matrix ring Noetherian, Artinian, semisimple?

Question

Let $k$ be a field and $\Lambda=\begin{bmatrix} k & 0 \\ k^2 & k[x]/(x^2) \end{bmatrix}$. This ring is an algebra over $k$.

(a) What is $\dim_k \Lambda$?

(b) Is $\Lambda$ a left Artinian ring? Is $\Lambda$ a left Noetherian ring?

(c) Is $\Lambda$ a semisimple ring?

I'm not so sure whether my solutions are correct.

(a) I would say that the dimension is 5 because $\Lambda$ has the basis

$$ \begin{align} \begin{bmatrix} 1 & 0 \\ (0,0) & \bar{0} \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ (1,0) & \bar{0} \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ (0,1) & \bar{0} \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ (0,0) & \bar{1} \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ (0,0) & \bar{x} \end{bmatrix}, \end{align} $$

where $1$ denotes the unity in $k$. Is this correct?

(b) The set $I=\begin{bmatrix} 0 & 0 \\ k^2 & 0 \end{bmatrix}$ is an ideal in $\Lambda$ and $\Lambda / I \simeq k \oplus k[x]/(x^2)$. $k$ is a field, hence Noetherian and Artinian. The submodules of $k[x]/(x^2)$ are $(0), \frac{(x)k[x]}{(x^2)}, k[x]/(x^2)$ and thus, $k[x]/(x^2)$ is also Noetherian and Artinian. So, $\Lambda/I$ is Noetherian and Artinian.

Since $I$ hasn't any non-trivial submodules, it is also Noetherian and Artinian and therefore, $\Lambda$ is Noetherian and Artinian.

(c) Suppose that $\Lambda$ is semisimple. Since $\Lambda$ is also Artinian, it cannot have any nilpotent ideals. But $I$ is nilpotent ($I^2=0$), a contradiction. So, $\Lambda$ is not semisimple.

Is this correct?

Answer 1

Your a) and c) are completely correct, and I can't really find any substantial improvement to recommend. Good job!

The approach for b) is OK, but could use some clarification. You need to carefully argue that $\Lambda /I$ and $I$ are Artinian and Noetherian $\Lambda$ modules. You've argued that $\Lambda /I$ is an Artinian/Noetherian ring (=$\Lambda /I$ module) and once you connect that with $\Lambda$, then your proof that $\Lambda$ is Artinian/Noetherian is fine.

But also, you can drastically simplify the argument for b). You've already concluded it's 5 dimensional over $k$. Considering that every ideal is also a $k$-subspace, having only $5$ dimensions greatly restricts how long a chain of ideals (ascending OR descending) can be in this ring. See?