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Let $A=M_2(K)$ be the algebra of all $2\times 2$ matrices over $K$. Let $e_1=\left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right)$ and $e_2=\left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right)$. Then it is easy to see that $P(1)=e_1A=\{ \left( \begin{matrix} a & b \\ 0 & 0 \end{matrix} \right) \mid a, b \in K\}$ and $P(2)=e_2A=\{ \left( \begin{matrix} 0 & 0 \\ c & d \end{matrix} \right) \mid c, d \in K\}$. Where $P(1), P(2)$ are all indecomposable projective right $A$-modules. Let $S(1), S(2)$ be all simple right A-modules and $I(1), I(2)$ be all indecomposable injective right A-modules. How to compute $S(1), S(2)$ and $I(1), I(2)$? Thank you very much.

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Hint: Show that $P(i)$ is simple, then $P(i) = S(i)$. The module $I(i)$ is the dual of $Ae_i$ and duals flip the submodule lattice upside-down, so prove that $Ae_i$ is simple. Then $I(i)$ is simple so $I(i) = S(i)$.

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  • $\begingroup$ thank you very much. Are there some explicite reference about the fact that duals flip the submodule lattice upside-down? $\endgroup$ – LJR Mar 22 '13 at 6:06
  • $\begingroup$ Don't bother with an explicit reference, it's easy to prove yourself. $U \subseteq M$ gets sent to the set of $\phi \in M^\ast$ such that $\phi(U) = 0$ and $U' \subseteq M^\ast$ gets sent to the set of $m \in M$ such that $\phi(m) = 0$ for all $\phi \in U'$. $\endgroup$ – Jim Mar 22 '13 at 14:59

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