# Commutative Algebra/Noetherian rings

## Rings as modules

Theorem 14.1:

We had already observed that a ring ${\displaystyle R}$ is a module over itself, where the module operation is given by multiplication and the addition by ring addition. In this context, we further have that the submodules of ${\displaystyle R}$ are exactly the ideals.

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.${\displaystyle \Box }$

## Transfer of the properties

Definition 14.2:

Let ${\displaystyle R}$ be a (commutative) ring. ${\displaystyle R}$ is called Noetherian if and only if every ascending chain of ideals of ${\displaystyle R}$

${\displaystyle I_{1}\subseteq I_{2}\subseteq \cdots \subseteq I_{k}\subseteq \cdots }$

eventually becomes stationary.

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

Theorem 14.3:

The following are equivalent:

1. ${\displaystyle R}$ is Noetherian.
2. Every ideal of ${\displaystyle R}$ is finitely generated.
3. Every set of ideals of ${\displaystyle R}$ has a maximal element with respect to inclusion.

In analogy to theorem 6.11, we further obtain

Theorem 14.4:

If ${\displaystyle R}$ is Noetherian, ${\displaystyle S}$ is another ring and ${\displaystyle \phi :R\to S}$ is a surjective ring homomorphism, then ${\displaystyle S}$ is Noetherian.

Proof 1: Proceed in analogy to theorem 6.11, using the isomorphism theorem of rings.${\displaystyle \Box }$

Proof 2: Use theorem 6.11 directly.${\displaystyle \Box }$

## 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.

Theorem 14.4:

Let ${\displaystyle R}$ be Noetherian and let ${\displaystyle M}$ be a finitely generated ${\displaystyle R}$-module. Then ${\displaystyle M}$ is Noetherian.

Theorem 14.5 (Hilbert's basis theorem):

Let ${\displaystyle R}$ be a Noetherian ring. Then the polynomial ring over ${\displaystyle R}$, ${\displaystyle R[x]}$, is also Noetherian.

Proof 1:

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

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

Since ${\displaystyle R}$ is Noetherian, ${\displaystyle J}$ as a finite set of generators; call those generators ${\displaystyle j_{1},\ldots ,j_{n}}$. All ${\displaystyle j_{k}}$ belong to a certain ${\displaystyle f_{j}\in R[x]}$ as a leading coefficient; let thus ${\displaystyle d_{k}}$ be the degree of that polynomial for all ${\displaystyle 1\leq k\leq n}$. Set

${\displaystyle d:=\max _{1\leq k\leq n}d_{k}}$.

We further form the ideals ${\displaystyle K:=\langle f_{1},\ldots ,f_{n}\rangle }$ and ${\displaystyle L:=\langle 1,x,x^{2},\ldots ,x^{d-1}\rangle \cap I}$ of ${\displaystyle R[x]}$ and claim that

${\displaystyle I=K+L}$.

Indeed, certainly ${\displaystyle K,L\subseteq I}$ and thus ${\displaystyle K+L\subseteq I}$ (see the section on modules). The other direction is seen as thus: If ${\displaystyle g(x)\in I}$, ${\displaystyle \deg g=m}$, then we can set ${\displaystyle a\in R}$ to be the leading coefficient of ${\displaystyle f}$, write ${\displaystyle a=r_{1}j_{1}+\cdots +r_{n}j_{n}}$ for suitable ${\displaystyle r_{1},\ldots ,r_{n}\in R}$ and then subtract ${\displaystyle h(x):=(r_{1}j_{1}f_{1}x^{m-d_{1}}+\cdots +r_{n}j_{n}f_{n}x^{m-d_{n}})}$, to obtain

${\displaystyle \deg(g-h)

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

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

### Exercises

• Let ${\displaystyle R}$ be a Noetherian ring, and let ${\displaystyle M}$ be an ${\displaystyle R}$-module. Prove that ${\displaystyle M}$ is Noetherian if and only if it is finitely generated. (Hint: Is there any surjective ring homomorphism ${\displaystyle R[x_{1},\ldots ,x_{n}]\to M}$, where ${\displaystyle n}$ is the number of generators of ${\displaystyle M}$? If so, what does the first isomorphism theorem say to that?)