ag.algebraic geometry - 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.

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