I'm confused and probably have a thinking error. Exercise [6 on page 420](https://books.google.com/books?id=6iLUBwAAQBAJ&pg=PA420) 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](https://books.google.sk/books?id=spmzW6-Pt-sC&pg=PA27) 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?



Where is my mistake in calculating duals?
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 $yx$.


