Simple submodule of modules
I know that a simple right module is a non-zero right module $M_R$ whose submodules are only $M_R$ and $0$.
Now fix a module $N_R$. If I want to show that a submodule $M_R$ of $N_R$ is simple what do I need to show?
Think about the following ring $R$:
\begin{pmatrix} \mathbb{Z} & 0\\ \mathbb{Q} & \mathbb{Q} \end{pmatrix} and consider the following submodule:
\begin{pmatrix} 0 & 0\\ \mathbb{Q} & 0 \end{pmatrix}
In my opinion it is not a simple module but it is a simple submodule of the left module $R$.
$\endgroup$ 81 Answer
$\begingroup$An equivalent condition for a nonzero (left) module $M$ to be simple is that, for each $x\in M$, $x\ne0$, we have $Rx=M$.
Now, for $a\in\mathbb{Z}$, $b,c\in\mathbb{Q}$, we have $$ \begin{bmatrix} a & 0 \\ b & c \end{bmatrix} \begin{bmatrix} 0 & 0 \\ q & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ qc & 0 \end{bmatrix} $$ This shows that the submodule generated by $x=\left[\begin{smallmatrix}0&0\\q&0\end{smallmatrix}\right]$ is the whole of $\left[\begin{smallmatrix}0&0\\\mathbb{Q}&0\end{smallmatrix}\right]$ for every $q\ne0$ (as a left module), because we can arbitrarily choose $c$.
Note that this is also a right module (being a two-sided ideal of the ring), but it is not simple as a right module.
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