# Is simreflexive left-right symmetric?

Call an Artin algebra (we can assume that it is basic) $A$ simreflexive in case every simple A-module is reflexive. Is A simreflexive iff the opposite algebra $A^{op}$ is simreflexive? Or equivalently formulated: Is every simple right A-module reflexive iff every simple left A-module is reflexive? An answer for intersting subclasses would also be good.

It was discussed in a previous thread wheter simreflexive implies selfinjective (and this question is very related, so probably the better question is to ask for which interesting subclasses of Artin algebras this holds). The left-right symmetry question is motivated by the following: I think when A and $A^{op}$ are simreflexive, $A$ has to be selfinjective.

Proof: Let $S$ be a simple A-module and $M:=Hom_A(S,A)$. We have two cases:

Case 1: $M$ is not semisimple. Then $M$ is torsionless and thus there is an embedding $M \rightarrow A^n$ for some $n$. Then there is a map $M \rightarrow A^n \rightarrow A$ where the first map is the inclusion and the second map a projection, whose image is not included in the socle of $A$. On the other hand there is a map $M \rightarrow A$ whose image is included in the socle of $A$ (by mapping the top of $M$ into some simple socle summand). Thus in this case $S$ is not isomorphic to $S^{**}$ because $S^{**}=Hom_A(M,A)$ has length at least two. A contradiction to our assumption of $A$ being simreflexive.

Case 2: $M$ is semisimple. Then $M$ has to be simple in fact or else $S^{**}=M^{*}$ is not indecomposable (here we use that $A^{op}$ is simreflexive and thus $U^{*}$ is non-zero for any simple $A^{op}$-module). This implies that $Hom_A(S,A)$ is simple for all $S$ and thus every simple module is exactly once a direct summand of $soc(A)$. This implies that $A/J=soc(A)$, which is equivalent to $A$ being selfinjective.

• It looks like the assumption that $A^{op}$ is simreflexive is used in Case 1, but not needed in Case 2. Since $M = S^*$ is always torsionless, so is any simple summand of $M$. Thus if $M$ is semisimple, $S \cong M^*$ implies $M$ is simple. For Case 1, you need the assumption on $A^{op}$ to know there is a map $M \rightarrow A$ with image in the socle, i.e., you need to know that the top of $M$ is torsionless. – Alex Dugas Mar 6 '18 at 18:15