If $R$ is noetherian then every family $\mathcal{C}$ of ideals in $R$ has a maximal element in $\mathcal{C}.$

Question

Here is the question I want to answer:

If $R$ is noetherian then every family $\mathcal{C}$ of ideals in $R$ has a maximal element in $\mathcal{C}$ relative to the partial order of set inclusion.

Here is a trial for the solution:

Is this trial correct or I need to use the axiom of choice somewhere

Answer 1

You are using the axiom of choice exactly here:

If every chain is finite, then there is a maximal element.

This, as explained in the comments, is a weak version of Zorn's lemma, also known as Dependent Choice. The reason being that we have a partial order where every chain is finite, and therefore every chain has an upper bound: its largest element (because it's finite, and finite linear orders have a maximum).

And indeed, if Dependent Choice fails, then there is a partial order, a tree if you want to think about it, without maximal elements, where every chain is finite. That means that a recursive process cannot "go on forever", but nevertheless, it can always go "one more step".

This is counterintuitive, since we are used to Dependent Choice so much, that we don't even notice it. The key point is that recursion is defined with a function, so it's not "there is some ideal $A_2$ extending $A_1$", but rather a function that specifies which ideal that is. If you're only using the fact that "there is some extension", then you're subtly sneaking the axiom of choice into your proof in order to turn this into an actual recursive process.