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.

See also my answer to this question.