's basis theorem is one of the most basic results in modern
. It makes it possible to study geometry
in an algebraic way.
It's about commutative Noetherian
s 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 f1 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
fi (for some i>0). Then choose
fi+1 to be a polynomial in I that is
not in f1R +...+ fiR and has least possible
degree with this property.
Suppose that fi has degree ni
and leading coefficient ai. Note that
we have ni <= ni+1, for all
i> 0, because of the way we chose the fs.
Now consider the chain of ideals of R
a1R <= a1R+a2R <= ...
We will show that this chain does not become stationary.
Well suppose that
a1R +...+ aiR = a1R +...+ ai+1R
Thus we can write
ai+1=a1r1 +...+ airi
for some r
s in R
. Now think about the polynomial
fi+1 -xni+1-n1f1 r1 -...-
This polynomial has the properties
it is in I
it is not in f1R +...+ fiR
it has lower degree than fi+1
This contradicts the way we chose fi+1
and the proof is finished.
If R is a commutative Noetherian ring then so
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.