# not locally of finite type implies not universally closed?

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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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. – Damian Rössler Jan 6 '12 at 22:23
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. – Dima Sustretov Jan 7 '12 at 17:49
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? – Dima Sustretov Jan 9 '12 at 14:16

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.