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A proper morphism is defined as separated, of finite type and universally closed. I wonder if the requirement of being of finite type is superfluous, i.e. if being not of finite type implies not universally closed. Recall that a morphism is of finite type if and only if it is locally of finite type and quasi-compact.

In an answer to this question Bjorn Poonen showed that not quasi-compact implies not universally closed.

Is it true that being not locally of finite type (plus possibly quasi-compact) implies not universally closed?

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    $\begingroup$ Let $L|K$ be an infinite algebraic extension. Then the morphism ${\rm Spec}\ L\to{\rm Spec}\ K$ is not locally of finite type but it is universally closed, because it is entire. See EGA II, 6.1.10. $\endgroup$
    – Damian Rössler
    Commented Jan 6, 2012 at 22:23
  • $\begingroup$ Thank you! I didn't realize untill seeing your comment that one doesn't use the finite type assumption when proving properness of finite morphisms, so in fact the proof works for integral morphisms too. $\endgroup$
    – Dima Sustretov
    Commented Jan 7, 2012 at 17:49
  • $\begingroup$ Now that I thought a bit about it, I have a follow up question. Is it true that an arbitrary extension of fields (not necessarily algebraic) gives rise to a universally closed morphism? $\endgroup$
    – Dima Sustretov
    Commented Jan 9, 2012 at 14:16

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Let $k$ be a field, $A=k[X_1,X_2,\dots]$ and $I=(X_1,X_2,\dots)$. Then $\mathrm{Spec}(A/I^2)\to\mathrm{Spec}(k)$ is a universal homeomorphism, but not locally of finite type.

added in edit: In particular, there is no purely topological condition which implies locally finite type.

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    $\begingroup$ To generalise both this example and Damian's: every integral morphism is universally closed (EGA II, 6.1.10). $\endgroup$
    – Laurent Moret-Bailly
    Commented Jan 7, 2012 at 8:34
  • $\begingroup$ Thank you! Why does the definition of a proper morphism include "of finite type" then? I thought that if the answer to the title question is positive, then it is natural to include the requirement of being of finite type in the definition. Since this requirement turned out to be not vacuous, it seems that there must be a reason to assume it. $\endgroup$
    – Dima Sustretov
    Commented Jan 7, 2012 at 17:57
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    $\begingroup$ Coherence of direct images fails if the finite type condition is dropped or weakened to being locally of finite type or to quasi-compactness. $\endgroup$
    – user2035
    Commented Jan 8, 2012 at 11:46

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