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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.

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