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.



Smoothness of hyperplane sections
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.


