Are $_{\Bbb Q}\Bbb Q$ and $_{\Bbb R}\Bbb R$ artinian left modules?

Question

My original problem is the following: given the ring $$ R:=\left\{\begin{bmatrix}q & 0\\ r & s\end{bmatrix}\;:\;q\in\Bbb Q,\;r,s,\in\Bbb R\right\} =:\begin{bmatrix}\Bbb Q & \Bbb 0\\ \Bbb R & \Bbb R\end{bmatrix} $$ I have to show that it is a left artinian ring, i.e. seen as a left module on itself, $_RR$, is artinian, i.e. the lattice of its submodules $\mathcal{L}(_RR)$, partially ordered by inclusion $\subseteq$ is an artinian poset.

So I tried to figure out how the submodules of $_RR$ are, and after some elementary computation I deduced that $$ \mathcal{L}(_RR)=\left\{\begin{bmatrix}A & 0\\ B & C\end{bmatrix}\;:\;A\;\mbox{is a $_{\Bbb Q}\Bbb Q$-submodule},A\subseteq B,\\ \;B,C\;\mbox{are $_{\Bbb R}\Bbb R$-submodules}\right\} $$ thus a descending chain of submodules of $_RR$ is of the type $$ \begin{bmatrix}A_0 & 0\\ B_0 & C_0\end{bmatrix}\ \supseteq \begin{bmatrix}A_1 & 0\\ B_1 & C_1\end{bmatrix}\ \supseteq\cdots\;\;\;\;\;\;\;\;\;(*) $$ where $\{A_i\}_{i\ge0}$ is a descending chain of $_{\Bbb Q}\Bbb Q$-submodules and $\{B_i\}_{i\ge0},\;\{C_i\}_{i\ge0}$ are descending chains of $_{\Bbb R}\Bbb R$-submodules. Thus if $_{\Bbb Q}\Bbb Q$ and $_{\Bbb R}\Bbb R$ were artinian, then $(*)$ would be stationary, thus my ring $R$ would be left artinian. Is my argument correct? How can I prove that $_{\Bbb Q}\Bbb Q$ and $_{\Bbb R}\Bbb R$ are artinian modules?

Answer 1

Are $_{\Bbb Q}\Bbb Q$ and $_{\Bbb R}\Bbb R$ artinian left modules?

In both cases, these modules have exactly two submodules, so of course they satisfy the DCC.

left/right ideals of that ring

Your characterization of left ideals of this ring is not quite correct. You can work by analogy from here or here to find a correct description.

The left ideals have the form $J_1+ J_2$ where $J_1\lhd \begin{bmatrix}\Bbb Q&0\\ 0&0\end{bmatrix}$ and $J_2$ is an $\Bbb R$ submodule of $\begin{bmatrix}0&0\\ \Bbb R&\Bbb R\end{bmatrix}$, and $\begin{bmatrix}0&0\\ \Bbb R&0\end{bmatrix} J_1\subseteq J_2$. The choice for $J_2$ has more possibilities than just $\begin{bmatrix}0&0\\ B&C\end{bmatrix}$ for two subspaces $B$, $C$.

Considering this characterization of the left ideals, you can see, by focusing on the two halves, that a chain can't be too deep. Any chain of the $J_2$ components could be no more than three deep, because $\Bbb R\times \Bbb R$ has $\Bbb R$ dimension $2$. The $J_1$ component is bounded above by $2$ since it has $\Bbb Q$ dimension $1$. The best case scenario is dropping $1$ dimension lower at each step, and that would only allow you to go $4$ steps at most.

For example, one composition series for the ring as a left module looks like this:

$\begin{bmatrix}0&0\\0&0\end{bmatrix}\subseteq\begin{bmatrix}0&0\\\Bbb R&0\end{bmatrix}\subseteq\begin{bmatrix}0&0\\\Bbb R&\Bbb R\end{bmatrix}\subseteq\begin{bmatrix}\Bbb Q&0\\\Bbb R&\Bbb R\end{bmatrix}$