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Let $R$ be a local or graded ring. (If it helps, can assume the ring is "good", e.g. $R=k[[x_1,..,x_p]]$, where $k$ is a field of zero characteristic.)

Let $M$ be a finitely generated $R$-module of finite length, i.e. it is supported only at the origin. How the ``generic" such module looks like?

More precisely, consider the modules of finite length, with presentation matrix of size $m\times n$, i.e. the minimal resolution begins as: $\cdots\rightarrow R^n\rightarrow R^m$. (Thus obviously $dim(R)\le n-m+1$ and the length of the resolution is $dim(R)$.) What are the "generic/typical" values of the invariants? What is the typical betti table? Castelnuovo-Mumford regularity?

(I guess the presentation matrix for the generic such module can be assumed as the matrix of linear forms. Much is known about such matrices for $dim(R)> n-m+1$, but here we have the opposite case.)

A related question: given a morphism of modules $R^n\stackrel{\phi}{\rightarrow} R^m$, what is the analogue of the Eagon-Northcott complex for the annihilator-of-cokernel ideal $ann.coker(\phi)$. (As $dim(R)\le n-m+1$, the ideal $ann.coker(\phi)$ is usually much larger than the Fitting ideal $F_0(\phi)$.)

What is known about the "non-generic" modules? (Some stratification by the ``degeneracy type"?)

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    $\begingroup$ If $p\geq 3$, in general the Hilbert scheme is not irreducible and so can have many generic modules and I do not think much is known. If $p=1$, little to be said and if $p=2$, the Hilbert scheme is irreducible and the generic ones, say of length $n$ are isomorphic to $k[[t]]/t^n$. $\endgroup$
    – Mohan
    Commented Jan 28, 2016 at 14:29
  • $\begingroup$ @Mohan: right, but I guess among the components of the Hilbert scheme there is one of the biggest dimension? (or some other "natural choice" of the component) $\endgroup$
    – Dmitry Kerner
    Commented Jan 28, 2016 at 16:13
  • $\begingroup$ The only natural choice I can think of is the closure of the ones of the form $k[[t]]/t^n$, but it does not have the largest dimension. In fact, if I remember correctly, the way one proves lack of irreducibilty is by showing a component of dimension larger than the natural one. $\endgroup$
    – Mohan
    Commented Jan 28, 2016 at 16:35
  • $\begingroup$ @Mohan: see the update. Say $R=k[[x_1,..,x_p]]$. If $m=1$, i.e the case of ideals, then $dim(R)\le n$. For $dim(R)=n$ then genericity means: $\phi$ is a tuple of mutually generic linear forms, then $coker(\phi)=k$, hence the ideal defines the reduced point. $\endgroup$
    – Dmitry Kerner
    Commented Jan 30, 2016 at 20:10

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