# Given a zero-dimensional ideal $(f_1,...,f_n)$, is the ideal $(f_1-\alpha_1,...,f_n-\alpha_n)$ also zero-dimensional?

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 $$\dim_K(S/J)$$.

• No. Try $(x_1-1, x_1x_2)$. Sep 27, 2019 at 1:59
• @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. Sep 27, 2019 at 14:16
• Expanding on the previous comment, in this example we have $G_I=\{x_1-1,x_2\}$, so $\dim_k(R/I)=1=\dim_K(S/J)$. Sep 27, 2019 at 19:35
• Sorry, I misread the question. The answer should be positive, by semicontinuity of the fiber dimension (EGA IV 13.1.3). Sep 30, 2019 at 18:25
• 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? Oct 16, 2019 at 17:01

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.)
• 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! Oct 17, 2019 at 12:15