Polynomial ring

Hilbert's basis theorem

Let be a Noetherian ring. Then the polynomial ring is also Noetherian. #m/thm/ring

Proof

Let . We prove is finitely generated. For , let denote the leading coƫfficient of . Consider the ideal

which is finitely generated since is noetherian. It follows there exist such that where

Let , and consider the -submodule

consisting of all polynomials of degree . Since is -module-finite, it is -module-noetherian by ^P2. Therefore

for some .

We claim

which proves the theorem. It suffices to show . To this end, let . If , let , whence there exist such that

so

has degree . By iterating this procedure one finds such that

has degree and is thus contained in , so

for some , whence

as required.

It a simple corollary of this that any Commutative -monoid of finite type for noetherian is a noetherian ring, applying ^P1


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