Why are flat morphisms "flat?" - MathOverflow

Why are flat morphisms "flat?"

2009-11-25T11:41:08Z

Of course "flatness" is a word that evokes a very particular geometric picture, and it seems to me like there should be a reason why this word is used, but nothing I can find gives me a reason!

Is there some geometric property corresponding to "flatness" (of morphisms, modules, whatever) that makes the choice of terminology obvious or at least justifiable?

Answer by Timo Schürg for Why are flat morphisms "flat?"

2009-11-25T12:33:01Z

One of the consequences of flatness of morphisms between projective schemes is that the dimension of the fibers stays constant. Maybe this is the reason for the term. I'm not sure whether this makes so much sense though. After all, the alps stay three dimensional all the time, but they don't really count as being "flat". But it probably would even much harder to climb them if they had one more dimension...

At least if you think about something that has 0-dimensional fibers all the time and suddenly aquires one two-dimensional fiber flatness really makes sense in the usual sense. I tried drawing a picture here, but mathoverflow always eats my ascii art.

Answer by Joel Fine for Why are flat morphisms "flat?"

2009-11-25T13:15:07Z

This is not really an answer, more of a suggestion of where to look for one. In a locally trivial fibre-bundle, the cohomology groups of the fibres with real coeffs naturally form a flat vector bundle over the base. This is because the integer-valued cohomology gives a lattice which one can use to define the parallel sections. (I guess this is called the Gauss-Manin connection?)

From my outsider's perspective on algebraic geometry, I always imagined that the cohomology groups (of the relevant theory) associated to members of a flat family behaved in a similar way. Perhaps to make this true, one should talk of the alternating sum of cohomology groups as an element of K-theory in the base, or something similar. I've no idea if this can be made rigourous, but various consequences of flatness seem to fit into this mould. E.g., the constancy of the fibre dimension mentioned in Timo Schürg's answer, the fact that the holomorphic Euler characteristic is constant etc.

I would love to hear from algebraic geometers whether this has any rigourous sense to it!

Answer by S1 for Why are flat morphisms "flat?"

2009-11-25T13:39:31Z

I remember the following two quotes about flatness (I forgot who said/wrote this):

For every geometric description of flatness there is a counterexample.
Flatness is one of the few notions in algebraic geometry that were motivated by algebra and not by geometry.

This does not answer your question, I know... :)

Answer by Daniel Erman for Why are flat morphisms "flat?"

2009-11-25T14:02:53Z

The key geometric meaning is that flat families are those families where the fibers vary "continuously". This notion allows one to talk about limits of families of algebraic varieties, which is particularly important in the study of deformation theory/moduli problems. Since the colloquial meaning of flatness also suggests a certain uniformity or lack of variation, one might imagine that this justifies its use in algebraic geometry.

For instance, if you have a flat family of projective varieties, then as Timo points out, the dimension of each fiber is the same. But more is true: the Hilbert polynomial of each fiber is also the same. This allows degeneration techniques. For instance, you can take a flat degeneration of your variety, compute a property about the degeneration, and then lift this information to your original variety.

I think that the geometric meaning of flatness is best understood via simple examples. Consider first $\text{Spec}(k[x,y,t]/(xy-t))\to \text{Spec}(k[t])$ via the natural map. This is a flat family. You can see this geometrically, as the fiber over t is a hyperbola when $t\ne 0$, and as $t$ approaches $0$, the hyperbola gets sharper and sharper and then it "breaks" into two lines when $t=0$.

Constrast this example with $\text{Spec}(k[x,y,t]/(txy-t))\to \text{Spec}(k[t])$. This is not a flat family. Here, when $t\ne 0$, the fiber is always the same hyperbola {xy-1=0}. But, when $t=0$, the fiber is an entire copy of $\text{Spec}(k[x,y])$. This pathological variation of the fibers is encoded by the fact that this is not a flat family.

Answer by Marc Nieper-Wißkirchen for Why are flat morphisms "flat?"

2009-11-25T14:53:21Z

As others have stated above, flatness of a family should mean that the fibres of the family vary somehow continuously. Let state this in terms of a module M over a ring R. Here a fibre of M over a prime P of R is M(P), the k(P)-vector space M_P/PA_P, where k(P) denotes the quotient field of R/P. If the fibres vary continuously, it should be possible to extend a basis of M(P) to nearby fibres, i.e. that the lift of a k(P)-basis wrt. the canonical map M_P -> M(P) should yield a basis of M_P over A_P, i.e. that the stalk M_P is a free module.

And in fact: If M is a finitely presented R-module it is flat if and only if M is locally free, i.e. that stalks are free.

(And that a notion may become less geometric when we turn to non finitely presented modules is something which one may expect anyway.)

Answer by Andrew Critch for Why are flat morphisms "flat?"

2009-11-25T21:03:04Z

A lot of people will tell you that flatness means "continuously varying fibres" in some sense, and that flatness was invented to have correspondingly nice consequences, which is true. But there is a way to expect this (vague) interpretation a priori from an alternative, equivalent definition:

An $A$-module $M$ is flat $\iff$ $I \otimes_A M \to IM$ is an isomorphism for every ideal $I$.

I would prefer to present this as the definition of flatness, and present the fact that tensoring with $M$ preserves exact sequences as a theorem. Why?

Thinking "geometrically", $I$ just corresponds (uniquely) to a closed subscheme $Z=Z(I)=$
$=Spec(A/I)\subseteq Spec(A)$. If we think of $M$ in the usual geometric way as a module of generalized functions on $X$ (like sections of a bundle), and $M/IM \simeq M\otimes_A A/I$ as its restriction to $Z$, then the above definition of flatness can be interepreted directly to mean that $M$ restricts nicely to closed subschemes $Z$.

More precisely, it says that what we lose in this restriction, the submodule $IM$ of elements which "vanish on $Z$", is easy to understand: it's just formal linear combinations of elements $i\otimes m$, with no surprise relations among them, i.e. the tensor product $I \otimes_A M$.

In topology, continuous functions "restrict nicely" to points and closed sets (by taking limits), so you can see, without much experience at all, how this definition corresponds in an intuitive way to continuity.

Having this motivation in place, the best thing to do is to check out examples along the lines of Dan Erman's answer to see the analogy with continuity and limits at work.

Answer by Sasha for Why are flat morphisms "flat?"

2009-11-26T15:31:55Z

As I understand, at least a part of the original question is about the WORD flat; Why this word is used. Then, if I am not wrong (but maybe I am), this word was introduced for modules first, and then for schemes just by extension of terminology. Hence, maybe, the choice of word, "flat", should not really contain some geometric intuition about the schemes (rather, it was chosen for some linear algebra intuition for modules, maybe as one can see from the discussion of module flatness in one of the books by S. Gelfand, Y. Manin).

Answer by Sándor Kovács for Why are flat morphisms "flat?"

2010-10-04T05:49:14Z

See the illustration on the 4th page of Miles Reid's Undergraduate Commutative Algebra (go to Amazon, click on look inside and click the right arrow 4 times).

This illustration shows a module $M$ that's flat over $A/{\rm ann}(M)$.