Hilbert's basis theorem

Let $𝑅$ be a Noetherian ring. Then the polynomial ring $𝑅 [𝑥]$ is also Noetherian. #m/thm/ring

Proof

Let $𝐼 ⊴ 𝑅 [𝑥]$ . We prove $𝐼$ is finitely generated. For $𝑓 (𝑥) \in 𝑅 [𝑥]$ , let $l e a d 𝑓$ denote the leading coëfficient of $𝑓$ . Consider the ideal

𝐴 = {0} \cup {𝑎 \in 𝑅 : (\exists 𝑓 (𝑥) \in 𝐼) [l e a d 𝑓 = 𝑎]} ⊴ 𝑅,

which is finitely generated since $𝑅$ is noetherian. It follows there exist ${𝑓 𝑖 (𝑥)} 𝑟 𝑖 = 1 \subset 𝐼$ such that $𝐴 = ⟨ {𝑎 𝑖} 𝑟 𝑖 = 1 ⟩ 𝑅$ where

𝑎 𝑖 = l e a d 𝑓 𝑖 .

Let $𝑑 = m a x {d e g 𝑓 𝑖} 𝑟 𝑖 = 1$ , and consider the $𝑅$ -submodule

𝑀 = ⟨ {𝑥 𝑖} 𝑑 - 1 𝑖 = 0 ⟩ 𝑅 \leq 𝑅 𝑅 [𝑥]

consisting of all polynomials of degree $< 𝑑$ . Since $𝑀$ is $𝑅$ -module-finite, it is $𝑅$ -module-noetherian by ^P2. Therefore

𝑀 \cap 𝐼 = ⟨ {𝑔 𝑖 (𝑥)} 𝑠 𝑖 = 1 ⟩ 𝑅 \leq 𝑅 𝑀

for some ${𝑔 𝑖 (𝑥)} 𝑠 𝑖 = 1 \subset 𝐼$ .

We claim

𝐼 = ⟨ {𝑓 𝑖 (𝑥)} 𝑟 𝑖 = 1 \cup {𝑔 𝑖 (𝑥)} 𝑠 𝑖 = 1 ⟩ 𝑅 [𝑥]

which proves the theorem. It suffices to show $𝐼 \subseteq ⟨ {𝑓 𝑖 (𝑥)} 𝑟 𝑖 = 1 \cup {𝑔 𝑖 (𝑥)} 𝑠 𝑖 = 1 ⟩ 𝑅$ . To this end, let $𝛼 (𝑥) \in 𝐼$ . If $𝑒 : = d e g 𝛼 \geq 𝑑$ , let $𝑎 = l e a d 𝛼$ , whence there exist ${𝑏 𝑖} 𝑟 𝑖 = 1 \in 𝑅$ such that

𝑎 = 𝑟 \sum 𝑖 = 1 𝑏 𝑖 𝑎 𝑖

𝛼 (𝑥) - 𝑟 \sum 𝑖 = 1 𝑏 𝑖 𝑥 𝑒 - d e g 𝑓 𝑖 𝑓 𝑖 (𝑥)

has degree $< 𝑒$ . By iterating this procedure one finds ${𝛽 𝑖} 𝑟 𝑖 = 1 \subset 𝑅 [𝑥]$ such that

𝛼 (𝑥) - 𝑟 \sum 𝑖 = 1 𝛽 𝑖 (𝑥) 𝑓 𝑖 (𝑥)

has degree $< 𝑑$ and is thus contained in $𝑀 \cap 𝑁$ , so

𝛼 (𝑥) - 𝑟 \sum 𝑖 = 1 𝛽 𝑖 (𝑥) 𝑓 𝑖 (𝑥) = 𝑠 \sum 𝑖 = 1 𝑐 𝑖 𝑔 𝑖 (𝑥)

for some ${𝑐 𝑖} 𝑠 𝑖 = 1 \subset 𝑅$ , whence

𝛼 (𝑥) = 𝑟 \sum 𝑖 = 1 𝛽 𝑖 (𝑥) 𝑓 𝑖 (𝑥) + 𝑠 \sum 𝑖 = 1 𝑐 𝑖 𝑔 𝑖 (𝑥)

as required.

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

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