# Finite maps and jacobian condition

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.

The map $R \to A$ is a special case of an étale morphism, and these are not always finite (merely quasi-finite, i.e. all fibres of $\operatorname{Spec} A \to \operatorname{Spec} R$ are finite).
Example. Let $R = k[Y]$, and let $A = R[X_1,X_2]/(X_1Y-1,X_1X_2-1)$. Then $$\det J = \det \begin{pmatrix}Y & 0 \\ X_2 & X_1 \end{pmatrix} = YX_1,$$ which equals $1$ in $A$. On the other hand, one easily computes $$A = R\left[\tfrac{1}{Y}\right],$$ which is evidently not finite over $R$. $\square$
• I used $\det J_f = 1 \in A$ instead of $\det J_f = 1 \in R[X_1,\ldots,X_n]$ because the latter is extremely rare. For $R = \mathbb C$, the Jacobian conjecture says that $\det J_f = 1 \in \mathbb C[X_1,\ldots,X_n]$ implies that $f \colon \mathbb A^n_{\mathbb C} \to \mathbb A^n_{\mathbb C}$ is an isomorphism. In particular, $A \cong \mathbb C$, which is not a very interesting case. – R. van Dobben de Bruyn Aug 13 '18 at 16:15
No. Take $f = tx^p + x$ over $k = \mathbf{F}_p[t]$.