Finding all simple $R$ modules of a ring. 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).
 A: 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.
