# Morphism of schemes with non-sober image

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.
• can this happen if both source and target are integral schemes of finite type over $\mathbb{C}$?
• 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. Apr 21 '19 at 19:19
• @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! Apr 21 '19 at 19:51