I am learning about rational maps at the moment, and the notes I am reading gives an example, the projection $\mathbb{P}_A^n \rightarrow \mathbb{P}_A^{n-1}$ given by $[x_0, ..., x_n] \rightarrow [x_0, ..., x_{n-1}]$.
I was wondering if someone could possibly explain to me how this is a rational map and what its domain of definition is? Thank you very much!
PS also what would be an easy way to describe this projection? Thank you!
PPS My definition of rational map from $X -\rightarrow Y$ here is, it is a morphism on a dense open set, with the equivalence relation $(f: U \rightarrow Y) \sim (g: V \rightarrow Y)$ if there is a dense open subset $Z \subseteq U \cap V$ such that $f|_{U \cap V} = g|_{U \cap V}$.
Okay, so first off, to be able to more concrete, note that even though by definition a rational map is an equivalence class of regular morphisms, we can always think of a rational map as just a regular map defined on some dense open subset. In fact, there is an exercise (not too difficult) in Hartshorne saying that there is always a unique open subset $U$ such that a rational map $X-->Y$ is represented by $U \to Y$. (just take the union of all the domain in the equivalence class)
Now for your question.
You're looking at the map $f:\mathbb P^n \to \mathbb P^{n-1}$ given by $[x_0,\ldots,x_n] \mapsto [x_0,\ldots,x_{n-1}]$. Okay, this is really not a map, because it is not defined when all $x_i=0$ for $i=0,\ldots,n-1$. But it is a map on the open subset where not all those $x_i$ are zero. In fact, the complement is just the point $[0,\ldots,0,1]$.
Thus we see that the restrion of $f$ to the open set $\mathbb P^n \backslash \{ [0,\ldots,0,1] \}$ is a well-defined map of sets. It is a regular map because it is a given by polynomials (here, variables).
In fact, it is easy to see that the pullback of a rational function $g:\mathbb P^{n-1} \to k$, to $\mathbb P^n$ is just $g(x_1,\ldots,x_{n-1},1)$, which is clearly regular. So by definition, $f$ is a regular map.