I'm working through Shafarevich's Basic Algebraic Geometry, and one of the problems is: Prove that the curve given by the equation $f_{n-1}(x,y)+f_n(x,y)=0$ is rational if it is irreducible. Here $f_{n-1}$ and $f_n$ denote homogeneous polynomials of degree $n-1$ and $n$ respectively.
My solution goes like this: Consider the points of intersection of the curve with the line $y=tx.$ The $x$-values of these are zeroes of the univariate polynomial of degree $n$ $$f_{n-1}(x,tx)+f_n(x,tx)=0.$$ Because $f_{n-1}(x,y)$ is homogeneous of degree $n-1$, each term of $f_{n-1}(x,tx)$ is of the form $a_it^ix^{n-1}$ for some $0 \leq i \leq n-1$ and likewise for for $f_n.$ Therefore, $n-1$ of the $n$ roots of the equation is given by $x=0,$ while the other one is a rational function of $t,$ say $\phi(t).$ Because this is a point on the line $y=tx,$ clearly the $y$-coordinate of this point of intersection must be $t\phi(t),$ yielding a rational parametrization of the curve: $$x=\phi(t), y=t\phi(t).$$
Now, I'm fairly confident in my proof, but I fail to see where the hypotheses of irreducibility is used. In my mind, it has something to do with the fact that if the equation were reducible, the zero set would be two "separate" curves, while the line is only one "piece," so obviously the curve would need to be irreducible. However, I don't see how my proof breaks down in the case of a reducible curve. This leads me to suspect that my proof is not entirely airtight, hence my inclusion of the {proof-verification} tag. If anyone could tell me that my proof works, that'd be greatly appreciated! It would also be appreciated, though slightly less, if anyone could point out a flaw in it.