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 non-empty 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 $𝒮$ be a non-empty family of left ideals in $R$. Since $𝒮$ is non-empty, take any left ideal $I_1\in 𝒮$. If $I_1$ is maximal, then we are done. If not, $𝒮-\{I_1\}$ must be non-empty, 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 $𝒮-\{I_1,I_2\}$ 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 $𝒮$ be the family of all finitely generated ideals of $R$ contained in $I$. $𝒮$ is non-empty since $(0)$ is in it. By assumption $𝒮$ has a maximal element $J$. If $J\ne I$, then take an element $a\in I-J$. Then $⟨J,a⟩$ is finitely generated and contained in $I$, so an element of $𝒮$, 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.