorem (Theorem B.7.9) and Corollary B.7.4 imply that in
a polynomial ring over a field one gets a unique factorization of ideals into
primary ideals whose associated prime ideals are unique.
(2) Prime ideals are irreducible. (Irreducible ideals are not necessarily prime but
they are primary in a Noetherian ring. A primary ideal is not necessarily
irreducible.)
We do not quite get the unique decomposition of an arbitrary ideal into irreducible
ideals (see Exercise 10.8.2), but we are close. We can prove
10.8.17. Theorem.
If k is algebraically closed, then every radical ideal in k[X
1
,X
2
,
...,X
n
] can be expressed as a unique finite irredundant intersection of prime ideals.
