On a Geometric Interpretation of the Local Criterion for Flatness in Eisenbud's

Question

The local criterion for flatness goes this way:

Let $\phi : (A,m)\rightarrow (B,m')$ be a local morphism of local Noetherian rings, and $M$ a finitely generated $B$-module. If $x\in m$ is a non-zero divisor on $A$ then $M$ is flat over $A$ iff $M/xM$ is flat over $A/xA$ and $x$ is a non-zero divisor on $M$.

One usual geometric interpretation (see for instance Eisenbud, Commutative Algebra with a View Towards Algebraic Geometry, chapter 6.4) is the following:

If we have a morphism of affine varieties $X\rightarrow Y$ over $\mathbb{A}^1$ such that the maps to $\mathbb{A}^1$ are flat and dominant, for any point $p$ in $\mathbb{A}^1$ choose a point $p'$ in $Y$ above $p$ and a point $p''$ in $X$ above $p'$. If the map of fibers $X_{p}\rightarrow Y_{p}$ is flat in a neighborhood of $p''$ in $X_{p}$, then the map $X\rightarrow Y$ is also flat in a neighborhood of $p''$ in $X$.

It is easy to see that using the local criterion for flatness we get the flatness of the map $X\to Y$ at the point $p''$, but I fail to see how we get this property on a neighborhood of $p''$ in $X$ without using a much stronger result, namely the openness of the flat locus (or, since we are dealing with irreducible variety, generic flatness (or freeness) type of results, see the discussion in the comments following @Eric Canton answer). Am I missing something here ??

Answer 1

This is a standard statement in the general principle of "true at a point implies true in a neighborhood" that can be applied to many situations involving f.g. modules over noetherian rings.

I'll show how to prove that free at a point implies free in a neighborhood for f.g. modules (see Hartshorne Algebraic Geometry, exercise II.5.7).

So suppose $A$ is a noetherian ring and $M$ is a finitely generated $A$ module free at some point $P \in Spec(A)$. Then there are elements $m_1, \dots, m_s \in M$ such that their localization at $P$ give a basis for $M_P$. Writing $N$ for the submodule of $M$ generated by these elements $m_i$, we have an exact sequence

$$ 0 \to C \to A^s \to M \to M/N \to 0 $$ The middle map here is the surjection of the free module onto the submodule $N$. The modules $C$ and $M/N$ are both zero after localizing at $P$ and are finitely generated (here's the only place I used noetherian) and so we know there is some $f \not\in P$ such that $f$ annihilates both $C$ and $M/N$. The basic open set $D(f) = V(f)^c$ of $Spec(A)$ is now a neighborhood of $P$ on which $M$ is free.

I think you can combine this idea with some flatness criteria to get the statement at the end of your question.

Edit: I'm including one of my comments below to more completely answer the question of how to go from local flatness to the statement in Eisenbud.

First, recall that for finitely generated modules, flat and projective are equivalent (since both are equivalent to locally free). Therefore, if $f: X \to Y$ is a finite morphism between schemes, we're done. The hard part (and indeed one of the very useful aspects of this flatness criterion) is proving openness for finite type morphisms, and here I don't see how to get around re-proving some part of generic freeness. Here's a sketch, which is in the spirit of Eisenbud's proof but is more overtly geometric.

Suppose we have a morphism of varieties $\phi: X \to Y$. Being varieties, both $X$ and $Y$ are of finite type over a field $k$, and so in particular $\phi$ is of finite type and we could locally factor $\phi$ in a neighborhood of $P''$ and $P'$ as $$ X \to \mathbb{A}^n_Y \to Y $$ where $X \to \mathbb{A}^n_Y$ is finite and $\mathbb{A}^n_Y \to Y$ is the projection. Here $n$ is the relative dimension of $A(X)_{P''}$ over $A(Y)_{P'}$. Flatness is easy to pass back and forth for $\mathbb{A}^n_Y \to Y$ (e.g. using the local flatness criterion on the coordinates of affine space), and now we can reduce to the case $\phi$ is finite and use the proof mentioned previously.

Answer 2

Let us interpret the following theorem.

Let $x$ be a regular element of a local ring $A$, and let $\phi:A\rightarrow B$ be a local morphism of Noetherian local rings. Then $B$ is flat over $A$ iff $B/xB$ is flat over $A/xA$.

What does it mean to take a regular element of $A$? An element $x$ of $A$ corresponds to a map: $A\rightarrow \mathbb{A}^1=\operatorname{Spec}k[z]$ (since a map from $k[z]$ is just defined by the image of $z$). A regular element of $A$ is given by a flat, dominant morphism to $\mathbb{A}^1$ (I let you check this). Take a point of $\mathbb{A}^1$. WLOG it can be the ideal generated by $z$ (we assume $k$ algebraically closed). Then $X_p=\operatorname{Spec} B\otimes_{k[z]} k[z]/z =B/xB$, similarly $Y_p=\operatorname{Spec}A/xA$. Voila, c'est la.