radical and composition series

Question

let V be a module with composition series $0 \subset V_1 \subset ... \subset V_{n-1} \subset V$. Then we know that $V_i \cong V_{i-1} \oplus V_i/V_{i-1}$. So $rad(V_i) \cong rad(V_{i-1}) \oplus rad(V_i/V_{i-1}) = rad(V_{i-1}) \oplus 0$ (since the factor module is simple), in other words, $rad(V_i) \cong rad(V_{i-1}$). If we apply this inductively we get $rad(V) \cong rad(0) = 0$. I'm not sure of this result, it would be great if someone could point of my mistake, thanks!

Answer 1

As has been pointed out in the comments, the mistake is assuming the isomorphisms $V_i\cong V_{i-1}\oplus V_i/ V_{i-1}$. This is not true in general. If this were true, it would imply in particular that over an arbitrary ring $R$, the only finite-length indecomposable modules are the simple ones, which is very rarely the case.

As an example, let $R=\mathbb{Z}$ and $V=\mathbb{Z}_4$. Then $0\subseteq 2\mathbb{Z}_4\subseteq V$ is a composition series for $V$ but the fundamental theorem of abelian groups tells us that $V$ is an indecomposable $\mathbb{Z}$-module so $V\ncong 2\mathbb{Z}_4 \oplus \mathbb{Z}_4/2\mathbb{Z}_4$.

If you are interested in finite-dimensional algebras over fields, there are plenty of examples in there too. One can consider for example, given a field $K$, $R=K[X]/(X^2)$ and $V=R_R$. In this case, $V$ is an indecomposable module of length $2$ (and so it is not simple).