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Is it true, that if $A$ is finitely generated commutatative algebra over a field $k$, not necessary algebraically closed, then prime ideal $p \subset A$ is maximal if and only if $k \subset Quot(A/p)$ is finite extension of fields?

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closed as off-topic by Steven Landsburg, Karl Schwede, Andrey Rekalo, Martin Brandenburg, Jack Huizenga Jul 15 '13 at 21:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Steven Landsburg, Martin Brandenburg, Jack Huizenga
If this question can be reworded to fit the rules in the help center, please edit the question.

I've deleted my earlier comment, which was based on a misreading of the question. My apologies. – Steven Landsburg Jul 15 '13 at 15:52
Yes. If $p$ is maximal, then $A/p$ is finite over $k$ by the Nullstellensatz. Conversely, if $A/p$ is finite over $k$, then $A/p$ must be a field, so $p$ is maximal. – Keenan Kidwell Jul 15 '13 at 16:00
Why finiteness of $Q(A/p)$ implies finiteness of $A/p$? – user46336 Jul 15 '13 at 16:06
This is reduces to a basic fact about integral domains and when finite extensions can be fields. A basic book on algebra (for example Dummit and Foote) has the references you want. – Karl Schwede Jul 15 '13 at 18:08
Voted to close because this well-known fact is contained in every good introduction to algebraic geometry or commutative algebra - definitely not research level. – Martin Brandenburg Jul 15 '13 at 20:41
up vote 1 down vote accepted

Suppose $k\subset Quot(A/p)$ is a finite extension of fields. Then every nonzero element $x$ of $A/p$ is algebraic over $k$ and so satisfies a minimal polynomial with non-zero constant term $a_0\in k$. Therefore $x$ divides $a_0$, so $x$ is a unit.

Because every nonzero element of $A/p$ is a unit, $A/p$ is a field, i.e. $p$ is maximal.

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