David Hilbert's basis theorem is one of the most basic results in modern
algebra. It makes it possible to study
geometry in an algebraic way.
It's about
commutative Noetherian
rings so you'll want to read the
Noetherian node first. The
basis
in the statement is not a
basis for a vector space it refers to
a finite generating set for an ideal.
The Hilbert Basis Theorem
If R is a commutative Noetherian ring then so is R[x].
Proof: The proof I give here is not Hilbert's original one but
a short proof due to Heidrun Sarges.
Suppose that R[x] is not Noetherian.
We will show that this implies R isn't either.
By assumption there is an ideal I of R[x] that is
not finitely generated. Let f_{1} be a polynomial
in I that has the smallest possible degree amongst
polynomials in I. We are going to inductively pick some more
polynomials from I. So suppose that we already have
f_{i} (for some i>0). Then choose
f_{i+1} to be a polynomial in I that is
not in f_{1}R +...+ f_{i}R and has least possible
degree with this property.
Suppose that f_{i} has degree n_{i}
and leading coefficient a_{i}. Note that
we have n_{i} <= n_{i+1}, for all
i> 0, because of the way we chose the fs.
Now consider the chain of ideals of R
a_{1}R <= a_{1}R+a_{2}R <= ...
We will show that this chain does not become stationary.
Well suppose that
a_{1}R +...+ a_{i}R = a_{1}R +...+ a_{i+1}R
Thus we can write
a_{i+1}=a_{1}r_{1} +...+ a_{i}r_{i}
for some
rs in
R. Now think about the polynomial
f_{i+1} x^{ni+1n1}f_{1} r_{1} ...
x^{ni+1ni}f_{i}r_{i}
This polynomial has the properties

it is in I

it is not in f_{1}R +...+ f_{i}R

it has lower degree than f_{i+1}
This contradicts the way we chose
f_{i+1}
and the proof is finished.
Corollary
If R is a commutative Noetherian ring then so
is R[x_{1},...,x_{n}].
This follows by induction from the theorem. In particular we see that
the polynomial ring in any number of variables over a field or the
integers is a Noetherian ring.
If you read the proof carefully you can see that if we drop
the commutative assumption from the theorem and replace Noetherian
by right Noetherian the result still holds.