---- > [!theorem] Theorem. ([[Hilbert's basis theorem]]) > Every [[subalgebra generated by a subset|finitely generated]] [[algebra]] over a [[Noetherian ring]] is [[Noetherian ring|Noetherian]] (as a ring). > > Since [[subalgebra generated by a subset|finitely generated algebras]] are, by definition, [[quotient ring|quotients]] of [[polynomial 4|polynomial algebras]], and [[module is Noetherian (resp. Artinian) iff submodule and quotient is]], an equivalent statement is as follows: if a [[ring]] $R$ is [[Noetherian ring|Noetherian]], then so is $R[T_{1},\dots,T_{n}]$. ^theorem > [!generalization] > Can in fact show the [[power series]] $R [ [ T_{1},\dots,T_{n}] ]$ is Noetherian when $R$ is. We do this in [[adic completions are well-behaved like localization is for Noetherian rings|power series over Noetherian ring is Noetherian]]. ^generalization Summary: - Enough to show $R$ Noetherian $\implies$ $R[T]$ Noetherian. So let $\mathfrak{a} \subset R[T]$ be an ideal; need to show $\mathfrak{a}$ is finitely-generated - Need to find a way to use Noetherianity of $R$, i.e., need to get from $\mathfrak{a}$ ideal(s) of $R$. Do this: for $i \geq 0$, define a certain ideal $\mathfrak{a}(i) \subset R$. Will have $\mathfrak{a}(i) \subset \mathfrak{a}(i+1)$. Since $R$ is Noetherian, this chain stabilizes and $\mathfrak{a}_{i}$ is f.g. This gives a table like this: ![[Pasted image 20250523223158.png|500]] - Find a way to turn the $r_{i,j} \in R$, $i=1,\dots,m$, $j=1,\dots,n_{i}$, into elements of $R[T]$. Then argue that these (finitely many) elements generate an ideal $\mathfrak{b}$. Argue that $\mathfrak{b}=\mathfrak{a}$; it will be useful to use the fact that $\mathfrak{b}(i)=\mathfrak{a}(i)$. > [!proof]+ Proof. ([[Hilbert's basis theorem]]) > We'll show $R[T_{1},\dots,T_{n}]$ is [[Noetherian ring|Noetherian]] when $R$ is. Since $R[T_{1},\dots,T_{n}] \cong R[T_{1},\dots,T_{n-1}][T_{n}]$, it suffices to prove the result for $R[T]$, then conclude the claim inductively. > > So let $\mathfrak{a}$ be an [[ideal]] of $R[T]$. We want to show $\mathfrak{a}$ is [[ideal generated by a subset|finitely generated]]. For $i \geq 0$, write $\mathfrak{a}(i)=\{ c_{0}: c_{0}T^{i} + \dots + c_{i}T^{0} \in \mathfrak{a}, c_{0},\dots,c_{i} \in R \} \subset R$ for the set of all leading coefficients of degree-$i$ polynomials in $\mathfrak{a}$ (and $0$). Then $\mathfrak{a}(i) \subset \mathfrak{a}(i+1)$ is an inclusion of ideals of $R$. Since $R$ is [[Noetherian ring|Noetherian]], the [[ascending chain condition|ascending chain]] $\mathfrak{a}(0) \subset \mathfrak{a}(1) \subset \dots$ stabilizes, say, $\mathfrak{a}(m')=\mathfrak{a}(m)$ for some $m \geq 0$ and all $m' \geq m$, and each ideal $\mathfrak{a}(i)$, $i \geq 0$, is generated by some finite subset $\{ r_{i,1}, \dots r_{i, n_{i}} \}$ of $R$. > > By the definition of $\mathfrak{a}(i)$, there is $f_{i,j} \in\mathfrak{a} \subset R[T]$ such that $f_{i,j}=r_{i,j}T^{i} + \{ \text{terms of degree } < i \text{ in }T \}$. Let $\mathfrak{b}$ be the ideal of $R[T]$ generated by the finite set $\{ f_{i,j} : 0 \leq i \leq m , 1 \leq j \leq n_{i}\} \subset R[T].$ > Can verify that $\mathfrak{b}(i)=\mathfrak{a}(i)$ for all $i \geq 0.$ > > We are done if we can show $\mathfrak{a}=\mathfrak{b}$. Certainly $\mathfrak{b} \subset \mathfrak{a}$. For the other inclusion, assume $\mathfrak{a} \neq \mathfrak{b}$. Then take $f \in \mathfrak{a} \setminus \mathfrak{b}$ of least degree $i$. Since $\mathfrak{b}(i)=\mathfrak{a}(i)$, there is $g \in \mathfrak{b}$ with the same leading coefficient as $f$, and so $\text{deg}(f-g)<i$. $f$ is of minimal degree, so $f-g$ must not be in $\mathfrak{a} \setminus \mathfrak{b}$, hence it must be in both $\mathfrak{a}$ and $\mathfrak{b}$... but then $f=\underbrace{ f-g }_{ \in \mathfrak{b} } +\underbrace{ g }_{ \in \mathfrak{b} }$ is in $\mathfrak{b}$, a contradiction. ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```