Hilbert's Nullstellensatz (idea)

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 a_i is in k }.

This is a geometric object, an an n-tuple a=(a₁,...,a_n) in kⁿ is a point in n-space.

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

m_a=(x₁-a₁)R + ... + (x_n-a_n)R

Lemma

1. Each m_a is a maximal ideal of k[x₁,...,x_n].
2. If a and b are distinct points then the ideals m_a and m_b are distinct

Proof: 1. Given any polynomial f=f(x)=f(x₁,...,x_n) we can evaluate it at a point a=(a₁,...,a_n) to form f(a)=f((a₁,...,a_n). Thus, we have a function e_a:R-->k defined by e_a(f)=f(a). Clearly this is a surjective ring homomorphism and its kernel contains m_a. By the first isomorphism theorem we see that the kernel is a maximal ideal. It follows from this that m_a 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,...,x_n) to f(x₁-a₁,...,x_n-a_n). This automorphism takes the ideal m₀ into m_a so WLOG we only need to prove the result in the case a=0=(0,...,0). Now consider some polynomial f=f(x₁,...,x_n). 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 m_a then, from the above f(a)=0. Thus to show that m_a and m_a are distinct we just have to find f in m_a that doesn't vanish at b. Since a and b are distinct it must be that for some i we have a_i is different to b_i. Consider then the polynomial x_i-a_i which is in m_a. This polynomial takes the nonzero value b_i-a_i at the point b and so we see that it is not in m_b.

Hilbert's Nullstellensatz Every maximal ideal of k[x₁,...,x_n] has the form m_a 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,...,x_n]. If a is a point the corresponding maximal ideal is m_a.

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.

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