Showing that the projection from $A^2$ to $A^1$ and $[x_0 : x_1] \to [x_0^l : x_1^l] $ are finite maps I am self-studying algebraic geometry mostly from these notes:
https://www.dpmms.cam.ac.uk/~cb496/ag2007-final.pdf
and this is the definition of finite map given:
Definition Let $X, Y$ be quasi-projective algebraic sets. A regular map
$$
f: X \rightarrow Y
$$
is finite, if for all $y \in Y$ there is some affine open neighbourhood $V \subset Y$ of $y$, such that $U:=f^{-1}(V)$ is affine and $\left.f\right|_U: U \rightarrow V$ induces a finite map of rings $f^*: k[V] \rightarrow k[U]$, i.e. the pullback map $f^*$ makes $k[U]$ a finitely generated $k[V]$-module.
Exercise Let $\phi: \mathbb{A}_k^2 \rightarrow \mathbb{A}_k^1$ be the regular map which sends $\left(x_1, x_2\right) \in$ $\mathbb{A}_k^2$ to $x_1 \in \mathbb{A}_k^1$. Prove that $\phi$ is not a finite map.
Now I know there are some additional lemmas that will make this problem trivial such as the fact that the preimage of a point of a finite map is finite, but I want to do this directly from the definition.
Now we know that open sets of $A^1$ are either the entire space of the complement of finite sets. However this is where I am a bit unsure of, as I don't know whether those open sets are in fact affine (isomorphic to some algebraic sets in A^n and possibly in A^2). If so I would have to check those neighborhoods too. But assuming none of those sets are affine, then I just have to check $A^1$ itself as the neighborhood.
So what I know from earlier theorems is that if a regular map between two affine sets is dense, then the pullback map is injective. Now, in that case, $f^* k[t]$ maps $t \to x$ and $1 \to 1$ and the reasoning as to why $k[x,y]$ cannot be finitely generated over $k[x]$ is just that elements in $k[x,y]$ contains polynomials with arbitrarily large degrees in $y$ while multiplying by elements in $k[x]$ cannot change the degree of $y$.
Ok so now, what do I do about the other open neighborhoods, and are those even affine neighborhoods?
Exercise: Let $\ell$ be a positive integer, then
$$
f: \mathbb{P}^1 \rightarrow \mathbb{P}^1, \quad\left[x_0: x_1\right] \mapsto\left[x_0^{\ell}: x_1^{\ell}\right]
$$
is finite.
So for this one, I think it is clear we want to use the standard affine cover. Now I have two questions about this:
So I don't think this map is surjective, so for the definition of a finite map, what do we do in the case where our point is not even in the codomain of the image of our map? Also I am having some difficulty imagining what the preimage of the affine covers are for this map. I feel like we ultimately want to somehow reduce to the case where $
\mathbb{A}^1 \rightarrow \mathbb{A}^1, t \mapsto t^{\ell} \text { is finite. }
$ but I'm not sure how we do what.
It would be great if someone can show me how to work out the details of this one using the above definition of finite maps as I am having difficulties going from the projective space to the affine charts.
 A: In fact every open subset of $\mathbb{A}^1$ is affine; one can check that $V(y(x-a_1)\cdots(x-a_n)-1)$ is isomorphic to $\mathbb{A}^1\setminus\{a_1,\ldots,a_n\}$. So unfortunately one has to consider every open neighborhood.
I think a concise way one can argue is as follows: as $\mathbb{A}^n$ is irreducible of dimension $n$, every open subset of it is also irreducible of dimension $n$ (this is a purely topological statement). Now let $V\subseteq \mathbb{A}^1$ be an affine open subset such that $U=f^{-1}(V)\subseteq \mathbb{A}^2$ is affine open as well (this is infact also true for every $V$), and such that $f^*:k[V]\to k[U]$ is an integral extension. It follows from the Going up and Going down theorems that then the Krull dimension of $k[U]$ is equal to the Krull dimension of $k[V]$. But the former must be $2$, and the latter must be $1$, so we have a contradiction. This can be generalized to see that a finite, dominant map can only happen between varieties of the same dimension.
Now for the second part: I think that $k$ is algebraically closed in the notes you are reeding, so the map is surjective. I will write $\mathbb{P}^1_{x,y}$ for the domain and $\mathbb{P}^1_{u,v}$ for the codomain, meaning that I write $x,y$ for the homogeneous coordinates on the former and $u,v$ on the latter. Now pick any point $y=[y_0:y_1]\in\mathbb{P}^1_{u,v}$ with $y_0\neq 0$, so that $y$ is in the standard affine open set $V=D(u)$. Now the preimage of $V$ is clearly just $U=D(x)$. Now we have to see what the pullback map does between the coordinate rings.
The regular functions of $V\subseteq \mathbb{P}^1_{u,v}$ are precisely $k[V]=k[v/u]$, and similarly we have $k[U]=k[y/x]$. Now precomposing the regular function $v/u$ by $f$ we see that $f^{*}(v/u)=y^l/x^l$. Hence if we identify $U$ and $V$ with $\mathbb{A}^1$, as you correctly guessed, we reduce to the map $\mathbb{A}^1\to\mathbb{A}^1$ mapping $t\to t^l$. On coordinate rings, this corresponds to the map $k[t]\to k[t]$ mapping $t$ to $t^l$. Hence the only thing to see is that $k[t]$ is finite over its subring $k[t^l]$. Can you see why this is the case?
