Help me please.


Let $k$ be an algebraically closed field (I am mainly interested in $k = \overline{\mathbb{Q}}, \overline{\mathbb{F}\_q}$). Consider a plane curve $C \subset \mathbb{A}^2$ of degree $d$ over the rational function field $k(t)$. Suppose that $C$ is absolutely irreducible, i.e., irreducible over $\overline{k(t)}$.


Am I right that there is only the finite number $n$ of elements $t \in k$ such that the reduction of $C$ to $k$ is reducible ? Is there an upper bound for $n$?


Thank you in advance.



Is the reduction of an absolutely irreducible plane curve still irreducible except for the finite number of cases?
This is an answer to the further question in the comments. The original question is answered in the comments (as was the further question while I was typing this :-)).


The vector of coefficients of a curve of degree $d$ gives a point in $\mathbb{P}^{N\_d}, N\_d = (d+1)(d+2)/2-1$. Multiplication of equations defines a map $\mathbb{P}^{N\_r} \times \mathbb{P}^{N\_{d-r}} \to \mathbb{P}^{N\_d}$. The union of the images of these maps for $1 \le r \le d-1$ is a proper closed subset $X$ of $\mathbb{P}^{N\_d}$ consisting of the reducible equations.


A curve over $k(t)$ can be viewed as a map $\mathbb{P}^{1} \to \mathbb{P}^{N\_d}$ over $k$. If its image is contained in $X$ then the curve is not absolutely irreducible. Otherwise it only meets $X$ in a finite number $n$ of points and $n$ can be bounded as a constant depending on $d$ times the maximum of the degrees in $t$ of the coefficients of the curve.


