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I am trying to classify all modules of length $4$ over the ring $A=\mathbb C[x,y]$, supported at the origin $0\in \mathbb C^2$, up to ($A$-linear) isomorphism. Let $\mathfrak m=(x,y)$ be the ideal of the origin and $k\simeq \mathbb C$ the residue field. Here is what I found so far:

  1. Structure sheaves of length $4$ subschemes $Z\subset \mathbb C^2$. These come into three types, $$B_1=A/(x,y^4),\qquad B_2=A/(x^2,y^2),\qquad B_3=A/(x^3,xy,y^2).$$
  2. $k^{\oplus 4}$,
  3. $k\oplus A/(x,y^3)$,
  4. $k\oplus A/\mathfrak m^2$,
  5. $k\oplus \textrm{Hom}_k(A/\mathfrak m^2,k)$,
  6. $k^2\oplus A/(x,y^2)$,
  7. $A/(x,y^2)\oplus A/(x,y^2)$,
  8. $\textrm{Hom}_k(B_3,k)$.

I think these are pairwise non-isomorphic.

Are these all the isomorphism types, or is any other module hiding? Also, I think this classification problem is not solved in general (for arbitrary length), does anyone know up to which length the classification is known?

Thanks!

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    $\begingroup$ Classification is solved at least for upto dimension 6, see a paper of Poonen. In general, of course this is difficult and not known. $\endgroup$
    – Mohan
    Feb 24, 2017 at 21:35
  • $\begingroup$ Thanks, this is good to know! But if I am looking at the correct paper, it classifies $k$-algebras, while I am interested in $A$-modules, or coherent sheaves on $\mathbb C^2$. $\endgroup$ Feb 25, 2017 at 0:05
  • $\begingroup$ They are the same except, the $k$ algebras will also include quotients of polynomial rings in more than 2 variables. $\endgroup$
    – Mohan
    Feb 25, 2017 at 1:04
  • $\begingroup$ I am not sure they are the same. For instance, if $n=3$, his classification is $\{A/x^3,A/\mathfrak m^2\}$, but there is also $\textrm{Hom}_(A/\mathfrak m^2,k)$, and that is not a structure sheaf. Also my point (8) is not listed. I am referring to the table on page 2 in math.mit.edu/~poonen/papers/dimension6.pdf. $\endgroup$ Feb 25, 2017 at 9:13
  • $\begingroup$ You can exchange the role of $x$ and $y$ and get modules not on your list. For instance $A/(x^2,y)$ can be a direct summand. $\endgroup$ Feb 25, 2017 at 18:46

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