Let $n \in \mathbb{N}$ and $f,g \in k[x_1,x_2]$ homogenous and coprime polynomials where $\deg (f)=n$ and $\deg(g)=n-1$. Let $C$ be the variety $V(f-x_0g) \subset \mathbb{P}^2$. Show that the linear projection $\phi : \mathbb{P}^2 \to \mathbb{P}^1$ with center $z = (1:0:0)$ is an isomorphism between $C \backslash \{ z \}$ and an open subset $U \subset \mathbb{P}^1$.
I am trying to solve this exercise right now. I followed a hint that said I should prove that $C$ is irreducible first, which I have done by proving that $f-x_0g$ is irreducible (by reduction). Now I am stuck and would appreciate some help. I am not that familiar with morphisms of varieties and am not sure how to tackle the problem.
Thx in advance!
Edit: I am not so sure anymore why $C$ is irreducible. Can't I just say that the polynomial $f-x_0g \in k[x_1,x_2][x_0]$ has degree $1$ and is therefore irreducible?
This linear projection is the map $\phi : \mathbb P^2 \setminus z \to \mathbb P^1$ sending $[x_0 : x_1 : x_2] \mapsto [x_1, x_2]$. So the image $\phi(C \setminus \{z\})$ consists of all $[x_1 : x_2 ] \in \mathbb P^1$ for which there exists some $x_0 \in k$ such that $f(x_1, x_2) = x_0 g(x_1, g_2)$. In other words, $\phi(C \setminus \{ z \})$ consists of all points in $\mathbb P^1$ except for points where $g$ vanishes but $f$ doesn't vanish. But since $g$ and $f$ are coprime, there are no points where $g$ and $f$ both vanish. So $\phi(C \setminus \{ z \})$ is simply the set of points in $\mathbb P^1$ where $g$ doesn't vanish, i.e. $\phi(C \setminus \{ z \}) = \mathbb P^1 \setminus V(g)$. This is an open set.
To show that $\phi$ is an isomorphism, we need to exhibit an inverse morphism from $\mathbb P^1 \setminus V(g)$ to $C \setminus \{z \}$. This inverse is given by $[x_1 : x_2 ] \mapsto [\tfrac{f(x_1, x_2) }{ g(x_1, x_2)} : x_1, : x_2]$.
As for irreducibility, the fact that $f - x_0 g \in k[x_1, x_2][x_0]$ has degree $1$ means that the only way that $f - x_0 g$ can be written as a product is for there to exist $p, q, r\in k[x_1, x_2]$ such that $f - x_0 g = (p + x_0 q)r$. But then, $r$ would be a common factor of $f$ and $g$, so since $f$ and $g$ are coprime, $r$ must be a constant. I'm not sure why irreducibility is relevant to this exercise though.