TITLE: What is the projection of hypersurface?
QUESTION [5 upvotes]: My question refers to the basic terminology in (algebraic) geometry: let $V(f)\subset \mathbb{A}^n$ be some hypersurface (i.e. the set of all zeros of some polynomial $f$) in the affine space. Let $p\notin V(f)$. What is the projection of $V(f)$ from $(p_1,...p_n)=p\in \mathbb{A}^n-V(f)$ into the hyperplane $\{x_1=0\}$? I think it's a map which sends each $(x_1,...x_n)\in V(f)$ into $(x_1+t(p_1-x_1),...x_n+t(p_n-x_n))$ for some $t\in \mathbb{k}$ such that $x_1+t(p_1-x_1)=0$, isn't it? But this is not regular map. It must not be regular or my definition is false?

REPLY [2 votes]: First of all you must suppose $p_1\neq0$, else the projection clearly does not make sense (you would always project onto $p$ !).
A  parametric description of your line is $L_x(t)=p+t(x-p)$.
It will hit the  hyperplane $H=V(x_1)$ for $t_0=-\frac{p_1}{x_1-p_1}$ and the hitting point  is thus the required projection $\pi(x)=L_x(t_0)=p-\frac{p_1}{x_1-p_1}(x-p)$.
This projection  $\pi:V(f)--\to H$ is, as you correctly stated,  not  regular  everywhere: it is a rational map with poles on the intersection of $V(f)$ with the hyperplane $x_1-p_1=0$ : this is obvious geometrically since for those points the line $L_x(t)$ and the hyperplane $H$ are parallel and you can thus not define their projection .