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How can we find the simple modules of this algebra $$ \begin{pmatrix} k & 0 &0 \\ k & k & 0 \\ k&0&k \end{pmatrix} $$ And why this algebra is not semisimple(i,e it is isomorphic to the algebra $k^{5}$????) I know that his radical is $$ \begin{pmatrix} 0& 0&0 \\ k& 0&0 \\ k&0&0 \end{pmatrix}$$

Thank you!

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How can we find the simple modules of this algebra

Denoting this algebra as $R$, the simple $R$ modules are exactly the simple $R/J(R)$ modules, where $J(R)$ is the Jacobson radical you have found. This is isomorphic to the product ring $k^3$. Now you just need to conclude what the isoclasses of simple modules over that ring are.

The same strategy was applied at this solution, for example: https://math.stackexchange.com/a/147195/29335

And why this algebra is not semisimple

You just said that the Jacobson radical is some nonzero set, but the Jacobson radical of a semisimple ring is zero.

i,e it is isomorphic to the algebra $k^5$????)

There is more than one semisimple $k$ algebra than $k^5$. It could have been $M_2(k)\times k$ or $Q\times k$ if $k$ admits a $4$-dimensional quaternion $k$-algebra other than $M_2(k)$.

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  • $\begingroup$ The simple modules of $k^{3}$ are isomorphic to $k^{3}/m$ where $m$ is a maximal ideal of $k^{3}$?? $\endgroup$ – Diego Havez Nov 24 '15 at 18:20
  • $\begingroup$ So we have just one simple module, $k$! $\endgroup$ – Diego Havez Nov 24 '15 at 18:23
  • $\begingroup$ @DiegoHavez just one simple module That is incorrect. There are, in fact, three nonisomorphic classes of simple modules. $\endgroup$ – rschwieb Nov 24 '15 at 18:25
  • $\begingroup$ @DiegoHavez The simple modules of $k^3$ are isomorphic to $k^3/m$ where $m$ is a maximal ideal of $k^3$? That is correct. $\endgroup$ – rschwieb Nov 24 '15 at 18:26
  • $\begingroup$ Because we have three maximal ideals? $\endgroup$ – Diego Havez Nov 24 '15 at 18:28

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