 Solution

The quotient $A$ is a field generated as a $k$-algebra by the images ${\alpha }_1,\dots ,{\alpha }_n$ of the variables. We prove the relevant form of Zariski lemma.

Choose a maximal algebraically independent subfamily, say ${\alpha }_1,\dots ,{\alpha }_r$, and putEvery remaining ${\alpha }_j$ is algebraic over $K$. After clearing the finitely many denominators in their monic equations, there is a nonzerosuch that all ${\alpha }_j$ with $j>r$ are integral overBecause $A$ is a field containing $R$ and is generated by these integral elements, $A$ is integral over $R$. If a field is integral over a subring, that subring is itself a field: for $0\ne c\in R$, an integral equation for $c^{-1}$ can be multiplied by a power of $c$ to express $c^{-1}$ in $R$.

If $r>0$, uncountability of $k$ lets us choose $a\in k$ such that ${\alpha }_1-a$ does not divide $h$. Then ${\alpha }_1-a$ is not invertible in the localization $R$, contradicting that $R$ is a field. Hence $r=0$, so every generator ${\alpha }_i$, and therefore every element of the finite algebraic extension they generate, is algebraic over $k$. Thus