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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.

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    $\begingroup$ I think you would get a better reception to questions like this if you included some discussion of how Choice is used in the standard proofs. $\endgroup$ – Eric Wofsey Apr 22 '14 at 23:12
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    $\begingroup$ Please explain the reason for the downvote. Otherwise I cannot improve the question. $\endgroup$ – Makoto Kato Apr 22 '14 at 23:39
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    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Apr 23 '14 at 1:24
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    $\begingroup$ @EricWofsey At that point, it would seem that one should just switch to "pointless topology" and work with locales instead of topological spaces. $\endgroup$ – Zhen Lin Apr 23 '14 at 7:16
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    $\begingroup$ You seem to be interested in whether the general theory of Noetherian schemes can be developed as usual without assuming the Axiom of Choice. I've shown that with your choice of definitions, it cannot. However, I'm suggesting that instead of just accepting this negative answer, you should use a different definition (which is equivalent if you assume Choice) for which I believe you will get a positive answer. This is similar to how in the absence of Choice, you are "asking the wrong questions" if you define "Noetherian ring" to mean "all ascending chains of ideals stabilize". $\endgroup$ – Eric Wofsey Apr 23 '14 at 21:46

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