I'm confused and probably have a thinking error. Exercise 6 on page 420 of Lam's Lectures on Modules and Rings says essentially:

Let $R$ be a ring and $C$ a cyclic right $R$-module: $C=R/A$ with some right ideal $A$ in R. Let $(-)^{*}$ denote the functor $\operatorname{Hom}_R(-,R)$. Then $C^{*} \cong \operatorname{ann}_l(A)$ as right $R$-modules.

I would think that this is correct.

Specialising to a $A=J$ being the Jacobson radical this gives: $C^{*} \cong \operatorname{ann}_l(J) \cong \operatorname{soc}(R)$ as the left annihilator of the Jacobson radical is the socle of the algebra; see Lemma 8.3 of Landrock's Finite Group Algebras and Their Modules. So Exercise 6 tells us that taking duals of simple modules gives again simple modules. In particular, the double dual of a simple module should again be the direct sum of simple modules?!

Now take $R$ to be a local non-selfinjective algebra and $C=R/J$ (which is the unique right simple $R$-module). Let $S$ be the simple left $R$-module. Then we get $$C^{*} \cong \operatorname{soc}(R) \cong \bigoplus_{k=1}^{n}{S}$$ for some $n$. Then we get doing the same again, $C^{**} \cong {\oplus}_{k=1}^{m}{C}$ for some $m$. But taking for example the algebra $$R=K\langle x,y\rangle / (x^2,y^2,xy),$$ my computer tells me that $C^{**}$ is not semisimple and has an indecomposable direct summand of dimension 2.

I do not see the mistake at the moment, so maybe Lemma 8.3. in the book of Landrock might be wrong?


If $C$ is a right module then $C^*$ is a left module.

In Landrock, $\text{soc}(R)$ is the right socle. As he proves, it is a two-sided ideal, but it may not be semisimple as a left module (it is not necessarily equal to the left socle). Your example illustrates this: the right socle is spanned by $x$ and $yx$, but the left socle is spanned by $y$ and $yx$.


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