Commutative Algebra/Noetherian rings

Rings as modules

Proof: Being a submodule means being an additive subgroup closed under the module operation. In the above context, this is exactly the definition of ideals.$◻$

Transfer of the properties

From theorems 6.7 and 14.1, we obtain the following characterisation of Noetherian rings:

In analogy to theorem 6.11, we further obtain

Proof 1: Proceed in analogy to theorem 6.11, using the isomorphism theorem of rings.$◻$

Proof 2: Use theorem 6.11 directly.$◻$

New properties in the ring setting

When rings are considered, several new properties show themselves in the noetherian case.

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1=Theorem 14.4:

Noetherian rings and constructions

In this section we will prove theorems involving Noetherian rings and module or localisation-like structures over them.

Proof 1:

Consider any ideal $I\le R[x]$. We form the ideal $J\le R$, that shall contain all the leading coefficients of any polynomials in $I$; that is

$a\in J:\iff \mathrm{\exists }f\in I:f(x)=a{x}^m+\text{(lower terms)}$.

Since $R$ is Noetherian, $J$ as a finite set of generators; call those generators $j_1,\dots ,j_n$. All $j_k$ belong to a certain $f_j\in R[x]$ as a leading coefficient; let thus $d_k$ be the degree of that polynomial for all $1\le k\le n$. Set

$d:=\underset{1\le k\le n}{\max }{d}_k$.

We further form the ideals $K:=⟨f_1,\dots ,f_n⟩$ and $L:=⟨1,x,x^2,\dots ,x^{d-1}⟩\cap I$ of $R[x]$ and claim that

$I=K+L$.

Indeed, certainly $K,L\subseteq I$ and thus $K+L\subseteq I$ (see the section on modules). The other direction is seen as thus: If $g(x)\in I$, $\mathrm{deg}g=m$, then we can set $a\in R$ to be the leading coefficient of $f$, write $a=r_1{j}_1+\cdots +r_n{j}_n$ for suitable $r_1,\dots ,r_n\in R$ and then subtract $h(x):=(r_1{j}_1{f}_1{x}^{m-d_1}+\cdots +r_n{j}_n{f}_n{x}^{m-d_n})$, to obtain

$\mathrm{deg}(g-h)<m$

so long as $m\ge d$. By repetition of this procedure, we subtract a polynomial $h^{\prime }$ of $g$ to obtain a polynomial in $L$, that is, $g\in K+L$.

However, both $K$ and $L$ are finitely generated ideals ($⟨1,x,x^2,\dots ,x^{d-1}⟩$ is finitely generated as an $R$-module and hence Noetherian by the previous theorem, which is why so is $L$ as a submodule of a Noetherian module). Since the sum of finitely generated ideals is clearly finitely generated, $I$ is finitely generated.$◻$

Exercises

• Let $R$ be a Noetherian ring, and let $M$ be an $R$-module. Prove that $M$ is Noetherian if and only if it is finitely generated. (Hint: Is there any surjective ring homomorphism $R[x_1,\dots ,x_n]\to M$, where $n$ is the number of generators of $M$? If so, what does the first isomorphism theorem say to that?)