I have asked this question in math.stackexchange but I did not have answers:

Let $k$ be an algebraically closed field and take $f_{1}, ..., f_{n} \in k[X_{1},..., X_{n}]$ with the jacobian condition: $\det J_{f} = 1$. Let $A:= k[X_{1},...,X_{n}]/(f_{1},...,f_{n})$ and consider the map induced:

$$f^{*}: Spec(A) \longrightarrow Spec(k)$$

In this case, $f^{*}$ is an finite map. Indeed, the jacobian condition implies that $A$ is an artinian $k$-algebra and so $l_{k}(A) = \dim_{k} A < \infty.$

My question is the following:

Assume that $k$ is only a domain. Is $f^{*}$ an finite map?

Let $k$ be an algebraically closed field and take $f_{1}, ..., f_{n} \in k[X_{1},..., X_{n}]$ with the jacobian condition: $\det J_{f} = 1$. Let $A:= k[X_{1},...,X_{n}]/(f_{1},...,f_{n})$ and consider the map induced:

$$f^{*}: Spec(A) \longrightarrow Spec(k)$$

In this case, $f^{*}$ is an finite map. Indeed, the jacobian condition implies that $A$ is an artinian $k$-algebra and so $l_{k}(A) = \dim_{k} A < \infty.$

My question is the following:

Assume that $k$ is only a domain. Is $f^{*}$ an finite map?

EDIT: In my original problem, $k$ is a DVR with characteristic $0$ and residue field finite.