Why does the algebraic condition of flatness on the structure sheaves give a good definition of family? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:24:05Z http://mathoverflow.net/feeds/question/35127 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35127/why-does-the-algebraic-condition-of-flatness-on-the-structure-sheaves-give-a-goo Why does the algebraic condition of flatness on the structure sheaves give a good definition of family? Bill Dubuque 2010-08-10T15:30:01Z 2010-08-26T20:07:28Z <p>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:</p> <blockquote> <p>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. </p> <p>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</p> </blockquote> http://mathoverflow.net/questions/35127/why-does-the-algebraic-condition-of-flatness-on-the-structure-sheaves-give-a-goo/35134#35134 Answer by David Lehavi for Why does the algebraic condition of flatness on the structure sheaves give a good definition of family? David Lehavi 2010-08-10T16:06:02Z 2010-08-10T18:55:19Z <p>There is a nice computational perspective in Bayer and Mumford's <a href="http://arxiv.org/abs/alg-geom/9304003" rel="nofollow">What Can Be Computed in Algebraic Geometry?</a> pages 4,5.</p> http://mathoverflow.net/questions/35127/why-does-the-algebraic-condition-of-flatness-on-the-structure-sheaves-give-a-goo/35148#35148 Answer by Hailong Dao for Why does the algebraic condition of flatness on the structure sheaves give a good definition of family? Hailong Dao 2010-08-10T17:23:16Z 2010-08-26T19:35:13Z <p>Here is an elementary and intuitive explanation. The fiber of a map is locally a <em>tensor product</em>: 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 $R_p/pR_p \otimes_R S$.</p> <p>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 <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.51.5741" rel="nofollow">Grothendieck localization problem</a>).</p> <p>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). </p> http://mathoverflow.net/questions/35127/why-does-the-algebraic-condition-of-flatness-on-the-structure-sheaves-give-a-goo/36804#36804 Answer by t3suji for Why does the algebraic condition of flatness on the structure sheaves give a good definition of family? t3suji 2010-08-26T20:07:28Z 2010-08-26T20:07:28Z <p>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 <code>families'. The problem with this approach is that<br> this is a family of</code>derived' objects. Here's an example:</p> <p>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.</p> <p>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.</p>