Graphics Reference
In-Depth Information
{
}
()
=Œ
n
()
=
V S
p
k
f
p
0
for all f
Œ
S
.
10.8.2. Lemma.
Let S be an arbitrary set of polynomials in k[X
1
,X
2
,...,X
n
].
(1) V(S) = V(<S>).
(2) V(S) is an algebraic variety, that is, V(S) is the set of zeros of a
finite
set of
polynomials.
Proof.
Part (1) is easy. Part (2) follows from part (1) and the Hilbert Basis Theorem,
which asserts that an arbitrary ideal has a finite basis.
It follows from Lemma 10.8.2 that what seemed like a new concept is actu-
ally nothing new. We could have defined a variety in the above more general way right
at the beginning of the chapter (some authors do that) but we did not in order to
emphasize the fact that a variety is defined by a finite set of polynomials because that
finiteness property is important. In any case, nothing would have been simplified
because one would have to appeal to the Hilbert Basis Theorem at some point no
matter what.
At any rate, we now have correspondences
I
¨æ
æ
n
[
]
subsets of k
ideals in k X
,
X
,...,
X
.
æ
12
n
V
Our main goal in this section will be to analyze these two correspondences. In par-
ticular, we are interested in the following two questions:
Question 1.
Given an ideal I of polynomials, how does its algebraic structure influ-
ence the topological structure of the set of points V(I)?
Question 2.
Given a set of points
A
, how does its topological structure influence the
algebraic structure of the ideal I(
A
)?
It is clear, however, that without some restrictions on the domain of the correspon-
dences and the field k we will not be able to say very much. For example, V(X) = V(X
2
).
If k =
R
and n = 1, then V(X
2
+ 1) = V(<X
2
+ 1>) = f.
We start with some needed preliminary results. The first lists some simple prop-
erties of the I and V operators.
Let
V
,
V
1
,
V
2
be varieties in k
n
10.8.3. Theorem.
and let I, I
1
, I
2
be ideals in
k[X
1
,X
2
,...,X
n
].
(1) If
V
1
Õ
V
2
, then I(
V
2
) Õ I(
V
1
).
(2) If I
1
Õ I
2
, then V(I
2
) Õ V(I
1
).
(3) For all
V
, V(I(
V
)) =
V
. In particular, I is one-to-one on varieties.
Proof.
The proofs of (1) and (2) are easy and left to the reader. To prove (3) we
need to show two inclusions. Let
V
= V(f
1
,f
2
,...,f
n
). Let
I
=<f
1
,f
2
,...,f
n
>. Clearly,
V
= V(
I
).
