Questions about socles and radicals.

Question

Let $M$ be a module of a finite dimensional algebra $A$ over an algebraically closed field $K$. Let $N=M/\operatorname{rad}M$ be the top of $M$ and suppose that $N$ is simple. Let $D=\hom_K(\cdot, K)$. Can we conclude that $DN$ is the simple Socle of $DM$ (that is, $DN$ is the Socle of $DM$ and $DN$ is simple)? If this is true, how to prove this result? Thank you very much.

Answer 1

Note that $D(-):mod-A\to mod-A^{op}$ is an (contravariant) equivalence. In particular it is exact, so $ 0\to rad(M)\to M\to S \to 0$ is sent to $0\to DS \to DM \to D(rad(M)) \to 0$.

Next thing you want simple sends to simple. Since simple module $S$ satisfy "if $S \to S' \to 0$ exact, then $S'\cong S$ or $S'=0$, so if we have $0\to N \to DS$, as $D(-)$ is equivalence, $N=DS'$ for some $S'$, which gets us $S'\cong S$ or $S'=0$, and so $N=DS$ or $N=0$. i.e. simple modules are sent to simple modules.

Therefore the simple top $S$ of $M$ is send to a simple submodule $DS$ of $DM$, hence the socle.

Alternatively, by your assumption, we have $P_S\to M\to 0$ for (indecomposable) projective cover of simple $S$, so we get $0 \to DM \to DP_S$. Note $D(-)$ sends (indec) projectives are sent to (indec) injectives, so $DM$ is a submodule of an indecomposable injective, which has a simple socle given by $DS$.