# Is a universally closed monomorphism a closed immersion?

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 ;)

• – user143116
Jul 16, 2019 at 10:00

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 $$B_\mathrm{red}{\simeq}C_\mathrm{red}$$.