Quasicompactness in terms of morphisms and $\operatorname{Spec} \mathbb{Z}$ In The Rising Sea, Vakil states the following right before exercise 8.3.B (p. 231):

Following Grothendieck’s philosophy of thinking that the important notions
are properties of morphisms, not of objects, we can restate the definition
of quasicompact (resp. quasiseparated) scheme as a scheme that is quasicompact
(resp. quasiseparated) over the final object $\operatorname{Spec}\mathbb{Z}$ in the category of schemes.

It is clear to me that if $X\to \operatorname{Spec}\mathbb{Z}$ is quasicompact (resp. quasiseparated), then X is quasicompact (resp. quasiseparated), since $\operatorname{Spec}\mathbb{Z}$ is affine. It is also clear that if $X$ is quasiseparated, then the morphism is quasiseparated, since every open subscheme of a quasiseparated scheme is also quasiseparated.
The problem is that I am not sure how to verify that if $X$ is quasicompact, then the morphism $X\to \operatorname{Spec}\mathbb{Z}$ is quasicompact. This is clear if $X$ is noetherian, but I can't see why it holds in general.
Related question that may be useful to solve my problem: how does $\operatorname{Spec}\mathbb{Z}$ behave? Is it true that every open subscheme of this scheme is affine? Is every affine open subscheme of the form $\operatorname{Spec}\mathbb{Z}_f$?

Lemma (Stacks 01K4). Let $f:X\to S$ be a morphism of schemes. The following are equivalent:

*

*$f$ is quasi-compact,

*the inverse image of every affine open is quasi-compact,

*there exists some affine open covering $S=\bigcup_{i\in I} U_i$ such that $f^{-1}(U_i)$ is quasi-compact for all $i$.

Proof. It's the standard strategy for this type of claim ("you can find one cover verifying the property" is equivalent to "all covers must verify the property"), see link for full details. $\blacksquare$
What this means is that if $f:X\to S$ is a morphism of schemes with affine target, $X$ is quasi-compact iff $f$ is.

As for your related question about $\operatorname{Spec} \Bbb Z$, everything you ask for is true, and this is because $\Bbb Z$ is a PID.
