Commutative Algebra/Noetherian rings

Rings as modules

Theorem 14.1:

We had already observed that a ring $R$ 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 $R$ 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.$\Box$

Transfer of the properties

Definition 14.2:

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

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

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

In analogy to theorem 6.11, we further obtain

Theorem 14.4:

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

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

Proof 2: Use theorem 6.11 directly.$\Box$

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 $R$ be Noetherian and let $M$ be a finitely generated $R$-module. Then $M$ is Noetherian.

Theorem 14.5 (Hilbert's basis theorem):

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

Proof 1:

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

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

Since $R$ is Noetherian, $J$ as a finite set of generators; call those generators $j_{1},\ldots ,j_{n}$. 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 $1\leq k\leq n$. Set

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

We further form the ideals $K:=\langle f_{1},\ldots ,f_{n}\rangle$ and $L:=\langle 1,x,x^{2},\ldots ,x^{d-1}\rangle \cap I$ 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 $g(x)\in I$, $\deg g=m$, 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 $r_{1},\ldots ,r_{n}\in R$ 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

$\deg(g-h)<m$

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

However, both $K$ and $L$ are finitely generated ideals ($\langle 1,x,x^{2},\ldots ,x^{d-1}\rangle$ 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.$\Box$

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 $R[x_{1},\ldots ,x_{n}]\to M$, where $n$ is the number of generators of $M$? If so, what does the first isomorphism theorem say to that?)