Is every finite representation completely reducible?

I have just began studying the lecture notes introduction to representation theory by Etingof for several days.

I learn that every finite representation $V$ of an associative algebra $A$ admits a filtration:

$0= V_0\subset V_1\subset \cdots \subset V_n =V$ such that $V_i/V_{i-1}$ is irreducible for all positive integer $i\leq n$

From that , I encounter a confusion.

Let $W = V_1\oplus V_2/V_1\oplus \cdots \oplus V/V_{n-1}$ , then $W$ has character as $\chi_W = \chi_V+\chi_{V_2/V_1} + \cdots +\chi_{V/V_{n-1}}$ which is the same as $\chi_V$. Hence $V \cong W$

However , $V = V_1\oplus V_2/V_1\oplus \cdots \oplus V/V_{n-1}$ shows $V$ is completely reducible. In other words , every finite representation is completely reducible.

I have already known there exists finite representation which is not completely reducible , but I don't know where I make mistakes.

For representations of arbitrary associative algebras it is not true that $\chi_V=\chi_W\Rightarrow V\cong W$.
• Oh, $V \cong W \Rightarrow \chi_V = \chi_W$ is true for associative algebra , but the inverse holds only for finite group , right? But that means two nonisomorphical finite representations can have the same character. It sounds odd. – Syuizen Nov 1 '15 at 14:53
• @Syuizen That's right. The fact that it holds for representations of finite groups (over $\mathbb{C}$ at least) is connected with the fact that in that case all representations are completely reducible. – Jeremy Rickard Nov 1 '15 at 16:44