Polynomial ring

Hilbert’s basis theorem

Let 𝑅 be a Noetherian ring. Then the polynomial ring 𝑅[𝑥] is also Noetherian. ring

Proof

Let 𝐼 ⊴𝑅[𝑥]. We prove 𝐼 is finitely generated. For 𝑓(𝑥) ∈𝑅[𝑥], let lead⁡𝑓 denote the leading coëfficient of 𝑓. Consider the ideal

𝐴={0}∪{𝑎∈𝑅:(∃𝑓(𝑥)∈𝐼)[lead⁡𝑓=𝑎]}⊴𝑅,

which is finitely generated since 𝑅 is noetherian. It follows there exist {𝑓𝑖(𝑥)}𝑟𝑖=1 ⊂𝐼 such that 𝐴 =⟨{𝑎𝑖}𝑟𝑖=1⟩𝑅 where

𝑎𝑖=lead⁡𝑓𝑖.

Let 𝑑 =max{deg⁡𝑓𝑖}𝑟𝑖=1, and consider the 𝑅-submodule

𝑀=⟨{𝑥𝑖}𝑑−1𝑖=0⟩𝑅≤𝑅𝑅[𝑥]

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

𝑀∩𝐼=⟨{𝑔𝑖(𝑥)}𝑠𝑖=1⟩𝑅≤𝑅𝑀

for some {𝑔𝑖(𝑥)}𝑠𝑖=1 ⊂𝐼.

We claim

𝐼=⟨{𝑓𝑖(𝑥)}𝑟𝑖=1∪{𝑔𝑖(𝑥)}𝑠𝑖=1⟩𝑅[𝑥]

which proves the theorem. It suffices to show 𝐼 ⊆⟨{𝑓𝑖(𝑥)}𝑟𝑖=1 ∪{𝑔𝑖(𝑥)}𝑠𝑖=1⟩𝑅. To this end, let 𝛼(𝑥) ∈𝐼. If 𝑒 :=deg⁡𝛼 ≥𝑑, let 𝑎 =lead⁡𝛼, whence there exist {𝑏𝑖}𝑟𝑖=1 ∈𝑅 such that

𝑎=𝑟∑𝑖=1𝑏𝑖𝑎𝑖

so

𝛼(𝑥)−𝑟∑𝑖=1𝑏𝑖𝑥𝑒−deg⁡𝑓𝑖𝑓𝑖(𝑥)

has degree <𝑒. By iterating this procedure one finds {𝛽𝑖}𝑟𝑖=1 ⊂𝑅[𝑥] such that

𝛼(𝑥)−𝑟∑𝑖=1𝛽𝑖(𝑥)𝑓𝑖(𝑥)

has degree <𝑑 and is thus contained in 𝑀 ∩𝑁, so

𝛼(𝑥)−𝑟∑𝑖=1𝛽𝑖(𝑥)𝑓𝑖(𝑥)=𝑠∑𝑖=1𝑐𝑖𝑔𝑖(𝑥)

for some {𝑐𝑖}𝑠𝑖=1 ⊂𝑅, whence

𝛼(𝑥)=𝑟∑𝑖=1𝛽𝑖(𝑥)𝑓𝑖(𝑥)+𝑠∑𝑖=1𝑐𝑖𝑔𝑖(𝑥)

as required.

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

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