Simple modules of an algebra

Question

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!

score 1 · Answer 1 · edited Apr 13 '17 at 12:20

1

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 $M_2(k)$.

edited Apr 13 '17 at 12:20

Community

1

answered Nov 24 '15 at 18:09

rschwieb

153,510

The simple modules of $k^{3}$ are isomorphic to $k^{3}/m$ where $m$ is a maximal ideal of $k^{3}$?? – Diego Havez Nov 24 '15 at 18:20
So we have just one simple module, $k$! – Diego Havez Nov 24 '15 at 18:23
@DiegoHavez just one simple module That is incorrect. There are, in fact, three nonisomorphic classes of simple modules. – rschwieb Nov 24 '15 at 18:25
@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. – rschwieb Nov 24 '15 at 18:26
Because we have three maximal ideals? – Diego Havez Nov 24 '15 at 18:28
$ K \textsf{X}0\textsf{X}K$ $0\textsf{X}K\textsf{X}K$ $K\textsf{X}K\textsf{X}0$ – Diego Havez Nov 24 '15 at 18:30
I don't understand why three nonisomorphic? – Diego Havez Nov 24 '15 at 18:33
rschwieb can you explain to me why? – Diego Havez Nov 24 '15 at 18:34
The maximal ideals of \begin{pmatrix} k & 0 &0 \ 0& k & 0 \ 0&0&k \end{pmatrix} are \begin{pmatrix} k & 0 &0 \ 0 & k & 0 \ 0&0&0 \end{pmatrix}, \begin{pmatrix} k & 0 &0 \ 0& 0 & 0 \ 0&0&k \end{pmatrix} and \begin{pmatrix} 0 & 0 &0 \ 0 & k & 0 \ 0&0&k \end{pmatrix}. But why three nonisomorphic simple modules? they have the same dimension non! – Diego Havez Nov 24 '15 at 18:53
@DiegoHavez Yes, those are the maximal ideals. You can plainly see they have different annihilators. They are not the same modules over the ring. Having the same $k$-dimension has almost nothing to do with their $R$ structure. – rschwieb Nov 24 '15 at 19:33
I finally accept lol – Diego Havez Dec 06 '15 at 23:51

Simple modules of an algebra

1 Answers1