This is Exercise 2.3.22 in Liu's book.
The goal is to show that the quotient in the category of schemes of $\mathbb{A}^1_k$ under the action of $\mathbb{Z}$ given by $n : T \to T + n$ (on the ring $k[T]$) is $\operatorname{Spec} k$. Here $k$ is a field of characteristic $0$.
We want to show that given a scheme $Z$ and a morphism $f : \mathbb{A}^1_k \to Z$ such that $f = f \circ n$ for all $n \in \mathbb{Z}$, there exists a unique $\phi : \operatorname{Spec} k \to Z$ such that $f = \phi \circ p$, where $p$ is the morphism from $\mathbb{A}^1_k \to \operatorname{Spec} k$ induced by the inclusion $k \to k[T]$.
If $Z$ is affine, then I'm able to show that this is true by turning the statement into a question about commutative rings, and using the requirement on $f$. But if $Z$ is not affine, I'm stuck trying to show that there exists an affine open subset of $Z$ that contains the entire image of $f$.
My only thought is to use that $\mathbb{A}^1_k$ is connected, and that $f$ is continuous, but I feel the solution should leverage the scheme structure of $Z$, although I am not sure how.
Any help is appreciated!
Best Answer
I think my hint was misleading, and in fact the sketch of a solution I had in mind turned out to be wrong, so here is an argument. The idea is that instead of looking at the images of closed points, we may look at the image of a generic point and see that it is actually the unique point of the image.
Let $\eta$ be the generic point of $\mathbb{A}^1_k$ and consider its image $z=f(\eta)$. Then the whole $\mathbb{A}^1_k$ is mapped to the closure of $z$, and we may replace $Z$ by this closure. So from now on we may assume that $\eta$ is mapped to the generic point $z$ of $Z$.
Consider a point $z'\in \mathrm{im}(f)$. Then the preimage of the closure $\overline{z'}$ is a nonempty closed $G$-invariant subset, which must therefore be all of $\mathbb{A}^1_k$. But we know that $f(\eta)=z$, so $z\in\overline{z'}$. As $z$ is the generic point of $z$, we have $\overline{z'}$ and therefore, $z=z'$.