Show $R$ is right-Artinian but not left-Artinian Let $R$ be the subring of $M_2(\mathbb{R})$ defined by
$$R=\left\{\begin{pmatrix}a&b\\0&d\end{pmatrix}\mid a\in\mathbb{Q},b,d\in\mathbb{R}\right\}$$
Show that $R$ is right-Artinian but not left-Artinian.
The "obvious" approach to show it is not left-Artinian is to find an $A\in R$ such that
$$RA\supset R(A^2)\supset R(A^3)\supset\dots$$
is an infinite decreasing chain of proper left submodules of $R$ (as a module over itself). However, after spending a lot of time in this endeavor I've come to the conclusion that no such element $A$ exists (I can post a proof if anybody is interested but it'd be long and not worth the time for me otherwise). Does anybody have an idea for another approach I can take to this? I have no idea what else submodules of $R$ would look like.
 A: Bad news
It is impossible to prove it with the strategy you chose: in fact $R$ is a perfect ring, and it is known that perfect rings satisfy the DCC on principal ideals. Your descending chain of ideals will therefore have to be more sophisticated.
Triangular rings
The ideal structure of this type of construction is explained in both of the following posts, but the second one is a slightly better fit:
Is there a nice way to classify the ideals of the ring of lower triangular matrices?
Left and right ideals of $R=\left\{\bigl(\begin{smallmatrix}a&b\\0&c \end{smallmatrix}\bigr) : a\in\mathbb Z, \ b,c\in\mathbb Q\right\}$
Following the description in the latter post, we think of the ring as $\Bbb Q\times \Bbb R\times \Bbb R$ with funny multiplication.
$J(R)$ is clearly $\begin{bmatrix}0&\Bbb R\\0&0\end{bmatrix}$ since this set is a nilpotent ideal and the quotient by this set is clearly ring isomorphic to $\Bbb Q\times \Bbb R$.
Homing in on where the chain is
Of course, basic module theory says that $R$ is left Artinian iff $R/J(R)$ and $J(R)$ are left Artinian $R$ modules. But $R/J(R)$ is just a direct sum of two simple left $R$ modules, so it is clearly Artinian. Therefore it's clear we just need to prove $J(R)$ is not a left Artinian $R$ module. 
Following this line of thought, we get the idea to look for an infinite descending chain in $\{0\}\times \Bbb R\times \{0\}$.
Apply the result on ideal structure
$\Bbb R$ is an infinite dimensional $\Bbb Q$ module, so we can find an infinite descending chain of $\Bbb Q$ submodules: index them $N_1\supset N_2\supset\ldots$. Reading from the second link, we need to make sure $MJ_2$ is a subset of our $J_1$, so to make things easy we'll just let $J_2=\{0\}$ so that this is satisfied. (The only other choice is $J_2=\Bbb R$, and that won't do.)
Now if we set $N'_i=\{0\}\times N_i\times\{0\}$, we clearly have an infinite descending sequence of $\Bbb Q$ submodules of $\Bbb Q\times \Bbb R$ containing $MJ_2=\{0\}\times\{0\}\times\{0\}$. Thus $N'_i$ furnishes an infinite descending chain of left ideals of $R$.
You could also make an infinite ascending chain of left ideals in the same way, showing it's not left Noetherian.
