Let $A$ be a finitely generated algebra over some field $k$, and assume that $A$ is an integral domain. By Noether normalization, we have a finite injective ring homomorphism $\varphi: k[x_1, \ldots, x_n] \hookrightarrow A$.
Is $A$ always a free module over $k[x_1, \ldots, x_n]$?
I know that this fails for arbitrary finite type $k$-algebras, as zero-divisors can produce torsion elements (see e.g. Can generators of Noether normalization be zero divisors?). But as $A$ is a domain, $A$ is torsion-free.
The limited amount of examples I tried (coordinate rings of hyperelliptic curves given by $y^2 = f(x)$, the free generators over $k[x]$ being $1$ and $y$) seem to suggest that $A$ is, in fact, free.