# Does unramified + radicial imply that a morphism is an immersion?

I know this is true with etaleness added (SGA I, 5.1; in that case the morphism is an open immersion) or if the morphism is in addition proper (since by ZMT the map is then finite and the fibers are all either isomorphic to the residue field or empty, so the morphism is a closed immersion). Is this true in general? I don't know if holds, for instance, that an unramified morphism is an open morphism onto its scheme-theoretic image (which would imply the result by the same argument as in SGA). I am interested in whether this can yield a "functorial" characterization of immersions (since the above remarks yield functorial characterizations of open and closed immersions: etale (resp. unramified) is equivalent to a nilpotent lifting property under finite type hypotheses, radicial is a condition on functors of points, and properness can be checked via the valuative criterion).

• Dear Akhil, open immersions are smooth monomorphisms when viewed as functors! Closed immersions are monos that are "locally a quotient of rings". I don't think anyone can do much better than that. – Harry Gindi Jan 19 '11 at 14:30
• Also, (locally of) finite type is strictly weaker than locally of finite presentation (which is the condition needed for smoothness, etaleness, and unramifiedness). Remember not to confuse them (unless you're in the Noetherian case, when they agree). – Harry Gindi Jan 19 '11 at 14:32
• Dear Harry: My understanding was that finite presentation was needed only for etaleness (and maybe smoothness), not unramifiedness (as in Raynaud's Anneaux locaux henseliens). "Smooth and monomorphic" is a nice functorial criterion. – Akhil Mathew Jan 19 '11 at 16:08
• Dear Akhil, at least as far as I can tell, it's finite presentation for all of them (definitely at the very least, smoothness and étaleness though). – Harry Gindi Jan 19 '11 at 16:15
• This is the first definition in $[\text{EGA 4}]$ §$17$ $\mathbf{N_\underline{o}}. 3$. – Harry Gindi Jan 19 '11 at 16:35

No, this is not true, things can go wrong with the topology. For example, consider a smooth curve $X$ with a point $p$, and the natural morphism $(X \smallsetminus \{p\}) \sqcup \{p\} \to X$.
• One can also construct an example which is even not a local immersion. Consider a curve $X$ (say over $\mathbb C$) with a doube point $p$, consider the normalization in which you remove one of the points above $p$. Then you have a radicial unramified morphism. It is not a local immersion because it would otherwise be an open immersion and hence an isomorphism. However if your morphism is moreover dominant from an integral scheme to an integral normal scheme, then it is an open immersion by ZMT. But this is not functorial. – Qing Liu Jan 19 '11 at 14:35