Is the following ring (artinian) semisimple? And are any of these attempts valid?

Question

I'm working though an exam set and came across the problem of showing that the ring

$$ R=\begin{bmatrix} \mathbb{R} & 0\\ \mathbb{R} & \mathbb{R} \end{bmatrix} = \{ \begin{bmatrix} a & 0\\ b & c \end{bmatrix} | a,b,c \in \mathbb{R} \} $$ is (Artinian) semisimple. I know that R is a noncommutative ring, and that it is both artinian and noetherian (Since $(R,+)$ is a three-dimensional vector space over $\mathbb{R}$, any ideal of $I$ must form a vector space over $\mathbb{R}$ of dimension $\leq 3$, so that any nested sequence of ideals contains at most 4 distinct ideals).

It can be shown that $I=\begin{bmatrix} 0 & 0\\ \mathbb{R} & 0 \end{bmatrix}, J=\begin{bmatrix} 0 & 0\\ 0 & \mathbb{R} \end{bmatrix} K=\begin{bmatrix} \mathbb{R} & 0\\ \mathbb{R} & 0 \end{bmatrix}$ are all ideals of $R$, but that $L = \begin{bmatrix} \mathbb{R} & 0\\ 0 & 0 \end{bmatrix}$ is not. My thought is that $R$ is not an artinian semisimple ring (=completely reducible module over itself) since any submodule of $R$ containing a nonzero element of $L$ must necessary contain $K$:

If a submodule $S$ contains $\begin{bmatrix} t & 0\\ 0 & 0 \end{bmatrix}, t\neq 0$, then $\begin{bmatrix} t^{-1} & 0\\ t^{-1} & 0 \end{bmatrix} \in R$, so that $\begin{bmatrix} t^{-1} & 0\\ t^{-1} & 0 \end{bmatrix} \begin{bmatrix} t & 0\\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix} \in S$ but then for any $\begin{bmatrix} a & 0\\ b & 0 \end{bmatrix} \in K$ we have $\begin{bmatrix} a & 0\\ 0 & b \end{bmatrix} \in R$ and accordingly $\begin{bmatrix} a & 0\\ 0 & b \end{bmatrix} \begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix}= \begin{bmatrix} a & 0\\ b & 0 \end{bmatrix} \in S$, showing that $K \subseteq S$. Thus: If $R$ is written as a sum of submodules, then one of the summands must contain $K$, which is not a simple/irreducible(submodule) since it has $I$ as a proper, nontrivial submodule.

$\textbf{Is my argument correct?}$

Also, i found a scribbled solution to the exam saying that $R$ is not simple since $I$ is a nilpotent ideal of $R$, specifically $I^2=(0)$.

$\textbf{Does this argument hold? And why?}$

Thanx, R :)

Answer 1

I think there is a big problem here:

If is written as a sum of submodules, then one of the summands must contain

That is not true, even in a semisimple ring. In $R=M_2(\mathbb R)$, which is certainly semisimple, $$ L_1=\left\{\begin{bmatrix}x&0\\y&0\end{bmatrix}\middle|\,x,y \in \mathbb R\right\} $$

$$ L_2=\left\{\begin{bmatrix}x&x\\y&y\end{bmatrix}\middle|\,x,y \in \mathbb R\right\} $$ are both left ideals with $L_1\oplus L_2=R$, but neither contains

$$ L_3=\left\{\begin{bmatrix}0&x\\0&y\end{bmatrix}\middle|\,x,y \in \mathbb R\right\} $$ even though all three submodules are simple.

It is also not clear on why you are concluding that since $K$ is not simple left $R$ module, the ring is not semisimple. You would need to elaborate on this. In principle it could be possible that $K$ is a sum of two simple submodules, which with a third would give a decomposition for $R$. Or perhaps you have another reason which just isn't coming out in the text you included.

In fact it is true that $K$ cannot be a semisimple left module, because $I$ is a submodule that has no complement in $K$, and that would be sufficient to show the ring isn't semisimple.

---

Also, i found a scribbled solution to the exam saying that $R$ is not simple since $I$ is a nilpotent ideal of $R$, specifically $I^2=(0)$.

No, that is not particularly lucid. But it does have a couple elements that are useful.

---

By far the easiest way to prove the claim is to comment that $I$ is a nonzero nilpotent ideal, and that the Jacobson radical contains nilpotent ideals, so it is not even Jacobson semisimple.

Another way is to quickly compute that it only has trivial central idempotents (the zero matrix and the identity matrix.) In a Wedderburn decomposition of the ring into simple components, this would mean there is only one simple (ring) component: but it is obviously not a simple ring since there's the nontrivial ideal $I$.

Answer 2

The ring is not semisimple, and your argument is correct. It shows that there is an indecomposable projective left module which is not simple.

The other argument goes as follows. A finite dimensional algebra, or more generally an artinian ring, $A$ is semisimple if and only if it is isomorphic to a (finite) product of full matrix rings over division rings, if and only if its Jacobson radical is zero. More generally, a two-sided ideal $I$ equals the Jacobson radical of $A$ provided $I$ is nilpotent and $A/I$ is semisimple.

For the example above we see that the ideal $I$ of lower triangular matrices is nilpotent ($I^2=0$), and $A/I$ is isomorphic to the diagonal matrices, so $\mathbb R^2$, so is semisimple. Thus $J(A)=I\neq0$, and $A$ is not semisimple.