Graphics Reference
In-Depth Information
n
[
]
k
Æ
set of
max
imal ideals in k X
,
X
,...,
X
12
n
Æ
()
pp
I
is one-to-one and onto.
What Corollary 10.8.9 shows is that the ring k[X
1
,X
2
,...,X
n
] completely deter-
mines k
n
when k is algebraically closed. Since one can define a topology on the set of
maximal ideals, k[X
1
,X
2
,...,X
n
] also determines the topology of k
n
. These facts are
actually not that surprising because they reflect similar facts about general topologi-
cal spaces. For example, a theorem of I. Gelfand and A. Kolmogoroff asserts that the
ring of continuous functions C(
X
) on a compact space
X
determines
X
. Specifically,
the map which sends a point
p
Œ
X
to the maximal ideal in C(
X
) consisting of the
functions that vanish on
p
is a homeomorphism. This theorem is false if
X
is not
compact, but generalizations are known.
Next, we look at Questions 1 and 2 above with respect to general varieties and
ideals. As a warm up, we show how the result in Theorem 10.5.13(1) regarding hyper-
surfaces translates into the language of ideals. First, given a polynomial f in k[X],
write f in the form
nn
n
k
fpp
=
1
...
p
,
12
k
where the p
i
are irreducible polynomials and p
i
and p
j
are nonassociates for i π j. Let
g = p
1
p
2
...p
k
be the minimal polynomial associated to V(f). Both f and g are poly-
nomials for the same hypersurface. What is the algebraic relation between f and g?
10.8.10. Lemma.
The ideal <g> is the radical of <f >.
Proof.
Easy.
10.8.11. Lemma.
If f is an irreducible polynomial in k[X], then the (principal) ideal
generated by f, <f >, is a prime ideal.
Proof.
Easy.
Lemma 10.8.11 generalizes to a characterization of the irreducibility of an arbi-
trary variety in terms of its ideal.
An affine variety
V
in k
n
10.8.12. Theorem.
is irreducible if and only if I(
V
) is a
prime ideal.
Proof.
First assume that
V
is irreducible and f
1
f
2
ΠI(
V
). Let
V
i
=
V
« V(f
i
). The sets
V
i
are varieties. Let
W
=
V
1
»
V
2
. Clearly,
W
Ã
V
. Furthermore, if (f
1
f
2
)(
p
) = 0, then
f
1
(
p
) = 0 or f
2
(
p
) = 0. This easily implies that
V
Ã
W
. Therefore,
V
=
W
. But
V
is irre-
ducible, so either
V
=
V
1
or
V
=
V
2
. In other words, either f
1
or f
2
vanishes on
V
, which
means that either f
1
or f
2
belongs to I(
V
). This shows that I(
V
) is prime.
Conversely, assume that I(
V
) is prime and
V
=
V
1
«
V
2
. Assume that
V
π
V
1
. We
shall show that
V
=
V
2
, which would prove that
V
is irreducible.
Claim.
I(
V
) = I(
V
2
).
