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.


2 Answers 2


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 $Y$.


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.

  • $\begingroup$ I had not thought about this one before, but in fact it is an instance of "my" second series of examples, the space $Y$ being the one-point compactification of $\mathbb{N}$. Indded, $X=\mathrm{Spec}(A)$ is the subset of $k^{\mathbb{N}}$ consisting of the origin $O$ plus the natural basis $B$ of $k^{(\mathbb{N})}$. It is easy to see that the Zariski topology on $B$ is discrete, $O$ is a limit point, and $A$ is canonically the ring of locally constant $k$-valued functions on $X$. $\endgroup$ Mar 25, 2011 at 7:50
  • $\begingroup$ Ah ! Très bien ! Merci Laurent. $\endgroup$ Mar 25, 2011 at 21:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.