# Factorize a morphism into a morphism locally of finite type and a quasi-compact morphism

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.

• 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.) Jul 14, 2019 at 14:12
• @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)?
– user143077
Jul 14, 2019 at 14:30
• Ah, that works; interesting. Jul 14, 2019 at 15:16

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 $$q \in S \setminus S'$$).