I'm reading Fulton's algebraic curves book on page 106 and I didn't understand this proof:
I didn't understand why can we assume $F_Y\neq 0$? (what $F$ irreducible has to do with this?). Afterwards Fulton says "$F$ doesn't divide $F_Y$" (why?), then he concludes that $F_Y(x,y)\neq 0$ (but he has already proved this!).
Why $d(F(x,y))=0$?
I'm sorry for these doubts, I'm almost sure these doubts are silly. Unfortunately, I really have problems with this subject.
Thanks
As long as one of $F_X$ or $F_Y$ is nonzero, we can suppose $F_Y\ne 0$, and if $F_X=F_Y=0$, then either $F$ is a constant, which is not the case, since $F$ is irreducible, or $k$ has characteristic $p$ and $F$ is a polynomial in $X^p$ and $Y^p$. Since $k$ is algebraically closed, the $p^{\text{th}}$ roots of all the coefficients exist, so $$F=\sum a_{ij}X^{pi}Y^{pj} = \left(\sum \sqrt[p]{a_{ij}} X^iY^j\right)^p.$$ Again since $F$ is irreducible, this is impossible. $F_Y$ has smaller $Y$ degree, so $F \nmid F_Y$. Actually he has proved $F_Y(X,Y)\ne 0$ not $F_Y(x,y)\ne 0$. You do need $F \nmid F_Y$ to conclude $F_Y(x,y)\ne 0$.
$F(x,y)=0$, so $d(F(x,y))=d(0)=0.$