TITLE: Commutative Algebra without the axiom of choice
QUESTION [15 upvotes]: It is well known that in a commutative ring with unit, every proper ideal is contained in a maximal ideal. The proof uses the axiom of choice. This fact, and others that are proved using essentially the same argument, anchor a large part of commutative algebra.
Suppose now that we disallow the use of the axiom of choice. My feeling is that this fact should still hold except for very pathological rings. I would find it odd if commutative algebra was entirely dependent on the axiom of choice. I also recall hearing that there was a workaround argument that did not use the axiom of choice for sufficiently nice rings, so this question is not entirely speculative.
So, assuming we work without the axiom of choice, for which rings can we prove that every proper ideal is contained in a maximal ideal? How is this done? And what characterizes the rings where we can't?

REPLY [2 votes]: [So, assuming we work without the axiom of choice, for which rings can we prove that every proper ideal is contained in a maximal ideal? How is this done?]
Most of rings you encounter in algebraic geometry.
For example, rings which are finitely generated over a field and their localizations.
For the proof, see this thread.