Let $f: X\to Y$ be a morphism between arbitrary schemes. It is proved in EGA IV Prop 2.4.6 that if f is flat and locally of finite presentation, then f is open. If one replaces locally of finite presentation by locally of finite type, I don't think that the statement is still true (unless X and Y are locally noetherian) but I don't know of any counterexample.
[Math] Is a flat, locally finite type morphism open
ag.algebraic-geometry
Best Answer
There are absolutely flat rings $A$ (i.e. every $A$-module is flat) with nonisolated (automatically closed) points $x$ in the spectrum. Then take the inclusion $x\to \mathrm{Spec}(A)$. Examples of such rings are ($k$ being any field):
• infinite products of fields, e.g. $k^\mathbb{N}$: you get a nontrivial maximal ideal from any nonprincipal ultrafilter on $\mathbb{N}$,
• the ring of locally constant $k$-valued functions on any (infinite) totally disconnected compact Hausdorff space $Y$: then it is a classical exercise that the spectrum is $Y$.