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
nspace consisting of all ntuples
k^{n}={ (a_{1},...,a_{1}) : each a_{i} is in k }.
This is a geometric object, an an
ntuple
a=(a_{1},...,a_{n}) in
k^{n}
is a
point in
nspace.
The corresponding algebraic object is
the polynomial ring R=k[x_{1},...,x_{n}].
For each point a in k^{n}
consider the ideal of R defined by
m_{a}=(x_{1}a_{1})R + ... + (x_{n}a_{n})R
Lemma

Each m_{a} is a maximal ideal of
k[x_{1},...,x_{n}].

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(x_{1},...,x_{n})
we can evaluate it at a point
a=(a_{1},...,a_{n})
to form
f(a)=f((a_{1},...,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(x_{1},...,x_{n})
to f(x_{1}a_{1},...,x_{n}a_{n}).
This automorphism takes the ideal m_{0}
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(x_{1},...,x_{n}).
Suppose that f is not in m_{0}.
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 m_{0}. We cannot have c=0
for then f would be in m_{0} after all.
Thus c is nonzero and any ideal containing m_{0}
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[x_{1},...,x_{n}] has the form m_{a}
for some point a in k^{n}.
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 k^{n}
and the set of maximal ideals of k[x_{1},...,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 nspace?
For example, if we consider the line in k^{2}
consisting of all points (a,b)
of the form 3a+2b=4
might this correspond to the ideal (not maximal this time)
(3x+2y4)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.
..