Using Krull-Remak-Schmidt to prove that upper-triangular matrix is not Semsimple. Again I wanted to know if $U_n(\mathbb{R})$ that is the space of upper triangular matrices over $\mathbb{R}$ is semisimple or not.
Now I was thinking if I can do it using the fact that a semisimple ring is indecomposable if and only if it is simple. I tried a lot but can't conclude.
I was studying Krull-Remak-Schmidt theorem and know that any module of finite length can be written as direct sum of finitely many indecomposable submodules.
Now if $U_n(\mathbb{R})$ is semisimple then it is of finite length. So it can be written as direct sum of indecomposable modules. Hence it is not indecomposable and hence $U_n(\mathbb{R})$ is not simple. But after this I cannot do anything. Is there any way to proceed further?
 A: Everything I write assumes the case where $U_n(\mathbb R)$ is not semisimple, namely when $n>1$.  (It's semisimple when $n=1$.)
I think maybe part of the problem is talking about indecomposability of modules and rings as if they were the same thing.
It's true that a nonzero semisimple ring is (ring) indecomposable iff it is simple. It is also true that a nonzero semisimple module is (module) indecomposable iff it is a simple module.  Only the second one would have a chance of being related to the Krull-Schmidt theorem.  But the Krull-Remak-Schmidt theorem holds for the finitely generated modules of all Artinian rings, and that includes both $U_n(\mathbb R)$ and semisimple rings, so it's not going to distinguish them.
Here is where it gets you into trouble:

Now if $_(ℝ)$ is semisimple then it is of finite length. So it can be written as direct sum of indecomposable modules. Hence it is not indecomposable and hence $_(ℝ)$ is not simple.

If $R$ is a finite direct sum of simple modules, it does not follow that it is (ring) decomposable, only that it is (module) decomposable. Even $M_n(\mathbb R)$, which is of course semsimple, is a direct sum of $n$ simple modules, so it is (module) decomposable while being (ring) indecomposable.  It turns out $U_n(\mathbb R)$ is likewise (module) decomposable and (ring) indecomposable, but unlike $M_n(\mathbb R)$ it is not simple.
Now with that sorted, you can go about this multiple ways. Let $J$ be the subset of strictly upper triangular matrices.

*

*Use A semisimple ring is (ring) indecomposable iff it is a simple ring. You can verify that the only nonzero central idempotent in $U_n(\mathbb R)$ is the identity and observe that it is NOT a simple ring, because the subset $J$ forms a nontrivial ideal.


*You can note that it is impossible for $J$ to be a (module) summand of $U_n(\mathbb R)$ because $J^n=\{0\}$, so it can't contain any nonzero idempotent.


*You can still use the method given as a solution to your earlier question like this one.
