I am looking at the ring $\left\{\begin{bmatrix}a&0\\b &c\end{bmatrix}: a,b,c\in k\right\}$, where $k$ is a field.
I think that it is both left and right Noethrian, but I do not know how to show it.
I looked at some questions about
$\left\{\begin{bmatrix}a&b\\0 &c\end{bmatrix}: a,b\in \mathbb{R}, c\in\mathbb{Q}\right\}$ and $\left\{\begin{bmatrix}a&b\\0 &c\end{bmatrix}: a,b\in \mathbb{Q}, c\in\mathbb{Z}\right\}$ for inspiration. They are both left Noetherian but not right Noetherian. It seems that the entry that is "different" (i.e. in $\mathbb{Q}$ or in $\mathbb{Z}$) is what makes it non-right-Noetherian, which is why I think that my ring might be both, since all entries are from the field $k$.
I could not seem to find inspiration to show that my examples is Noetherian in those examples.
Also the term Noetherian is very new to me, so I am not yet comfortable with it.
How would I go about showing this?
My idea was to look at every type of right ideal and every type of left ideal and show that they are finitely generated but that seems like a lot of work, so maybe there is an easier way?
If I have the right ideal $\left\{\begin{bmatrix}x&0\\y &0\end{bmatrix}: x,y\in k\right\}$ I might show that it is finitely generated by letting $(v_1,v_2)$, $(w_1,w_2)$ be a basis for $k^2$. Then there exist $a_1,a_2$ such that $(x,y) = a_1(v_1,v_2)+a_1(w_1,w_2) = (a_1v_1+a_2w_1, a_1v_2+a_2w_2)$, so I could write
$\begin{bmatrix} x&0\\y&0 \end{bmatrix} = \begin{bmatrix} v_1&0\\v_2&0 \end{bmatrix}\begin{bmatrix} a_1&0\\0&0 \end{bmatrix}+\begin{bmatrix} w_1&0\\w_2&0 \end{bmatrix}\begin{bmatrix} a_2&0\\0&0 \end{bmatrix}$
So the ideal is generated by $\begin{bmatrix} v_1&0\\v_2&0 \end{bmatrix}$ and $\begin{bmatrix} w_1&0\\w_2&0 \end{bmatrix}$, so finitely generated.
But what about e.g. the right ideal $\left\{\begin{bmatrix}0&0\\x &y\end{bmatrix}: x,y\in k\right\}$
Best Answer
Since it is an algebra over a field, every left ideal and every right ideal is, in particular, a subspace over that field.
Since the ring is $3$ dimensional to begin with, that limits every chain of left (or right) ideals, ascending or descending, to at most three proper containments.
So you can easily see, the upper triangular matrix ring over a field is always Artinian and Noetherian on both sides.
This does not occur in the other examples you cited when the entries are mixed (such as the one with $\mathbb Q$ and $\mathbb Z$) because those rings aren't algebras over fields.