Let $k$ be an algebraically closed field, let $\Bbb A^{n+1}$ be the $(n+1)$-dimensional affine space, let $\Bbb P^{n}$ be the $n$-dimensional projective space, and let $q:\Bbb A^{n+1}\setminus\{0\}\to\Bbb P^{n}$ be the quotient map, that is $q(a)=k^{\times}\cdot a$.
I want to show that each preimage $q^{-1}\{P\}\subseteq\Bbb A^{n+1}\setminus\{0\}$, for $P\in\Bbb P^{n}$, is irreducible. That is, if $U,V\subseteq\Bbb A^{n+1}\setminus\{0\}$ are open subsets such that $U\cap q^{-1}\{P\}\neq\emptyset$ and $V\cap q^{-1}\{P\}\neq\emptyset$ then $(U\cap V)\cap q^{-1}\{P\}\neq\emptyset$. We can assume that $U$ and $V$ are basic opens and so assume that $U=D(f)$ and $V=D(g)$ for some $f,g\in k[x_{0},x_{1},\dots,x_{n}]$. Suppose that $P=q(a)$, for some $a\in\Bbb A^{n+1}\setminus\{0\}$. Then, the above assumptions imply that there exist $\lambda,\mu\in k^{\times}$ such that $\lambda \cdot a\in D(f)$ and $\mu\cdot a\in D(g)$, or equivalently that $f(\lambda a)\neq 0$ and $g(\mu\cdot a)\neq 0$. Now, to prove that $D(f)\cap D(g)\cap q^{-1}\{P\}=D(fg)\cap q^{-1}\{P\}\neq \emptyset$ I must find an element $\nu\in k^{\times}$ such that $\nu\cdot a\in D(fg)$ ($\iff (fg)(\nu\cdot a)\neq 0)$.
I have no clue on how to find this element $\nu$. I have found a counterexample to the choice $\nu=\lambda\cdot \mu$.
I'm quite rusty when it comes to algebraic geometry but here's an idea:
The fiber over $p = [1 \colon a_1 \colon \cdots \colon a_n]$ consists of ponts $(x_1, \ldots, x_n)$ such that $(x_0, x_1,\ldots,x_n) = (\mu,\mu a_1,\ldots, \mu a_n)$ for some nonzero $\mu$. Equivalently, we want $x_0 \neq 0$ and
$$ x_j =x_0a_j $$
for each positive $j$, i.e. $x_j-x_0a_j = 0$ for all $j$.
Hence the preimage is
$$ q^{-1}(p) = V(x_j-x_0a_j : j > 0) \cap D_{x_0}. $$
The map $x \in \Bbb A^1 \mapsto (x,x a_1\ldots,x a_n) \in V(x_j-x_0a_j : j > 0)$ shoulld correstrict to an isomorphism
$$ \Bbb A^1 \setminus \{0\} \to V(x_j-x_0a_j : j > 0) \cap D_{x_0}. $$
So we're left with showing that $\Bbb A^1 \setminus \{0\} \simeq V(xy-1) \subset \Bbb A^2$ is irreducible, but indeed looking at its coordinate ring we see that
$$ k[x,y]/(xy-1) \simeq k[t,t^{-1}] $$
is a domain.