Zariski's lemma states that if $\mathbb{k}$ is any field and $m$ is maximal ideal in ring of polynomials $\mathbb{k}[x_1,\dots, x_n]$, then ${^{\displaystyle \mathbb{k}[x_1,\dots,x_n]}}\big/{_{\displaystyle m}}$ is a finite field extension of $\mathbb{k}$.
I want to know whether is true in somewhat sense the opposite statement:
Let $p$ be a prime ideal in $\mathbb{k}[x_1,\dots,x_n]$ such that the fraction field $\mathrm{Frac}\left({^{\displaystyle \mathbb{k}[x_1,\dots,x_n]}}\big/{_{\displaystyle p}}\right)$ is a finite field extension of $\mathbb{k}$. Then $p$ is maximal ideal.
I was trying to prove it by asserting that in this case for every $x_i$ there exists an irreducible polynomial $f_i(t) \in \mathbb{k}[t]$ of one variable such that $f_i(x_i)$ lies in $p$. Therefore $\left(f_1(x_1),\dots,f_n(x_n)\right) \subset p$. My next assumption was to check the maximality of $\left(f_1(x_1),\dots,f_n(x_n)\right)$ in $\mathbb{k}[x_1,\dots,x_n]$. But it seems to fail to be maximal, because after projectivizing we would get an intersection of $n$ degree $\mathrm{deg}~f_i$ hypersurfaces in $\mathbb{P}^n$, which is by Bezout's theorem a collection of $\mathrm{deg}~f_1 \cdot \ldots \cdot \mathrm{deg}~f_n$ points, while we want to get only one point.
However, the statement under consideration is vaguely true for me, because I couldn't find any counterexamples. Moreover, geometrically we consider field of rational functions on affine $\mathbb{k}$-variety. If this variety has positive dimension, it seems to have a transcendental extension of $\mathbb{k}$ as a field of rational functions on it.
Do you have any ideas?
P.S. If you have become interested, I am trying to show that a point $p$ of a locally finite type $\mathbb{k}$-scheme $X$ is a closed point if and only if $~{^{\displaystyle \mathcal{O}_{X,p}}}/{_{\displaystyle m_{X,p}}}$ is a finite field extension of $\mathbb{k}$.
