Partial cycles in projective resolutions of square-free algebra

Question

Short version: Over a square-free algebra must every projective resolution of a simple module eventually terminate or contain a shift of itself as a direct summand?

I suspect not, but have not found a counterexample and have checked a few thousand non-isomorphic algebras.

---

Let $R$ be a finite-dimensional, associative, unital, connected, basic algebra over an algebraically closed field such no projective indecomposable module has two isomorphic composition factors. In other words, $R$ is a square-free algebra.

For an (single-sided) finitely generated $R$-module $M$, define $\Omega^0(M) = M$ and $\Omega^{n+1}(M)$ to be the kernel of the projective cover of $\Omega^n(M)$.

Since $\Omega^n(M)$ has finite-length, it has a unique Krull-Schmidt decomposition involving some isomorphism classes of indecomposable modules $M_{n,i}$.

There are two mutually exclusive possibilities: $\Omega^n(M)=0$ for some $n$, or $M \cong M_{n,i}$ for some $n \geq 1$ and some $i$. In the former case, $M$ has finite projective dimension, and in the latter the projective resolution “repeats” insofar as $\Omega^{n+k}(M)$ contains $\Omega^k(M)$ as a direct summand.

Are those the only two possibilities?

If $R$ is a square-free algebra, and $M$ is a simple module then must there be $n>0$ and $i$ such that either $\Omega^n(M) = 0$ or $M_{n,i} \cong M$?

Even more strongly,

If $R$ is a square-free algebra, and $M$ is a simple module, then is $\{ [ M_{n,i} ] \}$ a finite set of isomorphism classes of indecomposable modules?

If so, then $\Omega$ can be represented as a non-negative integer matrix in the Burnside ring, and we are asking if a certain diagonal entry in every proper power is 0, then must the matrix be nilpotent. I haven't found a quick way to verify this, but in the cases I've checked it works, which strikes me as crazy.

Answer 1

If I understood your question correctly then the following provides a counterexample: $$1\to 2\to 3\to 4\to 2$$ where the two $2$'s should be identified. Take $A$ to be the path algebra of this quiver modulo $\operatorname{rad}^2 A$. If you take the projective dimension of $S_1$, then $P_1$ just appears in the beginning as $P_0$ and then $P_2, P_3$ and $P_4$ repeat.

Answer 2

Here is another more complicated example that shows Julian's ideas aren't just hiding the problem in a sneaky way, but that my dichotomy/theorem of the alternative is just completely wrong for simple modules:

Consider the (circular) quiver on 4 vertices with edges $1\to 2 \to 3 \to 4 \to 1$ and add the zero relations $1\to2\to3\to4$, $2\to3\to4\to1\to2$, $3\to4\to1\to2$, and $4\to1\to2\to3\to4$.

The set $\{M_{n,i}\}$ as $M$ ranges over the simple modules has 8 elements, all uniserial. The first four are the simple modules, and the rest are the radicals of the projective indecomposable modules, $J_1=“23”$, $J_2=“341”$, $J_3=“41”$, $J_4=“123”$.

$\Omega$ acts as the following transformation on modules: $$\begin{array}{cccc} \Omega^0 & \Omega^1 & \Omega^{2k} & \Omega^{2k+1} \\ \hline 1 & 23 & 41 & 23 \\ 2 & 341 & 0 & 0 \\ 3 & 41 & 23 & 41 \\ 4 & 123 & 0 & 0 \\ \end{array}$$

In particular, $\Omega^n(M)$ has no simple direct summands if $M$ is simple and $n \geq 1$, yet the 1st and 3rd simple modules all have infinite projective dimension.

It is true that the $8\times 8$ matrix of $\Omega$ is such that $\Omega^2$ has non-zeros on the diagonal, and it is true that the number of isomorphism classes of indecomposable modules required to write down $\Omega^n(M)$ is finite (8), so perhaps the matrix idea is not entirely crazy.