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The question is essentially in the title: $f\colon X \rightarrow Y$ is a monomorphism of schemes that is universally closed; does this imply that $f$ is a closed immersion? Any such $f$ is quasi-compact by https://stacks.math.columbia.edu/tag/04XU, so one may add this assumption if one wants.

In particular, is every integral monomorphism a closed immersion? (This would improve https://stacks.math.columbia.edu/tag/03BB)

If $Y$ is locally Noetherian, then the answer is positive and is Proposition 3.8 (i) in Ferrand's "Monomorphismes de schemas noetheriens". But is this perhaps true for any $Y$? Or are there known counterexamples? Of course, I do not want to assume that $f$ is of finite type ;)

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There is a non-surjective epimorphism $B\to C$ where $B$ and $C$ are zero-dimensional local rings (D. Lazard, see http://www.numdam.org/item/SAC_1967-1968__2__A8_0/). Then $\mathrm{Spec}\,(C)\to \mathrm{Spec}\,(B)$ is a monomorphism but not a closed immersion, and it is universally closed because $B_\mathrm{red}{\simeq}C_\mathrm{red}$.

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    $\begingroup$ I understand that an epimorphism of rings defines a monomorphism in the category of affine schemes. Does it necessarily define a monomorphism in the category of all schemes? Is every monomorphism in the category of affine schemes also a monomorphism in the category of all schemes? $\endgroup$
    – user143116
    Jul 16, 2019 at 8:44
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    $\begingroup$ @Ioki: Yes, indeed. $\endgroup$
    – Laurent Moret-Bailly
    Jul 16, 2019 at 9:19
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    $\begingroup$ Could you then kindly include a reference or an argument for that, for people who do not know it? $\endgroup$
    – user143116
    Jul 16, 2019 at 9:48
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    $\begingroup$ If two morphisms from a scheme agree on every affine open, they are equal. $\endgroup$
    – Laurent Moret-Bailly
    Jul 16, 2019 at 11:58
  • $\begingroup$ What happens if we assume that the schemes in question are reduced? Are universally closed monomorphisms of reduced schemes always closed immersions? $\endgroup$
    – Badam Baplan
    Feb 10 at 21:01

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