Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be a Dedekind ring, let $S = \mathrm{Spec} R$, and let us suppose that $f: X \to S$ is a finite morphism. Note that $X$ is not required to be connected. Does there exist a "numerical criterion" that will produce a closed subscheme $S_0 \subset S$ such that $f$ is flat when restricted to $f^{-1}(S - S_0)$?

Consider the following non-example: Let $F \in \mathbb{Z}[T]$ be a non-constant monic polynomial, and set \[ X = \mathrm{Spec} \mathbb{Z}[T] / (F(T)), \qquad f: X \to \mathrm{Spec} \mathbb{Z}. \] The structure morphism $f \ $ is finite and flat. A prime $p$ occurs as a factor of the discriminant $\Delta(F)$ if and only if the fiber of $f$ over $p$ contains a non-smooth point. So the discriminant can be viewed as a numerical criterion for detecting (non-)smoothness. All of the data needed to determine smoothness is contained in the discriminant. Does there exist a similar one-step gadget for detecting (non-)flatness? (I say "one-step" to mean that the vanishing of infinitely many Tor groups is not an adequate solution. Is there a single Tor group that captures what I'm after?)

share|improve this question
    
Your criterion for a map to $Spec~\mathbb{Z}$ is essentially a criterion for torsion-freeness. It so happens that over $\mathbb{Z}$ this is the the same thing as flatness. What is your $X$ a finite group scheme over? –  Keerthi Madapusi Pera Dec 7 '11 at 2:55
    
I think you just need to check that the dimension of the quotient by each maximal ideal is constant. Reason: If the morphism was projective than one would just have to use the Hilbert polynomial, but the Hilbert polynomial would always be constant and equal to the dimension. –  Will Sawin Dec 7 '11 at 3:09
    
In your example $X = \mathop{\mathrm{Spec}} \mathbb{Z}[T] / (F(T))$, if $F$ is not monic the map is not finite, but it may still happen to be flat (for example, take $F(T) = 2T+1$). If $f\colon X \to Y$ is a finite morphism, $Y$ is locally noetherian, you only need to check that the length of the fibers is locally constant. –  Angelo Dec 7 '11 at 6:16
    
@Keerthi - $X$ is a finite group scheme over the ring of S-integers in a number field. @Will and Angelo - It's true that one can check that the length of the fibers is locally constant, but that's not a "one-step" solution. I want an effective procedure for searching for bad fibers. @Angelo - Ack! You're right! That botches my example, but not my question thankfully. Good catch. –  Xander Faber Dec 7 '11 at 6:33
1  
I suppose you could find the primes over which the map is not smooth, and then compare the length of the fiber over each of this prime with the generic one. –  Angelo Dec 7 '11 at 6:49
show 5 more comments

1 Answer 1

up vote 1 down vote accepted

Take a free resolution of $X$ as an $S$-module, or, more importantly, the first two steps. Look at the map between them, which is given by a matrix over $S$. You need to compute how the rank of this matrix varies across various prime ideals, since this is a constant minus the dimension of the cokernel.

To compute the rank of a matrix, one of course uses the vanishing or nonvanishing of various determinants of $n$ by $n$ minors. These seem to be the kind of numerical criteria you are looking for.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.