Suppose you have a zero-dimensional ideal $I=(f_1,...,f_n)$ in a polynomial ring $R=k[x_1,...,x_n]$ over a field $k$, so that $\dim_k(R/I)<\infty$. Take indeterminates $\alpha_1,...,\alpha_n$ and consider the ideal $J=(f_1-\alpha_1,...,f_n-\alpha_n)$ of the ring $S=K[x_1,...,x_n]$ over the field $K=k(\alpha_1,...,\alpha_n)$. Is it true that $J$ is also a zero-dimensional ideal?

I'm especially interested in the relationship between $\dim_K(S/J)$ and $\dim_k(R/I)$. In all the examples that I tried (with my very limited Macaulay2 abilities), I found that $\dim_K(S/J)=\dim_k(R/I)$.

My thought is to compute (reduced) Gröbner bases $G_I$ for $I$ and $G_J$ for $J$, and then to compare the leading monomials that show up in $G_I$ and $G_J$. Since $I$ is zero-dimensional, there are finitely many standard monomials of $G_I$, and my hope is that the standard monomials of $G_J$ are a subset of these.

If it helps, you may assume that $k$ is algebraically closed of characteristic 0 and each $\alpha_i$ is of the form $\alpha_i=f_i(z_1,...,z_n)$ for indeterminates $z_1,...,z_n$.

**Edit:** If you assume that $f=(f_1,...,f_n):\mathbb{A}^n_k\to\mathbb{A}^n_k$ is quasi-finite, then $R/I$ has Krull dimension zero and hence $I$ is zero-dimensional. Since quasi-finiteness is stable under base change and composition, one can show that $J$ is a zero-dimensional ideal of $S$. However, this doesn't seem to tell us anything about the relationship between $\dim_K(S/J)$ and $\dim_k(R/I)$.

One idea is to note that $\dim_K(S/IS)\leq\dim_k(R/I)$ and try to compare $\dim_K(S/IS)$ and $\dim_K(S/J)$.