I have a very simple question that I seem not to be able to answer by myself. I want to understand the following:
"If the structure morphism $f: X \to \operatorname{Spec}\mathbb{Z}$ is not dominant, its image is finite".
Can someone explain this? Do we need any extra assumptions like $X$ is of finite type? Is the topology on $\operatorname{Spec}\mathbb{Z}$ the discrete topology?
Thanks for the help!
This is not using any scheme theory. It is just topology (of course after one has figured out the topology on $\operatorname{Spec}\mathbb Z$).
The topology on $\operatorname{Spec}\mathbb Z$ is the co-finite topology, hence every infinite subset is dense, because all proper closed subsets are finite.
The second assertion follows from the fact that the image of an irreducible is again irreducible. And a finite subset $\operatorname{Spec}\mathbb Z$ is of course irreducible iff it is a singleton.