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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$\begingroup$ I've deleted my earlier comment, which was based on a misreading of the question. My apologies. $\endgroup$– Steven LandsburgCommented Jul 15, 2013 at 15:52
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$\begingroup$ 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. $\endgroup$– Keenan KidwellCommented Jul 15, 2013 at 16:00
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$\begingroup$ Why finiteness of $Q(A/p)$ implies finiteness of $A/p$? $\endgroup$– user46336Commented Jul 15, 2013 at 16:06
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$\begingroup$ 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. $\endgroup$– Karl SchwedeCommented Jul 15, 2013 at 18:08
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$\begingroup$ Voted to close because this well-known fact is contained in every good introduction to algebraic geometry or commutative algebra - definitely not research level. $\endgroup$– Martin BrandenburgCommented Jul 15, 2013 at 20:41
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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.
answered Jul 15, 2013 at 16:35