Is the total space of tautological line bundle an affine variety?

Question

Let $E_n$ denote the total space of the tautological line bundle $\mathcal{O}_{\mathbb{CP}^n}(-1)$ over $\mathbb{CP}^n$. That is, $$E_n := \{(l,v) \in \mathbb{CP}^n \times \mathbb{C}^{n+1}: v \in l \}.$$ Then $E_n$ is a subvariety of $\mathbb{CP}^n \times \mathbb{C}^{n+1}$. Precisely speaking, $E_n$ is a quasi-projective variety sitting inside $\mathbb{CP}^n \times \mathbb{CP}^{n+1}$, cut out by $$x_iy_j=x_jy_i, \quad \text{for} \ \ i,j = 0,1, \dots, n, \quad \text{and} \quad z \neq 0,$$ where $[x_0, \dots, x_n]$ and $[y_0,\dots,y_n,z]$ are homogeneous coordinates for $\mathbb{CP}^n$ and $\mathbb{CP}^{n+1}$, respectively. Is $E_n$ also an affine variety?

Answer 1

The answer was already implied in the comment by loch, but I will post it here for completeness.

The total space $E_n$ is not affine for $n>0$. The tautological line bundle over $\mathbb{CP}^n$ has a zero section $s: \mathbb{CP}^n \to E_n$ by sending a line $l \in \mathbb{CP}^n$ to $(l,0) \in \mathbb{CP}^n \times \mathbb{C}^{n+1}$. It's clear $s$ is a morphism (i.e., an algebraic map described by polynomials) and its image is isomorphic to $\mathbb{CP}^n$. If $E_n$ is affine, then the morphism $s$ has to be a constant map, which is impossible unless $n=0$.