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I was hoping someone had an idea on how to go about solving the following;

Find (up to isomorphism) all simple R-modules where

i) $R = \begin{pmatrix} \mathbb{Z}/15 \mathbb{Z} & \mathbb{Z}/15 \mathbb{Z} \\ 0 & \mathbb{Z} \end{pmatrix} $;

ii) $R = \mathbb{H} \otimes_{\mathbb{R}} \mathbb{H} \hspace{5pt} $ $(\mathbb{H}$ is the quaternions).

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What have you tried? –  Alex B. May 19 '12 at 17:01
    
Dear Alex, The sentence "Let $R$ be a ring and $M$ and $R$-module" is completely superfluous; perhaps you should remove it. Regards, –  Matt E May 20 '12 at 1:08

1 Answer 1

up vote 2 down vote accepted

It's good to know the following facts:

  • the simple modules of $R$ correspond with the simple modules of $R/rad(R)$, where $rad(R)$ is the Jacobson radical.
  • the simple right modules of $R$ are exactly of the form $R/M$ for a maximal right ideal $M$ of $R$. (Likewise for simple left modules.)
  • The maximal right ideals of a $R\oplus S$ are of the form $A\oplus B$, where exactly one of $A$ and $B$ is its entire ring, and the other is a maximal right ideal in its ring.

For the first ring, there is a post which covers the ideal structure. Via that post, you will find that $rad(R)= \begin{pmatrix} 15\mathbb{Z}/15 \mathbb{Z} & \mathbb{Z}/15 \mathbb{Z} \\ 0 &0 \end{pmatrix}$ and so $R/rad(R)=\begin{pmatrix} \mathbb{Z}/15 \mathbb{Z} & 0 \\ 0 &\mathbb{Z} \end{pmatrix}\cong \mathbb{Z}/15 \mathbb{Z}\oplus \mathbb{Z}$.

So, the simple modules of your first ring look like $(\mathbb{Z}/15\mathbb{Z}\oplus \mathbb{Z})/K$ where $K$ is $3\mathbb{Z}/15\mathbb{Z}\oplus\mathbb{Z}$, or$5\mathbb{Z}/15\mathbb{Z}\oplus\mathbb{Z}$, or $\mathbb{Z}/15\mathbb{Z}\oplus p\mathbb{Z}$ for some prime $p$. Compute these quotients to get a feel for them.

The second ring is a little easier since $\mathbb{H}\otimes_\mathbb{R}\mathbb{H}$ is a simple, 16 dimensional $\mathbb{R}$ algebra. This means it is just isomorphic to the four-by-four matrices over the reals. As a result, there is only one isotype of simple right $R$ module, and in particular any minimal right ideal is a representative of all simple right modules.

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Sorry for the late reply and thanks alot for such an amazing answer. This has helped me so much! –  Alex Kite May 23 '12 at 2:38
    
@AlexKite glad it helped! –  rschwieb May 23 '12 at 10:57

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