Can we prove an open affine subscheme of a noetherian scheme is noetherian without Axiom of Choice?

I'm interested in proving basic results of algebraic geometry without Axiom of Choice. As for why I think this is interesting, please see Pete L. Clark's answer to this question. To state my problem, I need some basic definitions.

Definition 1 A ring $A$ is called noetherian if every nonempty set of ideals of $A$ has a maximal element.

Definition 2 A scheme $X$ is called noetherian if it is a finite union of affine open subschemes Spec $A_i$, where each $A_i$ is noetherian.

My Question Let $X$ be a noetherian scheme. Let $U =$ Spec $A$ be a nonempty affine open subscheme of $X$. Can we prove that $A$ is noetherian without Axiom of Choice?

Remark I came up with this problem when I tried to solve this problem.

If the answer is negative, what conditions are needed to make it affirmative? $X$ should be separated, of finite type over a noetherian ring, etc?

The usual proof uses the following fact.

Let $X =$ Spec $A$ be an affine scheme. Suppose $X$ is a finite union of open affine subschemes Spec $A_{f_i}$, where each $A_{f_i}$ is noetherian. Then $A$ is noetherian.

To prove this, the usual proof uses the fact that the set $\{f_i\}$ generates $A$, which can be easily proved using Axiom of Choice.

Remark 2 The following observations might help.

Let $X$ be a noetherian scheme. Let $U =$ Spec $A$ be an affine open subscheme of $X$.

It can be proved without Axiom of Choice that the underlying topological space of $X$ is noetherian. See my answer to this question.

It is easy to prove withhout Axiom of Choice that every subspace of a noetherian topological space is noetherian. Hence the underlying topological space of $U =$ Spec $A$ is noetherian.

• If $A$ is a (necessarily non-Noetherian) nonzero ring with no prime ideals, then $\operatorname{Spec} A$ is Noetherian (it can be covered by the affine open subscheme $\operatorname{Spec} 0$) even though $A$ is not. It seems to me like what you really want to do is change the definition of "open cover" to require that (in the affine case) the set $\{f_i\}$ generates $A$, rather than just looking at points. This formulation is much more natural in the absence of Choice, and I expect it makes the theory work as usual. – Eric Wofsey Apr 23 '14 at 1:24