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What is known about the structure of finitely generated modules over local artinian commutative rings $R$? Any information is appreciated. Let us denote by $\mathfrak{m}$ the maximal ideal and by $k$ the residue field. Let us assume that $k \subseteq R$. We may even restrict to the special case $R_{n,d}=k[T_1,\dotsc,T_d]/(T_1,\dotsc,T_d)^n$.

In the case $\mathfrak{m}^2=0$ there is a complete classification, see SE/623261. In the general case we have $\mathfrak{m}^n=0$ for some $n \geq 1$. Any finitely generated $R$-module $M$ has a filtration $0 \subseteq \mathfrak{m}^{n-1} M \subseteq \dotsc \subseteq \mathfrak{m} M \subseteq M$ in which each filtration quotient $\mathfrak{m}^k M / \mathfrak{m}^{k+1} M$ is a finite-dimensional $R/\mathfrak{m}$-vector space (hence is determined by its dimension). But how to classify the (iterated) extensions?

Actually for my purposes it would suffice to give a "basic" class of finitely generated $R$-modules $\mathcal{C}$ so that every finitely generated $R$-module embeds into a direct sum of modules in $\mathcal{C}$. I've read that we may take $\mathcal{C}=\{R^*\}$, where $R^* = \hom_k(R,k)$ with the obvious $R$-module structure. What about other examples?

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  • $\begingroup$ Interesting. If I had the time to absorb the solution at the link, I would have really liked to see why the case of $M^3=0$ is any different from the case $M^2=0$. Do you have a short demonstration of how this diverges from the proof for $M^2=0$? $\endgroup$ – rschwieb Jun 26 '14 at 14:07
  • $\begingroup$ Well, in the exact sequence $0 \to \mathfrak{m}^2 M \to M \to M/\mathfrak{m}^2 M \to 0$ the module on the right is just an $R/\mathfrak{m}^2$-module. Thus we may understand it via the extension of $R/\mathfrak{m}$-modules $0 \to \mathfrak{m} M / \mathfrak{m}^2 M \to M/\mathfrak{m}^2 M \to M/\mathfrak{m} M \to 0$. But how to reconstruct $M$ from that, explicitly? $\endgroup$ – Martin Brandenburg Jun 26 '14 at 15:27
  • $\begingroup$ @rschwieb: Actually, from a geometric point of view, it is not at all surprising that $M^3 = 0$ is quite different from $M^2 = 0$ (for a reference, see section II.3.2 in Eisenbud-Harris, The Geometry of Schemes) $\endgroup$ – zcn Jun 26 '14 at 18:32
  • $\begingroup$ @zcn No surprise here: I just thought it would be a sensible place to begin thinking about the problem. Thanks for the reference anyway. $\endgroup$ – rschwieb Jun 26 '14 at 18:54
  • $\begingroup$ Good question, i have wondered about the general case you ask here, but i thought i would first ask about $m^2=0$ case, and perhaps that will help me, but actually it did not. $\endgroup$ – user114539 Jun 27 '14 at 5:36
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This is more of a general answer to the first sentence.$\DeclareMathOperator{\m}{\mathfrak{m}}$ If $(R, \m, k)$ is an Artinian local ring (not just $k[x_1,\ldots,x_d]/(x_1,\ldots,x_d)^n$), then the $4$ categories of modules coincide: $\DeclareMathOperator{\Hom}{\operatorname{Hom}}$ $$\{ \text{f.g. $R$-modules}\} = \{\text{Noetherian $R$-modules}\} = \{\text{Artinian $R$-modules}\} = \{\text{finite-length $R$-modules}\}$$

Thus a first parametrization of this category is by length, a non-negative integer (in your "iterated extension", this corresponds to taking each successive quotient to be just $k$).

This category also enjoys a canonical duality, namely Matlis duality, given by $\Hom_R(\_, E(k))$, where $E(\_)$ denotes the injective envelope. Any dualizing functor on this category is isomorphic to this one, so in fact $R^* = \Hom_k(R, k) \cong E(k)$. However, $E(k)$ is "more natural" in that it only depends on $R$, and not the choice of the subfield $k \subseteq R$ (although in any case the residue field is a subfield $R/\m \hookrightarrow R$, since $\m \in \text{Ass}(R)$).

Also, that $E(k)$ is an injective cogenerator is clear, since it is the only indecomposable injective $R$-module, so any injective $R$-module is a direct sum of copies of $E(k)$. Finally, one has that $\Hom_R(\_, E(k))$ preserves lengths, and exchanges minimal number of generators with type.

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  • $\begingroup$ I already knew all this. This is not a classification. $\endgroup$ – Martin Brandenburg Jun 27 '14 at 18:54

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