I don't have a strong background in algebraic geometry, so my question could seems trivial, but I would like to know more about it. So if you have a book to recomend, thanks in advance!
Suppose you have a curve $C: f(x,y)=0$ over a finite field $K$.
Can I say that the affine curve $C$ is absolutely irreducible if it does not have singular points even if the homogenization of $C$ has singular points?
Knowing that the curve is absolutely irreducible and smooth. Can I apply the Hasse-Weil bound to the number of rational points of the affine curve $C$?
Unfortunately, no. The curve determined by $f(x,y)=x(x-1)$ is smooth in the plane, singular at infinity, and not absolutely irreducible. For something that's a little less cheat-y, consider $f(x,y)=x^2+x+1$ over $\Bbb F_2$ - this is smooth in the plane, irreducible, but not absolutely irreducible, and has a singular point at infinity.
Unfortunately, no. Removing the points at infinity can cause the Hasse-Weil bound to malfunction. The easiest example is $f(x,y)=x$: this is a curve of genus zero, so Hasse-Weil says it should have exactly $q+1$ points. But it only has $q$ points.