I'm looking for a characterization of modules that fit the following property:
(*) There's an infinite exact sequence with $\phi_i \neq 0 \space \forall i\in I $ $\require{AMScd}$
\begin{CD}\cdots @>{\phi_i}>>\mathbb{M} @>{\phi_i+1}>> \mathbb{M} @>{\phi_i+2}>> \cdots \\\end{CD}
I thought about the case of vector spaces and about factor rings $\mathbb{Z}/n\mathbb{Z}$:
- For vector spaces V with $dim(V)=1$ it's impossible, but possible for $dim(V)>1$ (Might hold for free modules in general)
- For $\mathbb{Z}/n\mathbb{Z}$ it's possible with 1 function, if n is square, with 2 functions as long as n ist not prime and impossible if n is prime.
But that's far from general... I'm also interested if the number of different functions has a deeper meaning.
Thanks in advance.
For any two nonzero $R$-modules $A$ and $B$, the maps
$$f(a,b)=(a,0), \quad g(a,b)=(0,b) $$
compose to zero, and moreover the sequence
$$ \cdots \to A \oplus B \xrightarrow{f} A \oplus B \xrightarrow{g} A \oplus B \xrightarrow{f} A \oplus B \to \cdots $$
is exact. This works for any ring, so indeed, if you allow rank at least $2$, this can always happen. Now if you want rank $1$, as you note, you can't allow $R$ to be a field. Indeed if $R$ is even an integral domain, there is no hope of composing two nonzero maps $R\to R$ to get zero. On the other hand, if $R$ is commutative but has zero divisors, you can still concoct an infinite exact sequence.