Let $R$ be a commutative ring with unity, then the following are equivalent
-1. Every ideal in $R$ is finitely generated
-3. Every nonempty collection of ideals of $R$ has a maximal element
I will also mention
-2. Every ascending chain of ideals in $A$ is stationary
which is also equivalent to 1 and 3.
I have proved $1\Rightarrow 3$ like this:
Let's suppose every ideal of $R$ is finitely generated and let $F$ be a nonempty family of ideals which does not have a maximal element. Let $I_1\in F$. As $I_1$ is not maximal there exists a $I_2\in F$ such that $I_1\subset I_2$. As $I_2$ is not maximal there exists $I_3$ such that...
And so we made the strictly ascending chain: $$I_1 \subset I_2 \subset \cdots$$ (Which clearly contradicts 2, but I will continue as if I didn't knew 2.)
Let's consider the ideal $$I=\bigcup\limits_{n=1}^\infty I_n$$ We know that $I$ is finitely generated, so $I=\langle x_1,\cdots,x_m\rangle$. We can easily note tht there exists some $n$ such that $x_1,\cdots,x_m\in I_n$, so $I_n=I$ which is a contradiction since the chain is not stationary.
I want to prove it ($1\Rightarrow 3$) in another way, without building an ascending chain of ideals, and without assuming 2.
Something like
Let's suppose every ideal in $R$ is finitely generated, and let $F$ be a nonempty family of ideals, and so ... ... ... $M$ is a maximal element of $F$.
Or
Let $F$ be a nonempty family of ideals which does not have a maximal element, and so ... ... (Not building ascending chains) ... ... we get a contradiction.
Is there any way to do it?
Note: I know I'm using some choice in the proof, I've no problem with that.
Note2: I usually see the proof of the equivalence as $1\Rightarrow 2\Rightarrow 3 \Rightarrow 1$, but I've found some proofs of $2\Rightarrow 1$, or $3\Rightarrow 2$ interesting, too.