contained in all irreducible components of $C^f$. Since $Z^f(k)=\emptyset$ we see that $P$ is the only $k$-point of $C^f$. The intersection of $C^f$ with any line in ${\mathbb{P}}^2_k$ that does not pass through $P$ is isomorphic to $Z^f$, hence [*the smooth locus of $C^f$ contains an infinite subset of $C^f({\mathbb{A}}_k)=C^f({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}$*]{}, where the equality is due to Proposition \[p-basic\] (3).

Now let $g(x,y,z)$ be a homogeneous polynomial over $k$ of the same degree as $f(x,y)$ such that the subset of ${\mathbb{P}}^2_k$ given by $g(x,y,z)=f(x,y)=0$ consists of $d^2$ distinct $\bar k$-points. Consider the projective surface $Y \subseteq {\mathbb{P}}^2_k \times {\mathbb{P}}^1_k$ given by the equation $$\lambda f(x,y) + \mu g(x,y,z) = 0,$$ where $(\lambda:\mu)$ are homogeneous coordinates on ${\mathbb{P}}^1_k$. One immediately checks that $Y$ is smooth. The fibre of the projection $Y\to {\mathbb{P}}^1_k$ over $\infty=(1:0)$ is $C^f$.

Let $E$ be a smooth, geometrically integral, projective curve over $k$ containing exactly one $k$-rational point $M$. By [@P2] such curves exist over any global field $k$. Choose a dominant morphism $\vphi: E \to {\mathbb{P}}^1_k$ such that $\vphi(M)=\infty$, and $\vphi$ is not ramified over all the points of ${\mathbb{P}}^1_k$ where $Y$ has a singular fibre (including $\infty$). Define $$X = E \times_{{\mathbb{P}}^1_k} Y. \label{e3}$$

\[aa\] The 