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Let $k$ be a field and $R$ an associative $k$-algebra suppose $R^2=R$. We say that $R$ is locally bounded if there exists a complete set of pairwise orthogonal primitive idempotents $\{e_x:x\in I\}$ such that $Re_x$ and $e_xR$ are finite dimensional over $k$ for all $x\in I$.

Complete set menas $R=\bigoplus_{x,y}e_x\;Re_y$. And primitive should mean that I cannot see $e_x$ as a sum of two other idempotents.

Call $P(x):= Re_x$, they are projective. Could you help me to prove that they are indecomposable? (this should follow from the fact that $e_x$ is primitive, but I cannot see why). And using Krull-Schimdt I proved that they are all the indecomposable projective modules.

Now I'm having problem with $I(x)=\mathrm{Hom}_k(e_xR,k)$. Could you tell me how to prove that they are injective, indecomposable and that they are all the injective indecomposable modules?

Finally could you tell me why $P(x)/\mathrm{rad}\;P(x)$ are all the simple modules?

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up vote 1 down vote accepted

The only part of your question that I didn't answer in the answer to your previous question is the part stating that the $P(x)$ are indecomposable (although you might have to change the arguments slightly because $R$ is not a finite-dimensional algebra here, if you need further help, comment on this answer).

So assume $P(x)$ was decomposable, write $P(x)=U\oplus V$. Then in particular $e_x=e_x^2\in Re_x$. Hence $e_x=u+v$, where $u\in U$ and $v\in V$. Therefore we have $u=ue_i=u(u+v)=u^2+uv$. Thus $u-u^2=uv$, where the left hand side obviously is in $U$, the right hand side obviously is in $V$ since these are left $R$-modules. But their intersection is zero, in particular $u=u^2$. Interchanging the rôles of $u$ and $v$ we get $v=v^2$. Thus $u$ and $v$ are idempotents in contradiction to $e_x$ being primitive.

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