Is this ring R both noetherian and artinian? Let the $R=\Big\{\begin{pmatrix}a & 0 \\ b & c \end{pmatrix} : a,b,c \in \mathbb{R}\Big\}$ be a ring.
Is it:
1) Artinian?
2) Noetherian?
If a ring $R$ is noetherian (artinian), then every nonempty set $S$ of left ideals of $R$ has a maximal (minimal) element. $R$ has only 5 ideals, as far as I can see, which are not $(0)$ or $R$ itself, so any $S$ of left ideals must have both a maximal and a minimal element since the number of ideals is finite. Is this correct, or have I misunderstood something? This part of the curriculum is probably what I'm struggling with the most.
 A: Hint: it is a finite dimensional $\Bbb R$ vector space, and left and right ideals are subspaces.
Argue by dimension that chains of left ideals or right ideals can't be longer than four steps.

But the way I argued in the post, is it too vague or is it just wrong?

The reason it isn't valid is this: for noncommutative rings you have to distinguish between right/left Artinian/Noetherian conditions, and convention says you call it "Artinian/Noetherian" if it satisfies the condition on both sides at once. In your solution, you only paid attention to two-sided ideals, and that alone is not enough to prove the ring is right/left Artinian/Noetherian. You have to look at the full sets of right and left ideals.
There are, in fact, infinitely many left ideals (in fact, infinitely many maximal left ideals) of the form $\begin{bmatrix}0&0\\  x&\lambda x\end{bmatrix}$ where $\lambda$ ranges over $\Bbb R$ (as Jeremy Rickard already demonstrated in another solution.)
While finiteness of a poset of left/right/two-sided ideals is certainly enough to guarantee chain conditions, it is really too strong. The conditions are more about the length of chains of submodules rather than the sheer number. As you saw here, infinitely many can exist next to each other, yet all chains must be very shallow.
Finally, but not importantly: you are right that there are five ideals. The set of strictly lower triangular matrices $T$ form a $1$-dimensional two-sided ideal, so there are no ideals below it. The remaining ideals must correspond to ideals of $R/T\cong \Bbb R\times \Bbb R$. Those four ideals, along with the only ideal contained in $T$ (the zero ideal) comprise the full set of $5$ two-sided ideals.
A: As explained in rschwieb's answer, the fact that $R$ is finite dimensional implies that it is both noetherian and artinian.
However, in your question, you try to prove this by claiming that $R$ has only finitely many (left) ideals, and this is not true. For any $\lambda\in\mathbb{R}$,
$$I_\lambda=\left\{\begin{pmatrix}0&0\\x&\lambda x\end{pmatrix}:x\in\mathbb{R}\right\}$$
is a left ideal.
