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Motivation This question came from my efforts to solve this problem presented by Andre Weil in 1951.

Can we prove the following theorem without Axiom of Choice?

Theorem Let $A$ be a commutative algebra of finite type over a field $k$. Let $I$ be an ideal of $A$. Let $\Omega(A)$ be the set of maximal ideals of $A$. Let $V(I)$ = {$\mathfrak{m} \in \Omega(A)$; $I \subset \mathfrak{m}$}. Let $f$ be any element of $\cap_{\mathfrak{m} \in V(I)} \mathfrak{m}$. Then there exists an integer $n \geq 1$ such that $f^n \in I$.

EDIT So what's the reason for the downvotes?

EDIT What's wrong with trying to prove it without using AC? When you are looking for a computer algorithm for solving a mathematical problem, such a proof may provide a hint. At least, you can be sure that there is a constructive proof.

EDIT To Martin Brandenburg, I think this thread also answers your question.

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    Take an arbitrary statement of non-trivial commutative algebra and ask if it holds in ZF - what's the point? Do you know that in ZF there might be nontrivial rings with no maximal ideals at all? So why should one expect that any "spectral" methods take over? Moreover, why do you answer your "question" yourself?2012-07-12
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    I can answer that last question, Martin: http://blog.stackoverflow.com/2011/07/its-ok-to-ask-and-answer-your-own-questions/2012-07-12
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    @MartinBrandenburg Before I answer your question, I'd like you to answer mine: http://math.stackexchange.com/questions/159699/the-group-of-invertible-fractional-ideals-of-a-noetherian-domain-of-dimension-12012-07-13
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    @MartinBrandenburg I edited my question. I hope you understand what the point is.2012-07-15

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