Let $R$ be a following subring of $M_2(\mathbb{C}):$ \begin{equation*} R = \left\{ \begin{bmatrix} a & r \\ 0 & s \end{bmatrix} ~:~ a\in \mathbb{Q} ~\mbox{and}~r,s \in \mathbb{C} \right\}. \end{equation*}
(a) Prove that $R$ is both right Noetherian and right Artinian.
(b) Prove that $R$ has uncountable many left ideals.
(c) Is $R$ left Noetherian? Is $R$ left Artinian?
According to me: The right ideals are \begin{equation*} \left\{ I_1 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}, I_2 = \begin{bmatrix} 0 & r \\ 0 & 0 \end{bmatrix}, I_3 = \begin{bmatrix} a & r \\ 0 & 0 \end{bmatrix}, I_4 = \begin{bmatrix} 0 & r \\ 0 & s \end{bmatrix}, I_5 = \begin{bmatrix} a & r \\ 0 & s \end{bmatrix} \right\}. \end{equation*} and the left ideals are \begin{equation*} \left\{ I_1 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}, I_2 = \begin{bmatrix} 0 & r \\ 0 & 0 \end{bmatrix}, I_3 = \begin{bmatrix} a & r \\ 0 & 0 \end{bmatrix}, I_4 = \begin{bmatrix} 0 & r \\ 0 & s \end{bmatrix}, I_5 = \begin{bmatrix} a & r \\ 0 & s \end{bmatrix} \right\}. \end{equation*} How from here I should say show (a) and (b) and (c) ? For (a) I think \begin{equation} I_1 \subseteq I_2 \subseteq I_3 \subseteq I_5 \quad or \quad I_1 \subseteq I_2 \subseteq I_3 \subseteq I_5 \end{equation} are the only possible chians and they stablizes and for Artinian I think reverse order works?? But I dont see why there should be uncountably many left ideals? What is that I am not seeing? I am not very good at Noetherian and Artinian rings. Thanks in advance.
For $r\in\Bbb C$ let
$$I_r=\left\{\begin{bmatrix}a&br\\0&0\end{bmatrix}:a,b\in\Bbb Q\right\}\;.$$
If
$$\begin{bmatrix}c&s\\0&t\end{bmatrix}\in R\quad\text{and}\quad\begin{bmatrix}a&br\\0&0\end{bmatrix}\in I_r\;,$$
then
$$\begin{bmatrix}c&s\\0&t\end{bmatrix}\begin{bmatrix}a&br\\0&0\end{bmatrix}=\begin{bmatrix}ac&(bc)r\\0&0\end{bmatrix}\in I_r\;,$$
so $I_r$ is a left ideal of $R$. For $w,z\in\Bbb C\setminus\{0\}$ write $w\sim z$ iff $\frac{w}z\in\Bbb Q$; $\sim$ is an equivalence relation on $\Bbb C\setminus\{0\}$, and $I_w=I_z$ iff $w\sim z$. Each $\sim$-equivalence class is countable, and $\Bbb C\setminus\{0\}$ is uncountable, so there are uncountably many (in fact $2^\omega=\mathfrak{c}$) $\sim$-classes and hence uncountably many left ideals of $R$.