"Random" modules of the same size over a polynomial ring seem to always have the same Betti table. By a "random" module I mean the cokernel of a matrix whose entries are random forms of a fixed degree.

For example if we take the quotient of the polynomial ring in three variables by five random cubics:
$S = \mathbb{Q}[x1,x2,x3]$
$M$ = coker random( S^1, S^{5:-3} )
then Macaulay2 "always" (e.g. 1000 out of 1000 times) gives the following Betti table

total: 1 5 9 5
0: 1 . . .
1: . . . .
2: . 5 . .
3: . . 9 5

It seems that the behavior can be explained by the fact that we can resolve the cokernel of a generic matrix of the given form and this resolution remains exact when specializing to any point in a Zariski open subset of some affine space. My question is whether anyone knows a slick proof of this fact.

To elaborate: we can adjoin a new variable to our original ring for each coefficient appearing in each entry of the matrix. So in the above example we would adjoin 10*5 = 50 new variables to $S$, say $y1..y50$. Call the new ring $T$. Consider the $1x3$ matrix $N$ over $T$ whose entries are cubic in the $x_i$ and linear in the $y_i$. Resolve the cokernel of $N$ over $T$ to get a complex $F$. We can then substitute any point in $\mathbb{Q}^{50}$ into the maps of $F$ to get a complex over the original ring $S$. The claim is that this complex is exact on a Zariski open set of the affine space $\mathbb{Q}^{50}$.

It seems like this must be well known but I'm having trouble finding references.

  • $\begingroup$ What does QQ mean? $\endgroup$ Jun 13, 2010 at 3:18
  • $\begingroup$ OK, I just looked up what QQ means in Macaulay 2, and apparently it means $\mathbb{Q}$. $\endgroup$ Jun 13, 2010 at 3:21
  • $\begingroup$ It does. I switched to regular notation. $\endgroup$ Jun 13, 2010 at 3:40

1 Answer 1


It sounds like the question you mean to ask is the following: if $X$ is an integral noetherian scheme with generic point $\eta$ and $C^{\bullet}$ is a finite complex of coherent sheaves on $X$ such that $C^{\bullet}_ {\eta}$ is exact, then does there exist a dense open $U$ in $X$ such that $C^{\bullet}|_U$ is exact and has exact fibers? If that is your question, then the answer is yes. It is instructive to make your own proof by using generic flatness considerations for finite modules over noetherian domains (applied to various kernels, cokernels, images, etc.). Alternatively, look at EGA IV$_3$, 9.4.2 and 9.4.3 for vast generalizations without noetherian hypotheses (e.g., 9.4.3 concerns compatibility of formation of homology sheaves with respect to passage to fibers over some dense open, not assuming exactness at the generic fiber). Even in this hyper-generality, the principle of using "generic flatness" remains the same.

  • $\begingroup$ I don't think that's the question I meant to ask. I used generic in a different sense: take polynomials with coefficients in a field and replace those coefficients with new variables. I think one of the problems that I'm having is that I don't see how generic in that sense relates to the notion of generic point of a scheme. $\endgroup$ Jun 13, 2010 at 2:43
  • 6
    $\begingroup$ So yes, it is the question you meant to ask. A finite collection of polynomials with indeterminate coefficients is parameterized by a big affine space, and at the generic point the coefficients are true indeterminates (whereas at the origin the polys are all zero, etc.). Working over a dense open in this affine space is precisely considering coefficients avoiding some nontrivial algebraic relations. There's a huge development in EGA IV3 showing how results at a generic fiber imply results over dense opens (thereby "justifying" the name "generic point"). $\endgroup$
    – BCnrd
    Jun 13, 2010 at 4:09

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