The Nullstellensatz (which is German for "zeroes theorem") is one of the cornerstones of algebraic geometry. It gives us a correspondence between geometric objects and algebraic objects. This allows us to use algebra to study geometry.

Before we state the theorem some preliminaries. Fix k an algebraically closed field, for example one might take the complex numbers k=C. Now consider n-space consisting of all n-tuples

kn={ (a1,...,a1) : each ai is in k }.
This is a geometric object, an an n-tuple a=(a1,...,an) in kn is a point in n-space.

The corresponding algebraic object is the polynomial ring R=k[x1,...,xn]. For each point a in kn consider the ideal of R defined by

ma=(x1-a1)R + ... + (xn-an)R
Lemma
1. Each ma is a maximal ideal of k[x1,...,xn].
2. If a and b are distinct points then the ideals ma and mb are distinct
Proof: 1. Given any polynomial f=f(x)=f(x1,...,xn) we can evaluate it at a point a=(a1,...,an) to form f(a)=f((a1,...,an). Thus, we have a function ea:R-->k defined by ea(f)=f(a). Clearly this is a surjective ring homomorphism and its kernel contains ma. By the first isomorphism theorem we see that the kernel is a maximal ideal. It follows from this that ma is proper. We will show in a minute that it coincides with this kernel.

Next notice that there is an automorphism of rings of R defined by mapping each polynomial f(x1,...,xn) to f(x1-a1,...,xn-an). This automorphism takes the ideal m0 into ma so WLOG we only need to prove the result in the case a=0=(0,...,0). Now consider some polynomial f=f(x1,...,xn). Suppose that f is not in m0. f can be written as f=c + g, the sum of a constant term c in k plus a term g which is a sum of monomials all of which have total degree at least one. Thus g is in m0. We cannot have c=0 for then f would be in m0 after all. Thus c is nonzero and any ideal containing m0 and f must also contain c and hence be all of R.

2. If f is in ma then, from the above f(a)=0. Thus to show that ma and ma are distinct we just have to find f in ma that doesn't vanish at b. Since a and b are distinct it must be that for some i we have ai is different to bi. Consider then the polynomial xi-ai which is in ma. This polynomial takes the nonzero value bi-ai at the point b and so we see that it is not in mb.

Hilbert's Nullstellensatz Every maximal ideal of k[x1,...,xn] has the form ma for some point a in kn.

Here is a proof of Hilbert's Nullstellensatz. This result has an immediate corollary which is also sometimes called the Nullstellensatz.

Corollary There is a bijection between the set of points of kn and the set of maximal ideals of k[x1,...,xn]. If a is a point the corresponding maximal ideal is ma.

This corollary opens up lots of interesting questions, such as, can we generalise it to apply not just to points but to appropriate higher dimensional subsets of n-space? For example, if we consider the line in k2 consisting of all points (a,b) of the form 3a+2b=4 might this correspond to the ideal (not maximal this time) (3x+2y-4)k[x,y] in k[x,y]? The answer is yes, see the correspondence between closed sets for the Zariski topology and radical ideals in the polynomial ring.