General Ring Theory/Noetherian rings

Definition (right-noetherian ring):

A commutative ring $R$ is called right-noetherian iff the set of all right ideals of $R$, ordered by inclusion, satisfies the ascending chain condition.

Left-noetherian rings are similarly defined.

Definition (noetherian ring):

A commutative ring $R$ is called noetherian iff the set of all ideals of $R$, ordered by inclusion, satisfies the ascending chain condition.

We will state and prove results only for right-noetherian rings, even though they are valid mutatis mutandis for left-noetherian and noetherian rings just as well.

Proposition (ideals of noetherian rings contain powers of their radicals):

Let $R$ be a noetherian ring, and let $I\leq R$ be an ideal. Then there exists $n\in \mathbb {N}$ such that

${\sqrt {I}}^{n}\subseteq I$.

Proof: Let $i_{1},\ldots ,i_{k}$ be a basis of $sqrt{I}$ considered as an $R$-module. Then choose $m\in \mathbb {N}$ sufficiently large so that

$\forall j\in [k]:i_{j}^{m}\in I$.

Then define $n:=km$ and observe that whenever $a=r_{1}i_{1}+\cdots +r_{k}i_{k}\in {\sqrt {I}}$, then by the pigeonhole principle, upon expanding the expression $a^{n}$ and considering each summand, we find that for each summand there is at least one $j\in [k]$ so that the corresponding power of $i_{j}$ is bigger than or equal to $m$. $\Box$

Proposition (elements of noetherian rings are products of irreducible elements):

Let $R$ be a noetherian ring, and let $a\in R$ be a non-unit. Then there exist irreducible elements $b_{1},\ldots ,b_{n}$ so that

$a=b_{1}\cdots b_{n}$.

(On the condition of the dependent choice.)

Proof: Indeed, a non-unit $a$ factors as $a=c_{1}c_{2}$, where $c_{2}$ is a non-unit and either $c_{2}$ is irreducible and $c_{1}$ is a unit, or $c_{1},c_{2}$ are both irreducible, or $c_{2}$ is not irreducible and a proper divisor of $a$. The same is true for $c_{2}=c_{3}\cdot c_{4}$, and proceeding inductively we gain an ascending chain

$\langle a\rangle \subseteq \langle c_{2}\rangle \subseteq \langle c_{4}\rangle \subseteq \langle c_{6}\rangle \subseteq \cdots$

which stabilizes by the noetherian assumption. But if $n\in \mathbb {N}$ is chosen large enough so that the sequence stabilizes after $c_{2n}$, $c_{2n}$ is irreducible. Hence, we may factor $a=d_{1}b_{1}$, where $b_{1}$ is irreducible, and continuing in the same fashion we obtain again an ascending sequence, whose stabilization implies the desired factorisation. $\Box$