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 $\dim\_K(S/J)$.



Given a zero-dimensional ideal $(f_1,...,f_n)$, is the ideal $(f_1-\alpha_1,...,f_n-\alpha_n)$ also zero-dimensional?
If $f$ is not generically finite-to-one then there are easy counterexamples (e.g. take $f\_1 = x\_1$, $f\_2 = x\_1 - 1 \in k[x\_1, x\_2]$). So assume $f$ is generically finite-to-one. 


 **Claim 1:**  If $f$ is generically finite-to-one and $\dim\_k(R/I) < \infty$, then $\dim\_k(R/I) \leq \dim\_K(S/J) < \infty$. 


 **Claim 2:**  It is possible in the situation of Claim 1 that $\dim\_k(R/I) <\dim\_K(S/J)$. 


For the first claim note that $\dim\_k(R/I)$ is the number (counted with multiplicity) of points in $f^{-1}(0)$ whereas $\dim\_K(S/J)$ is the number (counted with multiplicity) of points in a generic fiber of $f$. For the second claim you can take any $f: k^n \to k^n$ such that $f$ is generically finite-to-one, $f^{-1}(0)$ is finite, but $|f^{-1}(0)| < \deg(f)$ (where $\deg(f)$ is the cardinality of a generic fiber of $f$). For an explicit example, take $f\_1 = x\_1x\_2 - 1$, $f\_2 = x\_2(x\_1x\_2 - 1) + x\_1$. Then $\dim\_k(R/I) = 0$, whereas $\dim\_K(S/J) = 2$. (I knew of this map from a paper of Zbigniew Jelonek - it is one of the simplest maps from $k^2 \to k^2$ which is quasi-finite but non-surjective.) 


