| The results in this writeup are a sharpening of Hilbert's Nullstellensatz
so you might want to read that first. Also take a look at
Zariski topology.
Let k be an algebraically closed field (for example
C).
We are going to show there is a bijective correspondence between closed
subsets of kn (for the Zariski topology)
and radical ideals in the polynomial
ring k[x1,...,xn].
Thus we have a correspondence between geometric objects (the closed sets)
and algebraic objects (the ideals) and we can thus study geometry
using algebra.
Recall that if I is an ideal of
k[x1,...,xn] there is an associated
closed subset of kn
(for the Zariski topology) denoted by Z(I).
On the other hand if X is a subset of kn
then write
I(X) = { f in k[x1,...,xn] : f(x)=0, for all x in X }.
Note that I(X) is an ideal.
Theorem There is a bijective correspondence between radical ideals
of k[x1,...,xn] and closed subsets
of kn for the Zariski topology. In detail, if I
is a radical ideal then I=I(Z(I)) and if X is a closed set
then X=Z(I(X)).
Proof:
First the easy part.
I(X) vanishes on X so X<=Z(I(X)). On the other hand
suppose that X=Z({f1,...,fm}). Clearly
then f1,...,fm are in I(X)
and so X=Z({f1,...,fm})>=Z(I(X)).
This shows that the map X|--> I(X) is an injective map from
closed sets to ideals; in fact, to radical ideals.
For if fm is in I(X) then this means
that fm vanishes at all points of X. Obviously
then so does f and so f is in I(X).
To complete the proof we now have to show that for a radical ideal we have
I(Z(I))=I. In fact we prove the slightly more general fact
Lemma
For any ideal of k[x1,...,xn]
we have I(Z(I))=rad(I).
Proof:
Observe that we do at least have rad(I)<=I(Z(I)).
Since if fm is in I then f will
kill any point that is killed by fm.
The remainder of the proof uses a trick of Rabinowitsch. Let f
be a nonzero polynomial in I(Z(I)). we want to show that it is in the radical
of I. Introduce a new variable y and think about the ideal
J of the polynomial ring R=k[x1,...,xn,y]
that is generated by I and
fy-1. I claim that that J=R. Think about a point
p=(a1,...,an,b) in kn+1.
If this point is a zero of J then (a1,...,an)
is a zero of I. Thus, by assumption, f(a1,...,an)=0.
It follows that (fy-1)(p)=-1 This is absurd since p is
supposed to be a zero of fy-1. This contradiction shows that
J has no zeroes. If J is proper then it
is contained in a maximal ideal. But, by Hilbert's Nullstellensatz,
such a maximal ideal has a zero. it follows that J=R.
In particular there exist
fi in I and polynomials hi,h
in R such that
1 = f1h1 + ... + ftht + h(fy-1)
Now define a ring homomorphism
F:R-->k(x1,...,xn)
by mapping each xi to itself and y to 1/f.
Here k(x1,...,xn) denotes the field
of rational functions, i.e. the field of fractions of
k[x1,...,xn].
Applying F to the equation above we get
1 = f1F(h1) + ... + ftF(ht)
Of course each F(h1) has the form p/fr
with p in k[x1,...,xn].
Clearing the denominators we see that a power of f is in the ideal
I. The result is proven. | 
