Take the 2-minute tour ×

Hartshorne remarks that is is something of a mystery as to why the algebraic condition of flatness on the structure sheaves gives a good definition of a family (see below). Are there any known enlightening explanations that help serve to unravel this mystery? Below is Hartshorne's introductory motivation to flat families containing said remark:

For many reasons it is important to have a good notion of an algebraic family of varieties or schemes. The most naive definition would be just to take the fibres of a morphism. To get a good notion, however, we should require that certain numerical invariants remain constant in a family, such as the dimension of the fibres. It turns out that if we are dealing with non- singular (or even normal) varieties over a field, then the naive definition is already a good one. Evidence for this is the theorem (9.13) that in such a family, the arithmetic genus is constant.

On the other hand, if we deal with nonnormal varieties, or more general schemes, the naive definition will not do. So we consider a flat family of schemes, which means the fibres of a flat morphism, and this is a very good notion. Why the algebraic condition of flatness on the structure sheaves should give a good definition of a family is something of a mystery. But at least we will justify this choice by showing that flat families have many good properties, and by giving necessary and sufficient conditions for flatness in some special cases. In particular, we will show that a family of closed subschemes of projective space (over an integral scheme) is flat if and only if the Hilbert polynomials of the fibres are the same. -- Hartshorne, Algebraic Geometry, 1977, III.9.5, p. 256

share|improve this question
Here's something "technically useful" (& underlies Hilbert schemes, ur-example of flatness for moduli problems): if $X \rightarrow S$ is map of schemes and $j:Z \hookrightarrow X$ is closed immersion defined by q-coh ideal $I$ such that $Z$ is flat over $S$, then for any $f:S' \rightarrow S$ the exact sequence $0 \rightarrow I \rightarrow O_X \rightarrow j_{\ast}O_Z \rightarrow 0$ has $S$-flat right term, so $f^{\ast}I \rightarrow O_{X'}$ is injective. Thus, ideal of $Z' \hookrightarrow X'$ is the "abstract" pullback $f^{\ast}I$ treating $I$ as "abstract" $O_X$-module. This is important. –  BCnrd Aug 10 '10 at 16:11
To "explain" ubiquity of flatness, use Theorem 23.1 in Matsumura's Commutative Ring Theory book: if $f:X \rightarrow Y$ is map from regular scheme to Cohen-Macaulay scheme (e.g., smooth varieties over field) then "dimension formula" ${\rm{dim}} O_ {X,x} = {\rm{dim}} O_ {Y,f(x)} + {\rm{dim}} O_ {X_ {f(x)}, x}$ is equivalent to flatness of $f$ at $x$. A great non-trivial example is q-finite map between smooth varieties of same dimension. Serre told me that the geometric significance of flatness was entirely due to Grothendieck (Serre invented it as a purely algebraic device for GAGA). –  BCnrd Aug 10 '10 at 16:18
Correction: in 2nd comment above, $Y$ should be regular, and $X$ Cohen-Macalay (or just assume both regular so can save memory for other things...but the CM case is useful for surfaces...) –  BCnrd Aug 10 '10 at 16:19
Section II.3.4 in Eisenbud/Harris contains an elaborate introduction to flat families of schemes. Especially Proposition II.29 might help you. It says roughly that for a flat family + other conditions the fiber of a closed point 0 is the limit of the fibers of points b where b goes to 0. –  C. Jost Aug 11 '10 at 8:25
Thanks to all for the many interesting remarks, here and below. –  Bill Dubuque Aug 17 '10 at 18:54

3 Answers 3

Here is an elementary and intuitive explanation. The fiber of a map is locally a tensor product: if $X=\text{Spec} S$ and $Y=\text{Spec} R$ and the ring map is $R \to S$, then the fiber at a point $p \in Y$ is the Spec of $R_p/pR_p \otimes_R S$.

Flatness is exactly the condition that makes tensor products behave like a dream (almost by definition), it preserves a lot of useful structures. Many algebraic results with geometric consequences go like this: let $(P)$ be a reasonable property and $f: R\to S$ a flat local homomorphism. Then $S$ satisfies $(P)$ if and only if $R$ and the fiber at the closed point satisfy $(P)$ (these are called Grothendieck localization problem).

I am not a historian, but I suspect that was how flatness arised: people wanted certain nice things to be true, and were naturally lead to flatness (see BCnrd's comment below for the precise history).

share|improve this answer
Hailong, the history is that for his GAGA theorems, Serre needed an algebraic mechanism to explain how passing from algebraic to analytic settings (such as for vector bundles, coherent sheaves, etc.) preserved various exactness properties. In this case the algebraic and analytic local rings at a common point were known to be noetherian with the same completion, and completion was known to be preserve exactness under tensoring (Artin-Rees). This led Serre to isolate flatness as an interesting algebraic concept (incorporating exactness of localization), and then it took on a life of its own. –  BCnrd Aug 10 '10 at 19:12
Dear Brian, thanks for the precise history. –  Hailong Dao Aug 10 '10 at 19:27
$Quot(R/p) \otimes_R S$, right? –  Martin Brandenburg Aug 25 '10 at 9:39
Oh, I forgot to localize! –  Hailong Dao Aug 26 '10 at 19:36

There is also the following (probably unhistorical) point of view (it is a version of Hailong Dao's answer). Namely, you don't have to work with flat families at all, so if you want, you can just declare all morphisms to be families'. The problem with this approach is that
this is a family of
derived' objects. Here's an example:

Let $S$ be a scheme, and let $F$ be a coherent sheaf on S. When is it a `family' of its fibers? If it is flat, it definitely deserves to be called a family of vector spaces (a vector bundle). But even if it is not flat, you can still view it as a family, but the family of what? The (derived) fibers of $F$ are no longer vector spaces, they are complexes of vector spaces (precisely because $F$ fails to be flat), so we can view $F$ as a nice family of complexes of vector spaces, even though $F$ itself is a sheaf, not a complex.

To summarize: by all means, let's forget about flatness and declare any morphism to be a family... of some kind of derived objects. If we now want members of the family to be actual objects (schemes, vector spaces, sheaves, or whatever it is we are trying to include in a family), flatness is forced on you more or less by definition.

share|improve this answer

There is a nice computational perspective in Bayer and Mumford's What Can Be Computed in Algebraic Geometry? pages 4,5.

share|improve this answer
Could you turn the pdf link into an abs link? –  S. Carnahan Aug 10 '10 at 16:47
@Scott Carnahan - Done. –  David Lehavi Aug 10 '10 at 18:55

Your Answer


