3

An ordered pair $\ \mathbf X := (X\ d)\ $ is called a chain group $\ \Leftarrow:\Rightarrow\ X\ $ is an abelian group, $\ d:X\rightarrow X\ $ is an abelian group endomorphism, and $\ d\circ d= 0$.

A chain homomorphism $\ f:\mathbf X\rightarrow \mathbf X'\ $ of chain groups $\ \mathbf X := (X\ d)\ $ and $\ \mathbf X' := (X'\ d')\ $ is any abelian group homomorfizm $\ f:X\rightarrow X'\ $ such that $\ d'\circ f=f\circ d$. These are morphisms of the category of the chain groups, and chain monomorphisms (epimorphisms) are chain homomorphisms which are monomorphisms (resp. epimorphisms) as group homomorphisms.

A chain group $\ (X\ d)\ $ is chain generated by a set $\ A\subseteq X\ \Leftarrow:\Rightarrow\ $ for every chain group $\ (Y\ d')\ $ such that $\ Y\ $ is subgroup of group $\ X,\ d' = d|Y,\ $ and $\ A\subseteq Y,\ $ we have $Y=X$. A chain group is finitely generated if it admits a finite set of generators.

Problem   Provide a full classification of finitely generated chain groups.

Of course, the classification of finitely generated abelian groups is a classical result. I think that the above problem is open (is it?), and I feel that--in the view of the said classical classification, it should not be extremely hard.

Remark   A set $\ A\ $ generates a chain group $\ (X\ d)\ \Leftrightarrow\ $ the set $\ A\cup\{d(a) : a\in A\}\ $ generates the abelian group $\ X.\ $ Thus, it follows that the above definition of finitely generated chain groups is equivalent to the definition which says that $\ X\ $ should be finitely generated as an abelian group.

Wlod AA
  • 4,686
  • 16
  • 23
  • 1
    A "chain group" is the same as a $A=\mathbf{Z}[d]/d^2$-module $M$. A classification should involve indecomposables. The only f.g. indecomposable with $dM$ infinite is the free module of rank one, if I'm correct. Those with $dM$ finite are indeed more complicated: they can be of arbitrary rank as $\mathbf{Z}$-module, I guess (mod out the free module $A^k$ by some subgroup of finite index of $(dA)^k\simeq\mathbf{Z}^k$, which can be prescribed "modulo $p$" for many values of $p$). Possibly there are also infinite indecomposables that are not torsion-free as $\mathbf{Z}$-module. – YCor Jun 09 '17 at 16:48

1 Answers1

3

The general classification problem seems to be a wild problem (i.e., contains the classification of pairs of square matrices over a field up to simultaneous conjugacy as a subproblem), and so is probably intractable.

Fix a prime $p$, and consider only those "chain groups" $(X,d)$ where $X=(\mathbb{Z}/p^4\mathbb{Z})^n$ is a direct product of cyclic groups of order $p^4$, and where $p^2X\leq\ker(d)$.

Such an $(X,d)$ is determined by $n$ and the map $d':X/p^2X\to p^2X$ induced by $d$. Multiplication by $p^2$ gives an isomorphism $\alpha:X/p^2X\to p^2X$, independent of $d$.

So $(X,d)$ is determined by $n$ and the endomorphism $d'\alpha^{-1}:p^2X\to p^2X$, which is described by some $n\times n$ matrix $A$ over $\mathbb{Z}/p^2\mathbb{Z}$. Let $X_A$ denote the chain group corresponding to the matrix $A$.

Two chain groups $X_A$ and $X_B$ are isomorphic if and only if the matrices $A$ and $B$ are conjugate by an element of $\text{GL}(n,\mathbb{Z}/p^4\mathbb{Z})$, or equivalently by the image of that element in $\text{GL}(n,\mathbb{Z}/p^2\mathbb{Z})$.

I've not found a copy of the original paper, but several references claim that

Nagornyı, S. V. "Complex representations of the general linear group of degree three modulo a power of a prime." Zap. Naucn. Sem. Leningrad. Otdel. Mat. Inst. Steklov.(LOMI) 75 (1978): 143-150.

contains a proof that the classification of square matrices over $\mathbb{Z}/p^2\mathbb{Z}$ is a wild problem. More precisely, that the classification of $4n\times4n$ matrices over $\mathbb{Z}/p^2\mathbb{Z}$ up to conjugacy would suffice to classify pairs of $n\times n$ matrices over $\mathbb{Z}/p\mathbb{Z}$ up to simultaneous conjugacy.

Jeremy Rickard
  • 33,905