Irreducible polynomials and algebraic geometry I was reading Dummit and Foote and this was one of statements stated (without any proof), "An irreducible curve have finitely many singular points" I would like to know why is this true. Shouldn't it depend on what polynomial we choose? (here the irreducible polynomial is in two variables)
(Here we say "s" is a singular point if the partial derivatives of the irreducible polynomial $f(x,y)$ with respect to $x$ and $y$ is 0). Is this also true for polynomial with arbitrary number of variables?
 A: I don't know your background in algebraic geometry, but one way of seeing this is the following: It is well-known that on a variety (the vanishing set of a set of polynomials) over an algebraically closed field, the singular points form a proper closed set of the original variety, where "closed" means closed in the Zariski topology. 
In your case, it is easy to see that the singular points at least form a closed subset of your curve, since they are precisely the vanishing set of the partial derivatives and the polynomial that defines your curve. Moreover, using a bit of dimension theory, it is easy to see that on an irreducible curve, the only closed subsets are a finite set of points or the curve itself (and the empty set of course). So why can't every point on the curve be singular? 
Assume that $f$ is not constant and that if $f(x,y)=0$ then $f_x(x,y)=f_y(x,y)=0$, where $f_x$ and $f_y$ denote the partial derivatives of $f$. Fix $y_0$ such that $f(x,y_0)$ is not constant (assuming it exists); then the roots of $f(x,y_0)$ are contained in the roots of $f_x(x,y_0)$. However the degree of the latter polynomial is less than that of the former, and so we conclude that $f_x(x,y_0)=0$ for all $x$. This argument works even if $f(x,y_0)$ is constant. This implies that $f(x,y)$ only depends on the variable $y$. Using the same argument with the partial derivative with respect to $y$, we conclude that $f$ must be constant, a contradiction.
Using a similar argument, you can show that if $f$ is a polynomial in more than 2 variables, then its singular points cannot be the whole variety. However singular points need not be finite (and usually aren't).
For example: Let $f(x,y,z)=x^2+x^3+y^2$. The singular points are defined by the equations $2x+3x^2=0$, $y=0$, and so each point of the form $(0,0,z)$ is singular.
Clarification: As I said above, for any variety the singular points form a proper closed subset of the variety; the proof above just shows an easy proof for a hypersurface.
