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?
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
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.