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Let $\mathbb{A}^{n}$ be the $n$-dimensional affine space over a field $k$ and $H$ an arbitrary hypersurface of $\mathbb{A}^{n}$.

Q. Does there exist a hyperplane $P$ such that $P \cap H =\emptyset$?

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    $\begingroup$ For instance, take as $H$ the union of two non-parallel hyperplanes. Then no hyperplane can be disjoint from $H$. $\endgroup$ – Francesco Polizzi Jan 11 '18 at 15:54
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    $\begingroup$ @DonuArapura. Just to clarify: you are saying that for every $n\geq 2$, there exists $H$ a hypersurface such that for every $P$, the intersection is nonempty (i.e., answering the question for "arbitrary" hypersurfaces). Of course for some special $H$, there can exist $P$ with empty intersection, i.e., if $H$ is the zero scheme of $1+x_nf(x_1,\dots,x_n)$ and $P$ is the zero scheme of $x_n$. $\endgroup$ – Jason Starr Jan 11 '18 at 15:55
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    $\begingroup$ I think my example shows that the answer is negative, or am I missing something trivial? $\endgroup$ – Francesco Polizzi Jan 11 '18 at 15:58
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    $\begingroup$ @zeraouliarafik: how is this related to the OP's question? $\endgroup$ – Francesco Polizzi Jan 11 '18 at 16:10
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    $\begingroup$ Any general hypersurface (of degree at least 2) has no non-trivial units and then no hyperplane can be disjoint from it. $\endgroup$ – Mohan Jan 11 '18 at 17:26
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Let me expand my comment into a (negative) answer, so the question will not appear unanswered anymore.

Let us take as $H$ the union of two non-parallel hyperplanes of $\mathbb{A}^n$. Then, clearly, there exists no hyperplane $P$ disjoint from $H$.

In fact, as remarked by Mohan, a general affine hypersurface of $\mathbb{A}^n$ of degree at least $2$ intersects every hyperplane.

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