fraction field of polynomial ring that is a finite extension of the base field

Question

Let $k$ be a field. Let $P$ be a prime ideal of $k[x_1, ..., x_n]$. Let $K$ be a field of fractions of $k[x_1, ..., x_n]/P$. Suppose $K$ is a finite extension of $k$. Does it then follow that $P$ is a maximal ideal?

Answer 1

We first prove a separate result.

Let $R$ be an integral domain which contains a field $k$. If $R$ has finite dimension as a vector space over $k$, then $R$ is a field.

To see this, define, for each nonzero $v\in R$, a function $T_v:R\to R$ as $T_v(x)=vx$ for all $x\in R$.

It is clear that $T$ is a linear transformation. Since $R$ is an integral domain, we must have that $T_v$ is injective. Now by the fact that $R$ is finite dimensional over $k$ it follows that $T_v$ is in fact surjective and thus there exists $x_0\in R$ such that $T_v(x_0)=1=vx_0$. Therefore $v$ is invertible, and consequently $R$ is a field.

Now.

If $P$ is a prime ideal in $S=k[x_1,\ldots,x_n]$, we know that $R=k[x_1,\ldots,x_n]/P$ is an integral domain. Now if we further assume that the field of fractions of $R$ is a finite extension of $k$, we get that $R$ itself is finite dimensional over $k$. By the above proved theorem, $R$ itself is a field. This gives us that $P$ is a maximal ideal.