Let $P\in K[X_1,\dots,X_n]$ be a homogeneous polynomial. It then defines an affine variety $X=V(P)\subset \mathbb{A}^n_K$ like any polynomial, but also a projective variety $Y\subset \mathbb{P}^{n-1}_K$ by virtue of being homogeneous.
The restriction of $\mathbb{A}^n_K\setminus \{0\}\to \mathbb{P}^{n-1}_K$ is a surjective morphism $X\setminus \{0\} \to Y$. My question is: what is the geometric nature of this map? It is not quite a vector bundle because we took out $0$. Should I call it a fibration of some sort?
And more importantly to me, what does this yield on the level of functions fields? It seems rather clear to me that $K(X)$ is an extension of $K(Y)$ of transcendance degree $1$, but is it a purely transcendental extension in general? Even for the case of conics (so $X\subset \mathbb{A}^3_K$ and $Y\subset \mathbb{P}^2_K$ have degree $2$) I'm not quite sure I can answer that.
It suffices to analyze the situation affine-locally for both of these questions. Over $D(X_i)$, the map $\Bbb A^n\setminus\{0\} \to \Bbb P^{n-1}$ is the spectrum of $k[\frac{x_1}{x_i},\cdots,\frac{x_n}{x_i}]\to k[x_1,\cdots,x_n,x_i^{-1}]= k[\frac{x_1}{x_i},\cdots,\frac{x_n}{x_i},x_i,x_i^{-1}]$, so the fiber over each point is a copy of $\Bbb G_m$ and the function field extension is purely transcendental. This work just as well for any closed subvariety $X\subset \Bbb P^{n-1}$ with preimage $\pi^{-1}(X)\subset\Bbb A^n\setminus\{0\}$: if $I_i$ is the ideal of $X\cap D(X_i)$, then our map of rings over $D(X_i)$ becomes $k[\frac{x_1}{x_i},\cdots,\frac{x_n}{x_i}]/I_i \to k[\frac{x_1}{x_i},\cdots,\frac{x_n}{x_i},x_i,x_i^{-1}]/I_i[x_i,x_i^{-1}] \cong (k[\frac{x_1}{x_i},\cdots,\frac{x_n}{x_i}]/I_i)[x_i,x_i^{-1}]$.
As far as what to call this, a fibration is appropriate, or maybe you could call it a $\Bbb G_m$-bundle.