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Suppose $X\subset \mathbb{P}^n$ is a smooth hypersurface defined over $\mathbb{Q}$. For a "generic" prime $p$, what can be said about the set of hyperplanes $H$ in $\mathbb{P}^n(\mathbb{F}_p)$ for which $H \cap X$ is smooth over $\mathbb{F}_p$? For $p$ fixed and $X$ varying, by contrast, the situation can be arbitrarily bad: in fact, every hyperplane section of

${\sum_{i=1}^{n+1}X_i X_{i+n+1}^p=0} \subset \mathbb{P}^{2n+1}$ over $\mathbb{F}_p$ is singular.

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3 Answers 3

up vote 9 down vote accepted

Spread out $X$ over some $R=\mathbf{Z}[1/n]$ to a hypersurface $\mathcal{X} \subseteq \mathbf{P}^n_R$ that is smooth and projective over $R$. The standard proof of the Bertini smoothness theorem (as given in Hartshorne, Algebraic geometry, for instance) works over $R$: there is a Zariski dense open subscheme $U$ of the dual projective space $\mathbf{P}^n_R$ such that for $p \nmid n$, the hyperplanes $H$ in $\mathbf{P}^n_{\mathbf{F}_p}$ such that $H \cap \mathcal{X}_p$ is smooth are exactly those corresponding to $\mathbf{F}_p$-points of $U$. The complement of $U$ has at most $O(p^{n-1})$ points over $\mathbf{F}_p$ as $p \to \infty$, but $\#\mathbf{P}^n(\mathbf{F}_p)= p^n+p^{n-1}+\cdots+1$, so when $p$ is large enough, most hyperplanes over $\mathbf{F}_p$ will intersect the fiber $\mathcal{X}_p$ in something smooth.

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Thanks, Bjorn! Are there situations where the compliment of U has $O(p^{n-i})$ points for $i>1$? –  David Hansen Feb 28 '10 at 4:20
    
Yes, for instance if X itself is a hyperplane. But for most X, one cannot improve the exponent in O(p^{n-1}). It depends on the dimension of the generic fiber of Z:=P^n-U, since all but finitely many fibers above primes will have the same dimension. Usually this dimension will be n-1. –  Bjorn Poonen Feb 28 '10 at 5:17
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By Robinson's theorem in model theory, the reduction $X_p$ is smooth over $\bar{F_p}$ for almost all $p$. By Bertini's theorem, $X_p \cap H$ is thus smooth for H ranging over a dense subset of the hyperplane sections possible in $P^n(\bar{F_p})$.

I think the statement of nonsingularity can be stated within first-order logic. Nonsingularity is local, so assume $X$ affine. Let the ideal of $X$ be generated by $f_1, \dots, f_{n-d}$. Then nonsingularity means that each point where $f_1, \dots, f_{n-d}$ vanish, a certain Jacobian determinant is nonzero.

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Everything you wrote is correct, and in particular your remark about the good H's forming a dense subset of the dual P^n is key. The only reason I added an answer at all was to word it to say something about F_p-points and not just F_p-bar points, and to make it uniform as p varies. –  Bjorn Poonen Feb 28 '10 at 5:51
    
Good to know! I wasn't all that sure. –  Akhil Mathew Feb 28 '10 at 13:15
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You mey be interested in the paper: "Bertini Theorems over Finite Fields"(2002) Bjorn Poonen.

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