Let $H \simeq \mathbb{P}^n$ be a hyperplane in $\mathbb{P}^{n+1}$, and $p$ a point not lying on $H$. For every point $x \in \mathbb{P}^{n + 1} \setminus \{p\}$, the line $\overline{px}$ in $\mathbb{P}^{n+1}$ meets $H$ in a single point $\pi_p(x)$. This gives rise to a mapping $\pi_p$. We can restrict this mapping onto a variety $X$. Specifically, I have a problem about the fibre of this mapping.
Let $X = Z_p(f) \subset \mathbb{CP}^{n + 1}$ be an irreducible hypersurface, where $f$ is an irreducible homogeneous polynomial of degree $d$. Let $p$ be a point outside $X$ and $\pi_p \colon X \to \mathbb{CP}^{n}$ be the linear projection defined as above. Consider the fibre $\pi_p^{-1} (y)$ for $y \in \mathbb{CP}^n$. My lecture notes claim that there exists a nonempty open set $U$ such that for every $y \in U$, $\pi_p^{-1} (y)$ has exactly $d$ points.
However, I suspect the correctness. Since we can do an appropriate base change to set $p = [0:\ldots:0:1]$ and therefore $\pi_p \colon [x_0 : \ldots :x_{n+1}] \mapsto [y_0 : \ldots :y_{n}]$. Therefore $\pi_p^{-1} (y)$ contains all points in $X$ with coordinate $[y_0 : \ldots :y_{n}:x_{n+1}]$. This reduce to a polynomial equation of $x_{n+1}$. By the fundamental theorem of algebra, there exists a root, so the fibre $\pi_p^{-1} (y)$ is well-defined. And there are at most $d$ roots, corresponding to at most $d$ points in the fibre. However, what happens if there isn't an $x_{n+1}^d$ term in $f$? Provided without multiple roots, less than $d$ roots cannot generate $d$ points in the fibre.
Moreover, how to prove this statement if we assume the polynomial always assume $d$ roots (not necessarily distinct)? I guess we can use the proposition that multiple roots exist if and only if the discriminant is zero.
Thanks for any comment.
You assumed that $p\notin X$, so $f(0,\cdots,0,1)\neq 0$, and therefore $f$ has a nonzero coefficient on $x_{n+1}^d$. So the fiber is always $d$ points counted with multiplicity.
To see that these fibers are generically $d$ distinct points, consider $f'$, the derivative of $f$ with respect to $x_{n+1}$. For a point $[y_0:\cdots:y_n]\in H$, if $f(y_0,\cdots,y_n,x_{n+1})$ and $f'(y_0,\cdots,y_n,x_{n+1})$ considered as polynomials in $x_{n+1}$ have a common root, then $f(y_0,\cdots,y_n,x_{n+1})$ has a multiple root. So the projection of $V(f,f')$ to $H$ consists of all the points where the fiber is not $d$ points, and this projection will be a proper closed subset since $V(f,f')\subsetneq V(f)$. (This last step relies on being in characteristic zero - $f'\notin (f)$ since $f'\neq 0$ and it has lower degree, but $f'\neq 0$ only holds in characteristic zero.)