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Let $H_1, \ldots, H_n$ be $n$ linearly independent hyperplanes in $k^n$, for some arbitrary field $k$. Let $X = H_1 \cup H_2 \cup \cdots \cup H_n$. Is it true that if $F=(f_1, \ldots, f_n)$ is a polynomial map from $k^n$ to $k^n$, such that $F(X) \cap F(k^n - X) = \emptyset$, then $\sum \deg(f_i) \ge n$?

This holds under the stronger condition that for all $a \in X$, $F(a)$ has at least one coordinate equal to zero, and for all $a \notin X$, $F(a)$ has all coordinates nonzero: $\prod f_i$ then cuts out $X$, but since any polynomial cutting out $X$ has degree at least $n$, the conclusion follows.

More generally, for a variety $X \subseteq k^n$, define $C(X)$ to be the minimum of the sum of the degrees of the coordinate functions over all polynomial maps $F$ where $F(X) \cap F(k^n - X) = \emptyset$. Is this quantity equivalent to something that's well known? You trivially have $C(X) \le n$ by taking $F$ to be the identity map. Also, if $X$ is defined by equations whose degree-sum equals $m$, $C(X) \le m$. Is one of these inequalities always sharp?

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    $\begingroup$ With regard to "Is one of these inequalities always sharp?": For variety defined by $x^{100} +y^{100}= 1$ , the polynomial map $(x,y)$ satisfies this property, even though its degree $m$ is much greater than $2$, and the number of variables $n$ can be arbitrarily large. $\endgroup$
    – Will Sawin
    Commented Mar 15, 2022 at 17:30

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Edit: The following argument does not work, as pointed by OP in the comments.

Proceed by induction on $n$. By your reasoning it holds for $n = 1$. For arbitrary $n \geq 2$, take a generic hyperplane $H$ distinct from the $H_1, \ldots, H_m$, and apply induction on the restriction of $F$ to $H$.

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  • $\begingroup$ I'm having trouble understanding the induction. Say $n=m=2$. Restricting $F$ to a generic line through the origin, by the $n=1$ case the sum of the degrees of the restriction is at least 1. But you want to say this sum is at least 2. How do you get the +1? $\endgroup$
    – Kevin
    Commented Jun 18, 2020 at 0:40
  • $\begingroup$ $m$ does not change. Since $H$ is generic, $H'_j := H \cap H_j$ will be nonempty for each $j$. Linear independence of $H_1, \ldots, H_m$ in $k^n$ "should" imply linear independence of $H'_1, \ldots, H'_m$ in $k^{n-1}$. In particular if $n = m = 2$, after restricting to a generic line you will have $n = 1$ and $m = 2$. $\endgroup$
    – pinaki
    Commented Jun 18, 2020 at 15:47
  • $\begingroup$ How can you have 2 linearly independent points in $k$? $\endgroup$
    – Kevin
    Commented Jun 18, 2020 at 16:00
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    $\begingroup$ Aah, I understand. I was assuming by "linear independence" you actually meant "in general position." You are right, with "linear independence" there is a problem with induction. But with "in general position" there is not. But then there is a problem with the base case. I did not think it through - my apologies. $\endgroup$
    – pinaki
    Commented Jun 18, 2020 at 16:21

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