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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 ??

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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.

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  • $\begingroup$ Thank you for taking the time to try to answer my question. Unfortunately I think that it is much more difficult to prove that your principle is working in the case of flat modules. For free or projective modules it is not difficult, like you showed, but flat modules are trickier ... In any case, I wanted to be sure that Prof. Eisenbud did forget to mention he is using this strong hidden hypothesis to get his geometric interpretation working. I hate when they do that :( $\endgroup$ – brunoh Jun 13 '17 at 18:33
  • $\begingroup$ For finitely generated modules, flat and projective are equivalent (since both are equivalent to locally free). Looking more closely at the statements in section 6.4, I perhaps see why you are pausing in that the module $M$ is finitely generated over $S$, which is not assumed to be finitely generated over $R$ (if it is, then the proof I showed goes through without problem). $\endgroup$ – Eric Canton Jun 13 '17 at 18:59
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    $\begingroup$ In the case of varieties $\phi: X \to Y$, they are both of finite type over a field $k$, and so in particular we could locally factor $\phi$ in a neighborhood of $P''$ and $P'$ as $$ X \to \mathbb{A}^n_Y \to Y $$ where $X \to \mathbb{A}^n$ 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 $\mathbb{A}^n_Y \to Y$, and now we can reduce to the case $\phi$ is finite and use the proof I mentioned. This is really getting a lot messier than I thought it would, at first... $\endgroup$ – Eric Canton Jun 13 '17 at 19:05
  • $\begingroup$ @Eric_Canton Yes, you are obviously right, flat and projective are equivalent in this noetherian setting when the modules are finitely generated, but the whole point of the local criterion for flatness (and 6.4) is precisely to handle cases without this finiteness hypothesis, like you said in your second comment. That is why I suspected powerful results like the openness of the flat locus (works in the setting of finite type algebras) or generic flatness were used ... (see next comment for a following) $\endgroup$ – brunoh Jun 14 '17 at 8:16
  • $\begingroup$ @Eric_Canton ... The sketching for a proof you did in your second comment (+1) actually reminds me exactly of the steps of the proof of the generic flatness result ... by using Noether normalization first and then openness of flat (or free) locus in the finitely generated module setting ... (see next comment for the last part) $\endgroup$ – brunoh Jun 14 '17 at 8:26
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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.

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  • $\begingroup$ thank you very much for your answer. I had figured out the reasoning you clearly described. My interrogation came out from the fact that Eisenbud (and others ...) seemed to deduce from here that the flatness of the map of the fibers extend from the point considered to a neighborhood of it (in the fiber and then in the whole $X$) without mentioning they use the result on the openness of the locus of points where the map is flat, which is a result not proved in chapter 6. So the whole point of my question was: is there a way to prove it without using this result ?? $\endgroup$ – brunoh Aug 22 '13 at 19:28

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