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Let $k$ be a field. Under what assumptions on $k$ is the following true?

There exists a hypersurface $H$ of $\mathbb P_k^n$ such that $H$ does not contain any line?

By a hypersurface, I mean an algebraic hypersurface, i.e., "cut out" by a homogeneous polynomial in $k[x_0,\dots,x_n]$ of positive degree. I don't care about the degree of the hypersurface.

Such hypersurfaces exist if $k = \bar k$, as discussed in this post. But I am specifically interested in the case $k = \mathbb F_p$ for a prime $p$. If such hypersurfaces do not exist for this choice of $k$, the following weaker statement would be sufficient for my purposes.

For every large enough $n$, there exists a closed subscheme $X$ of $\mathbb P_k^n$ of dimension $n - O(1)$ such that $X$ does not contain any linear subspace of codimension $\mathrm{dim} X$.

EDIT (2024-06-13): Here is a version of the second statement avoiding big-$O$ notation since that is ambiguous in this context.

There exists a constant $c \in \mathbb N$ such that for every large enough $n$, there exists a closed subscheme $X$ of $\mathbb P_k^n$ of dimension $n - c$ such that $X$ does not contain any linear subspace of dimension $c$.

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  • $\begingroup$ Did you mean to write dimension $n-h^0(O(1))$ or something like that in your last paragraph? $\endgroup$
    – Tabes Bridges
    Commented Jun 12 at 15:23
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    $\begingroup$ @TabesBridges: I think $n-O(1)$ means a function of $n$ which differs from $n$ by a constant not depending on $n$ (nothing to do with the line bundle $O(1)$). $\endgroup$
    – Lazzaro Campeotti
    Commented Jun 12 at 15:27
  • $\begingroup$ @LazzaroCampeotti I suppose that does make more sense, and did cross my mind, but I cannot recall the last time I saw big O notation used in an AG context! $\endgroup$
    – Tabes Bridges
    Commented Jun 12 at 16:01
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    $\begingroup$ Over a finite field this can probably be done by counting, assuming that by a line you mean a line defined over $k$: Given a line, count the number of hypersurfaces of degree $d$ over $k$ containing it and then multiply this number by the number of lines (over $k$). If $d$ is sufficiently large this will be greater than the number of all hypersurfaces of degree $d$. $\endgroup$
    – naf
    Commented Jun 13 at 2:26
  • $\begingroup$ @LazzaroCampeotti is exactly right about the interpretation of O(1). Sorry for the ambiguous notation. $\endgroup$
    – ffx
    Commented Jun 13 at 8:32

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