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Does there exist a scheme not admitting a morphism locally of finite type to a quasi-compact scheme?

The reason I am asking this is that being locally of finite type and being quasi-compact are respectively most common local and global finiteness hypotheses in algebraic geometry and I think it is natural enough to wonder how they interact with each other.

A strictly stronger question has been asked and answered on MO (open immersions are, rather vacuously, locally of finite type).

Here is what I have tried. There exists a non-empty scheme $X$ having no closed points. Let $f:X\rightarrow Y$ be a morphism locally of finite type to a quasi-compact scheme. It is not too hard to show that there exists a closed point $p\in Y$. Consider the base change $f_p:X\times_Y \mathrm{Spec}\:k(p)\rightarrow X$ of $\mathrm{Spec}\:k(p)\rightarrow Y$. Being the base change of a closed immersion, $f_p$ is a closed immersion so it is enough to find a closed point in $X\times_Y \mathrm{Spec}\:k(p)$. The latter is a scheme locally of finite type over a field and as such has to have a closed point (first argue in the affine case by Zorn's lemma and then note that in this context, a point closed in an affine open is closed in the whole scheme).

The argument above is obviously incomplete because a morphism locally of finite type does not have to hit a closed point (consider e.g. $\mathrm{Spec}\:\mathbb{Q}_p\rightarrow \mathrm{Spec}\:\mathbb{Z}_p$). So it seems that one has to look for another obstruction.

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  • $\begingroup$ What if you take the disjoint union $\coprod_n \operatorname{Spec} \mathbf Q(x_1,\ldots,x_n)$? (Or if you're happy to allow 'essentially of finite type', make a union of spectra of fields that grow much faster.) $\endgroup$ – R. van Dobben de Bruyn Jul 14 '19 at 14:12
  • $\begingroup$ @R.vanDobbendeBruyn is there an open immersion from the disjoint union of these spectra to the Spec of the direct product of the fields (which is an affine scheme, so quasi-compact)? $\endgroup$ – user143077 Jul 14 '19 at 14:30
  • $\begingroup$ Ah, that works; interesting. $\endgroup$ – R. van Dobben de Bruyn Jul 14 '19 at 15:16
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The following is a special case of Example 3 in Laurent Moret-Bailly's answer here.

Let $P$ be the set of primes of $\mathbb{Z}$ and let $S \subseteq P$ be an infinite subset, let $X$ be the gluing of $\operatorname{Spec} \mathbb{Z}_{(p)}$ for $p \in S$ along $\operatorname{Spec} \mathbb{Q}$. Suppose there is a morphism $f : X \to Y$ where $Y$ is quasi-compact and $f$ is locally of finite type. Let $Y = \bigcup_{i=1}^{n} Y_{n}$ be a covering of $Y$ by affine open subschemes. Then there exists some $i$ such that $f^{-1}(Y_{i})$ contains infinitely many closed points of $X$, so after replacing $X$ by $f^{-1}(Y_{i})$ we may assume that $Y$ itself is affine. Let $Z \to Y$ be the scheme-theoretic image of $f$; then $X \to Z$ is locally of finite type by Tag 01T8 and dominant by Tag 056B. After replacing $Y$ by $Z$, we may assume that $Y = \operatorname{Spec} A$ for an integral domain $A$ and that $f$ is dominant. This means that $A$ is a subring of $\mathbb{Z}_{(p)}$ for all $p \in S$, in other words $A$ is a subring of the ring $B \subset \mathbb{Q}$ consisting of fractions $\frac{a_{1}}{a_{2}}$ such that $a_{2}$ is not divisible by any prime in $S$. This is a contradiction since $\mathbb{Z}_{(p)}$ is not finitely generated as a $B$-algebra (if $x_{1},\dotsc,x_{m} \in \mathbb{Z}_{(p)}$ generate $\mathbb{Z}_{(p)}$ as a $B$-algebra, let $S' \subset P$ be the (finite) set of primes appearing in the factorizations of the denominators of $x_{i}$, then $B[x_{1},\dotsc,x_{m}] \subseteq (S')^{-1}B$ which does not contain $\frac{1}{q}$ for $q \in S \setminus S'$).

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