Let $P_0,P_1,\ldots,P_r$ be distinct points in $\mathbb{P}^n$. Why there is a hyperplane $H$ in $\mathbb{P}^n$ passing through $P_0$ but not through any of $P_1,\ldots,P_r$?

# [Math] Hyperplane in projective space

algebraic-geometry

#### Related Solutions

Any non-constant morphism $\Phi\colon X\to Y$ between projective smooth curves has a degree $d$ and the morphism $\Phi$ will be an isomorphism if and only if that degree is $1$.

The incredibly good news is that you can calculate that degree by just looking at the fibre $\Phi^{-1}(y)$ of $\Phi$ at just one point ( any point!) of $y\in Y$: the degree is the dimension $d=dim_k \Gamma (\Phi^{-1}(y),\mathcal O)$ .

It is not terribly difficult to explain what the right-hand side means, but this is not even necessary here: we have $d=1$ as soon as for some non-empty $V\subset Y$ the restricted morphism $\Phi^{-1}(V)\to V$ is an isomorphism.

Since this is true in your case, we are done.

**Edit**

A more elementary proof (maybe the one Hartshorne had in mind at the level of Chapter 1, before "degree" is introduced) would be to consider the inverse isomorphism $\psi:V\to U$, to complete it to a morphism $\bar \psi:\mathbb P^1\to \mathbb P^1 $ (just as you did for $\phi$) and realize that $\bar \psi $ is an inverse to $\bar \phi $, which is thus an isomorphism.

The algebraic projective variety $V\subset \mathbb{P}^n$ is given by the zero locus of homogeneous polynomials $f_i\in \mathbb{C}[x_0,\dots,x_n]$.

Take now the open subset $U$ of $\mathbb{P}^n$ where $x_0\not=0$, which is an affine space. You can assume that $x_0=1$ and obtain then coordinates $x_1,\dots,x_n$. This gives an isomorphism $$\begin{array}{rcl}\mathbb{C}^n&\to& U\\ (x_1,\dots,x_n)&\mapsto & [1:x_1:\dots:x_n]\end{array}$$

Then $U\cap V$ is the locus of points of the form $[x_0:\dots:x_n]$ such that $x_0=1$ and $f_i(x_0,\dots,x_n)=0$. This shows that you obtain an affine variety given by the polynomials $f_i(1,x_1,\dots,x_n)$.

This process is very classical and can be founded in any course of algebraic geometry.

## Best Answer

By projective duality, your question is equivalent to asking why, given a finite collection of distinct hyperplanes in $\mathbb P^n$, there is a point lying on exactly one of them. Does that make it any easier?