Let $A$ be a commutative ring with identity. Then, the following are equivalent:
- Every set of ideals of $A$ has a maximal element;
- Every ascending chain of ideals in $A$ is stationary;
- Every ideal in $A$ is finitely generated.
I'd like my proof to be checked:
(1) $\Rightarrow$ (2): If $(J_i)_{i\in \mathbb{I}}$ is a chain of ideals in $A$ then $\{J_i: i \in \mathbb{I}\}$ has a maximal element $J_k$ and therefore is stationary.
(2) $\Rightarrow$ (3): Let $J$ be an ideal of a ring $A$. If $a \in J$, let $I_a = (a)$. If $I_a=J$ then $J$ is finitely generated. Otherwise, let $b \in J \setminus I_a$ and let $I_b = I_a + (b)$. Notice that $I_a \subset I_b \subset J$. Following by induction we have an ascending chain of ideals $(J_i)_{i \in I}$ (for some indexing family $I$), each one finitely generated and by hypothesis this chain is stationary. On the other hand, by construction $\bigcup_{i\in I} J_i = J$, therefore $J = J_k$ is finitely generated.
(3) $\Rightarrow$ (1): Let $X$ be a set of ideals which has no maximal element. Extract from this set an ascending chain $C$ of ideals that is not stationary. Now, as $C$ is a chain, $\cup C$ is an ideal. Because it is finitely generated, there is an $I \in C$ that contains all of its generators and thus $C$ is stationary. Absurd.