I know someone already asked about flatness over non-reduced schemes, but I think my question is different.

I'm reading Bosch, Lütkebohmert and Raynaud's "Néron models", and in the second chapter, they swiftly discuss some background material of algebraic geometry, notably flatness in section 2.4.

They mention first on page 52 that when the base has nilpotents : "there exists no criterion to test flatness by geometric properties".

On page 53, again they tease : "It is impossible to characterize the flatness of an S-scheme X of finite type by geometric properties when the base S is not reduced." Then they go on pointing EGA IV(2) 6.9.1 for the reduced case.

What had they in mind exactly?

  • 2
    $\begingroup$ To complement Sándor's answer, the "identity map" $X_{red}\to X$ is generally not flat. In fact generic flatness (EGA IV 6.9.1) will fail for the simplest example $Spec k\to Spec k[\epsilon]/(\epsilon^2)$. $\endgroup$ – Donu Arapura Oct 23 '11 at 14:28
  • $\begingroup$ @Donu Arapura: counter-examples will tell "This isn't a geometric characterisation of flatness" ; they won't give "There is no geometric characterisation of flatness"... (And I did mention they were referencing EGA IV(2) 6.9.1) $\endgroup$ – Julien Puydt Oct 23 '11 at 14:48
  • $\begingroup$ I'm aware of both of those facts. It's possible you're asking the wrong people. $\endgroup$ – Donu Arapura Oct 23 '11 at 15:52
  • $\begingroup$ @Donu Arapura: you sound like you feel offended ; that was definitely not the goal : I'm sorry if my comment went through as this. I'm asking the question precisely because I have the impression there was something deep I was missing. In fact, I'm not exactly sure what they mean by "geometric property"... it must not be about topology (or they would have used that word), it must not be purely algebraic (geometry is often opposed to algebra... as two sides of the same coin). $\endgroup$ – Julien Puydt Oct 23 '11 at 16:39
  • $\begingroup$ The internet seems to be an imperfect medium for communication. Thank you for clarifying. $\endgroup$ – Donu Arapura Oct 23 '11 at 17:12

I am not entirely sure what they had in mind, but one possible interpretation of this is that you cannot "see" the difference between a non-reduced scheme and its induced reduced scheme structure geometrically, because they have the same set of points.

A non-reduced scheme is the same topological space as a reduced scheme with extra functions, so you have to take these extra functions into account and that is inherently algebraic.

In other words, you cannot expect to have a condition that says something about the dimension of the fibers or components (associated points) dominating the base like in criteria over reduced bases.

OK, let me try to give a more complex answer. Take the simple flatness criterion that


Let $f:X\to B$ be a morphism such that $B$ is integral, regular, and of dimension $1$. Then $f$ is flat if and only if every associated point of $X$ dominates $B$.

This is not a topological characterization, because it allows $X$ to have embedded points, yet it requires $B$ to be reduced. I suppose you could argue that the part of the condition of the theorem that applies to $B$ is indeed topological, but I would not expect anybody call this criterion topological.

Most other flatness criteria require something about the fibers of the map, which is again only topological on the base, but more than that on the total space and/or fibers.

Also, one might argue that when someone says topological with respect to an algebraic scheme they might mean the euclidean or the étale topology. In some sense to work with the Zariski topology is already a little bit geometric rather than topological.

I realize that this may not sound satisfactory to you, but I think many remarks authors make do not hide some big secret. If they had some complex picture in mind when they wrote it, I bet they would have said more. Then again, I do not claim to know what was/is on their mind. As Donu suggests you could ask the authors. If you do, please share the answer with us!!

| cite | improve this answer | |
  • 1
    $\begingroup$ Indeed, a scheme isn't just a topological space ; but they chose to write "geometric" not "topologic"... that is why I'm expecting a more complex answer somehow :-/ $\endgroup$ – Julien Puydt Oct 23 '11 at 14:46
  • $\begingroup$ I guess Y=B? I'm also surprised that this remark came twice, but they didn't give references : they're so generous about pointers everywhere else... $\endgroup$ – Julien Puydt Oct 24 '11 at 11:09
  • $\begingroup$ Yes, thank you. I wanted to use $B$ instead of $Y$ to show that it is different than in your question. Clearly I immediately forgot. It is corrected now. $\endgroup$ – Sándor Kovács Oct 24 '11 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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