Let $f:X\rightarrow Y$ be a morphism of schemes. Can the image of $f$ endowed with the subspace topology not be sober?
Let $X$ be a disjoint union of $\operatorname{Spec}\mathbb F_p$ over primes $p$ and consider the obvious map $X\to\operatorname{Spec}\mathbb Z$. The image $\operatorname{Spec}\mathbb Z\setminus\{(0)\}$ is not sober.

$\begingroup$ can this happen if both source and target are integral schemes of finite type over $\mathbb{C}$? $\endgroup$ – user138661 Apr 21 '19 at 18:02

$\begingroup$ Which scheme do you mean? As a space, the disjoint union of the $\operatorname{Spec}{\mathbb F}_p$ is discrete. It is the underlying space of which scheme? But it is not e. g. $\operatorname{Spec}$ of the product of all ${\mathbb F}_p$, is it? Because not seeing what is $X$ I do not see why $(0)$ is not in the image. $\endgroup$ – მამუკა ჯიბლაძე Apr 21 '19 at 19:19

$\begingroup$ This scheme is not a spectrum of any ring (since it's not quasicompact), if that's what you are talking about. I honestly don't know how to describe this scheme any better than just as a disjoint union of ringed topological spaces. $\endgroup$ – Wojowu Apr 21 '19 at 19:25

3$\begingroup$ @Wojowu You can describe it as the open subscheme of $\mathrm{Spec}(\prod_p\mathbb{F}_p)$ whose complement is defined by the ideal $\bigoplus_p\mathbb{F}_p$. I do think your definition is better! $\endgroup$ – Laurent MoretBailly Apr 21 '19 at 19:51

1$\begingroup$ @schematic_boi Your question would be more appropriate for math.stackexchange so this is a welcoming present: you easily reduce to observing that if a space is noetherian, irreducible, and a finite union of finitely many irreducible locally closed subspaces, then one of them must be dense, hence open. $\endgroup$ – Laurent MoretBailly Apr 22 '19 at 6:40