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?



Finite extension of a field
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.


