equivalent defining conditions on a Noetherian ring
Let $R$ be a ring. Then the following are equivalent^{}:

1.
every left ideal^{} of $R$ is finitely generated^{},

2.
the ascending chain condition^{} on left ideals holds in $R$,

3.
every nonempty family of left ideals has a maximal element^{}.
Proof.
$(1\Rightarrow 2)$. Let ${I}_{1}\subseteq {I}_{2}\subseteq \mathrm{\cdots}$ be an ascending chain of left ideals in $R$. Let $I$ be the union of all ${I}_{j}$, $j=1,2,\mathrm{\dots}$. Then $I$ is a left ideal, and hence finitely generated, by, say, ${a}_{1},\mathrm{\cdots}{a}_{n}$. Now each ${a}_{i}$ belongs to some ${I}_{{\alpha}_{i}}$. Take the largest of these, say ${I}_{{\alpha}_{k}}$. Then ${a}_{i}\in {I}_{{\alpha}_{k}}$ for all $i=1,\mathrm{\dots},n$, and therefore $I\subseteq {I}_{{\alpha}_{k}}$. But ${I}_{{\alpha}_{k}}\subseteq I$ by the definition of $I$, the equality follows.
$(2\Rightarrow 3)$. Let $\mathcal{S}$ be a nonempty family of left ideals in $R$. Since $\mathcal{S}$ is nonempty, take any left ideal ${I}_{1}\in \mathcal{S}$. If ${I}_{1}$ is maximal, then we are done. If not, $\mathcal{S}\{{I}_{1}\}$ must be nonempty, such that pick ${I}_{2}$ from this collection^{} so that ${I}_{1}\subseteq {I}_{2}$ (we can find such ${I}_{2}$, for otherwise ${I}_{1}$ would be maximal). If ${I}_{2}$ is not maximal, pick ${I}_{3}$ from $\mathcal{S}\{{I}_{1},{I}_{2}\}$ such that ${I}_{1}\subseteq {I}_{2}\subseteq {I}_{3}$, and so on. By assumption^{}, this can not go on indefinitely. So for some positive integer $n$, we have ${I}_{n}={I}_{m}$ for all $m\ge n$, and ${I}_{n}$ is our desired maximal element.
$(3\Rightarrow 1)$. Let $I$ be a left ideal in $R$. Let $\mathcal{S}$ be the family of all finitely generated ideals of $R$ contained in $I$. $\mathcal{S}$ is nonempty since $(0)$ is in it. By assumption $\mathcal{S}$ has a maximal element $J$. If $J\ne I$, then take an element $a\in IJ$. Then $\u27e8J,a\u27e9$ is finitely generated and contained in $I$, so an element of $\mathcal{S}$, contradicting the maximality of $J$. Hence $J=I$, in other words, $I$ is finitely generated. ∎
A ring satisfying any, and hence all three, of the above conditions is defined to be a left Noetherian ring. A right Noetherian ring is similarly defined.
Title  equivalent defining conditions on a Noetherian ring 

Canonical name  EquivalentDefiningConditionsOnANoetherianRing 
Date of creation  20130322 18:04:27 
Last modified on  20130322 18:04:27 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  6 
Author  CWoo (3771) 
Entry type  Derivation 
Classification  msc 16P40 