Prove directly that a regular function on $\mathbb{P}^n$ is a constant map.

Question

Exercise 3 of section 1.5.5 of Shavarevich's Basic Algebraic Geometry I asks us to prove directly that regular functions on $\mathbb{P}^n$ are constant. I understand how one proves this statement for irreducible projective varieties using the theorem that the image of a projective variety under a regular map is closed. However I am not sure how to prove it directly, which I assume means without using that theorem. Does anyone have any useful hints or tips?

Answer 1

For $n=1$, $\mathbb P^1$ can be obtained by gluing two open sets $U_1$ and $U_2$ which are both isomorphic to $\mathbb A^1.$ So a regular function on $\mathbb P^1$ would mean a regular function $f_1$ on $U_1$ and a regular function $f_2$ on $U_2$ whose restriction to $U_1 \cap U_2$ is identical. Let's write $f_1 = p(x)$ and $f_2= q(y).$ Then we need $f_1$ and $f_2$ to satisfy $p(T) = q(T^{-1})$ in the ring $k[T, T^{-1}],$ where $k$ is the base field. This implies that $p$ and $q$ should both have degree $0$ as polynomials and in fact should agree as constants. The general case follows a similar idea.