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

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 $m$, 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?

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?

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

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 $m$, 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 $m$, $C(X) \le m$. These two cases don't cover everything though; for example, if $X$ is a point in $\mathbb{R}^n$ then $C(X) \le 2$.

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

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 $m$, 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?

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

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 $m$, the conclusion follows.

I am particularly interested in the case when $\deg(f_i) = 2$.

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

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 $m$, 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 $m$, $C(X) \le m$. These two cases don't cover everything though; for example, if $X$ is a point in $\mathbb{R}^n$ then $C(X) \le 2$.