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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 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 rs in R. Now think about the polynomial
fi+1 -xni+1-n1f1 r1 -...- xni+1-nifiri
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.

Corollary If R is a commutative Noetherian ring then so is R[x1,...,xn].

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.

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