TITLE: Is a flat, locally finite type morphism open?
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: Over a field $k$, another related (and somehow more algebraic) example is the "universal" $k$-algebra $A$ generated by countably many orthogonal idempotents. In other words $A$ is the $k$-algebra obtained as the quotient of the polynomial ring $k[X_1,X_2,...]$ in countably many variables, by the ideal generated by all polynomials $X_i^2-X_i$ and $X_iX_j$ for $i\ne j$. It is easy to see that the maximal ideal $m=(X_1,X_2,...)$ gives a nonisolated closed point of $Spec(A)$. Moreover the local ring $A_m$ is just $k$, so that the closed immersion $i:Spec(A/m)\to Spec(A)$ is flat (the extension of local rings is an isomorphism). Thus it is a flat, finite type, morphism which is not open.