On a certain morphism of schemes from affine space to projective space. I'm currently reading Ravi vakil's notes on algebraic geometry, and I'm completly stuck on a question and I would love for you all to shed some light on it. The question is:
"Make sense of the following sentence: '' $$\mathbb{A}^{n+1}_k \setminus \{0\} \rightarrow \mathbb{P}^n_k$$ given by $$(x_0,x_1,..x_n) \rightarrow [x_0,x_1,..,x_n]$$ is a morphism of schemes.Caution: you can’t just say where points go; you have
to say where functions go. So you may have to divide these up into affines, and
describe the maps, and check that they glue"
OK, so I've tried to divide this up into affines, but I'm not sure exactly what this map is on these affina schemes to be honest. I feel that I have a clue, but I feel a bit uneasy about my arguments and my biggest problem is to check that they agree on overlaps. How could I do that?
 A: The nice thing about this map is that it is defined globally.  So it suffices to find an open cover $\mathcal{U} = \{ U_i : i \in I \}$ of $\mathbb{P}^n$ such that $\pi \vert_{\pi^{-1}(U_i)} :\pi^{-1}(U_i) \rightarrow U_i$ is a morphism.  (Here, $\pi : \mathbb{A}^{n+1} \setminus \{0\} \rightarrow \mathbb{P}^n$ denotes the given map.)  The fact that these maps agree on overlaps comes from the global definition of the map!  What has to be shown is that these maps are given by polynomial equations, i.e. that $\pi$ is locally given by polynomial maps.
The open cover to take is the standard one for projective space.  Let's check that you get a morpishm for $U = U_0 := \{ x = [x_0 : x_1 : \dots : x_n] \in \mathbb{P}^n : x_0 \neq 0 \}$.  It is easy to see that $\pi^{-1}(U) = \{ (x_0, x_1, \dots, x_n) \in \mathbb{A}^{n+1} \setminus \{0\} : x_0 \neq 0 \}$.  I claim that $U$ and $\pi^{-1}(U)$ are both affine varieties.  The coordinate ring of $U$ is $k[t_1,\dots,t_n]$, where $t_i = x_i / x_0$, while the coordinate ring of $\pi^{-1}(U)$ is $k[x_0,x_1,\dots,x_n]_{x_0}$.  (This latter ring is the localization of $k[x_0,x_1,\dots,x_n]$ at the multiplicatively closed subset $S = \{x_0^n : n \in \mathbb{N} \}$.)
Let $\psi$ denote $\pi$ restricted to $\pi^{-1}(U)$.  To be a morphism, the map $\psi : \pi^{-1}(U) \rightarrow U$ must correspond to a ring homomorphism $\psi^* : k[t_1, \dots, t_n] \rightarrow k[x_0,x_1,\dots,x_n]_{x_0}$.  I leave it as an exercise to check that $\psi^*(t_i) = x_i / x_0$.
