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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 $\dim_K(S/J)$.

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    $\begingroup$ No. Try $(x_1-1, x_1x_2)$. $\endgroup$ – Angelo Sep 27 '19 at 1:59
  • $\begingroup$ @Angelo Thank you for the response! Correct me if I'm wrong, but if I take deglex to be my monomial order, then $G_J=\{x_1-1-\alpha_1,x_2-\frac{\alpha_2}{1+\alpha_1}\}$, so $J$ is also zero-dimensional. $\endgroup$ – Stephen McKean Sep 27 '19 at 14:16
  • $\begingroup$ Expanding on the previous comment, in this example we have $G_I=\{x_1-1,x_2\}$, so $\dim_k(R/I)=1=\dim_K(S/J)$. $\endgroup$ – Stephen McKean Sep 27 '19 at 19:35
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    $\begingroup$ Sorry, I misread the question. The answer should be positive, by semicontinuity of the fiber dimension (EGA IV 13.1.3). $\endgroup$ – Angelo Sep 30 '19 at 18:25
  • $\begingroup$ If I'm thinking about this correctly, then generically the fibers will be zero-dimensional. But is there any reason why semicontinuity prevents the fiber dimension from jumping up at a prescribed point? $\endgroup$ – Stephen McKean Oct 16 '19 at 17:01
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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.)

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  • $\begingroup$ I hadn't made the connection that $\dim_K(S/J)$ was the cardinality of the generic fiber, and that's exactly what I needed. Thank you for both the explanation and the counter-example! $\endgroup$ – Stephen McKean Oct 17 '19 at 12:15

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