TITLE: Projections are finite morphisms
QUESTION [6 upvotes]: Let $X$ be a variety in $\Bbb{P}^n$. I would like to see as simply as possible why the projection of $X$ from a point is a finite map. Suppose $p=(1:0:\ldots:0)\notin X$ and let $\pi:X\rightarrow\Bbb{P}^{n-1}$ be defined as
$$ \pi(x_0:\ldots:x_n)=(x_1:\ldots:x_n)$$
To show finiteness we may consider $U_j=\{x\in X: \ x_j\neq0\}$ and show that the restriction
$$\pi:U_j\longrightarrow \pi(U_j)=Y\subset\Bbb{A}^{n-1}$$
is finite, i.e. any $g\in k[U_j]$ is an integral element over $k[Y]$. 
Edit: the question really is: how can we practically construct an integral equation for $g$, i.e.
$$g^k+b_{k-1}g^{k-1}+\cdots+b_0=0, \quad b_j\in k[Y]. $$

REPLY [3 votes]: Here's one way to see this. First, the structure morphism  $X \to \operatorname{Spec} k$ is proper since $X$ is projective. Furthermore because $\Bbb{P}^{n}_k$ is separated over $\operatorname{Spec} k$, the diagonal $\Delta$ of the structure morphism $\Bbb{P}^n_k \to \operatorname{Spec} k$ is a closed immersion, in particular proper. It follows that the map $X \to \Bbb{P}^{n-1}$ is proper. 
Now for any point $q \in \Bbb{P}^{n-1}$, the fibre $\pi^{-1}(q)$ is a proper closed subset of a line (because $p\notin X$) and hence is a finite set. In conclusion, $\pi$ is a proper quasi-finite morphism and thus is finite by Stacks, Tag $020G$.