Reflexive modules up to multiplicity

Call an indecomposable module $M$ over a ring $A$ (restrict to finite dimensional algebras if you like or if it helps) $n$-almost reflexive in case $M^{**} \cong nM$, when $(-)^{*}=Hom_A(-,A)$ and $nM$ is the direct sum of $M$ $n$ times for a natural number $n \geq 1$.

Questions:

1. Is $n$ a square?

2. Can $n$ be any natural number?

3. Can there be infinitely many different $n$ that can appear for a given ring?

Example: Take $A$ to be the quiver algebra with quiver having one point and $n$ loops and radical square zero. Then the simple $A$-module has $S^{**} \cong n^2 S$.

1 Answer

Consider the quiver algebra with quiver $\require{AMScd}$ \begin{CD} \bullet@>>>\bullet@>>>\bullet@<<<\bullet@<<<\bullet\\ @.@.@VVV@.@.\\ @.@.\bullet@.@. \end{CD} and radical square zero.

The simple module $S$ at the central vertex has $S^{**}\cong 2S$, and by taking $n$ paths of length two into the central vertex, instead of two, any $n$ can be obtained.

For a finite dimensional algebra $A$ of dimension $d$, then at least for finite dimensional indecomposables it is clear that $n$ is bounded by $d^2$, since $$\dim\text{Hom}_A(X,A)\leq \dim\text{Hom}_k(X,A)=d\dim X$$ for any finite dimensional $A$-module $X$.

• Thanks, the example I did were for commutative algebras so maybe the square thing has to do with the algebra being commutative? But I did not find many examples anyway.
– Mare
Dec 21 '17 at 14:46