Let $k$ be a field, and consider an irreducible polynomial $f \in k[x,y]$. Let $S(f)$ denote the singular points of $f$ (points that are simultaneously zero on $f$, the $x$-derivative of $f$, and the $y$-derivative of $f$.)


If $k$ is algebraically closed, then I can prove $S(f)$ is finite. Also, I can prove that if the field has characteristic $0$, then $S(f)$ is finite.


But what if the field has characteristic $p$ and is not algebraically closed? Is it true that $S(f)$ is finite?


I asked this question to my algebraic geometry professor last semester and stumped him! Hopefully one of you can think of a counterexample or proof.



Singular points of an irreducible polynomial
What do you mean by irreducible and what do you mean by $S(f)$? 


Does irreducible mean absolutely irreducible (ie irreducible over the algebraic closure of $k$)? Is $S(f)$ considered as a scheme or as a set of rational points? If the latter, then is $S(f) := \{ (a,b) \in k^2: f(a,b) = 0 = \frac{\partial f}{\partial x}(a,b) = \frac{\partial f}{\partial y}(a,b) \}$? Or is it the set of singular points over the algebraic closure of $k$?


If by irreducible, you mean absolutely irreducible, then as Douglas Zare suggests, you can pass to the algebraic closure and prove that $S(f)$ is finite.


If irreducible is to be read over $k$, but you are considering $S(f)$ scheme theoretically or are evaluating the points in the algebraic closure of $k$, then the assertion is false. Consider for instance $k$ of characteristic $p$ with $a \in k$ a non-$p^\mathrm{th}$ power and $f(x,y) = x^p + y^p + a$. 


Finally, if by $S(f)$ you mean the $k$-rational points, then if $S(f)(k)$ were infinite, then the set of $k$-rational points on the curve defined by $f$ would be infinite and $f$ would then be absolutely irreducible so that by the first case considered, $S(f)$ would be finite.


