Finitely many singular points of an irreducible polynomial let $k$ be a field, and consider an irreducible polynomial $f∈k[x,y]$. 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, How can I prove that $S(f)$ is finite? 
Thanks,
Alex

The singularities of the curve $V(f)$ form the closed set $S(f)=V(f,\partial f/\partial x, \partial f/\partial y)\subset V(f)$  . 
Closedness of $S(f)$ is obvious but finiteness is not, contrary  to what you often read on this site and elsewhere.     
The basic (non-trivial!) result  is that two affine plane curves $V(f),V(g)\subset \mathbb A^2(k)$ intersect in finitely many points if $f,g$ have no common irreducible factor.
So in our case, since $f$ is irreducible, it suffices to prove that for example $f$ does not divide one of $\partial f/\partial x, \partial f/\partial y$,  which looks pretty obvious for reasons of degree.
 End of story? No, because $\partial f/\partial x$ (for example)  might be zero if $char. k=p$ !
For example $f=x^py+1$ is irreducible but nevertheless satisfies $\partial f/\partial x=0$ in characteristic $p$.
However we are saved because $\partial f/\partial y=x^p\neq 0$
This will always be the case: an irreducible polynomial $f(x,y)$  with $\partial f/\partial x=\partial f/\partial y=0$ is a  polynomial $f(x,y)=\phi(x^p,y^p)$ in $x^p,y^p$ and thus over an algebraically closed field is  a $p$-th power $f(x,y)=\phi(x^p,y^p)=\psi(x,y)^p$  and so is not  irreducible.
Contrapositively, if $f$ is irreducible then it is guaranteed to not divide at least one of $\partial f/\partial x,\partial f/\partial y$ and so, indeed, $S(f)=V(f,\partial f/\partial x, \partial f/\partial y)\subset V(f)$ is a finite (maybe empty) set.     
Conclusion
Yes, an irreducible plane curve has only finitely many singularities but you have to work to prove it!
