Assume $A$ is an algebra over a field $K$ with composition length $n$ and $A$ has $n$ pairwise non-isomorphic simple modules. Is it true that this implies that $A$ is semisimple?
The answer seems to be "Yes" (see the proof below), but I wonder whether my proof is correct because the proof seems to be quite simple but the exercise 4.14 in Erdmann, Holm's "Algebras and Representation Theory" (pdf) is similar to the above statement with the additional requirements that $A$ is commutative and $n$-dimensional.
Assume $A$ is a commutative $K$-algebra of dimension $n$. Show that if $A$ has $n$ pairwise non-isomorphic simple modules then $A$ must be semisimple.
If the proof is indeed correct, that would mean that the condition "commutative" is unnecessary and "dimension $n$" can be relaxed to "composition length $n$", and then it is unclear why these additional conditions were picked in the wording of the exercise.
Proof: Assume the contrary. Since $A$ is not semisimple but has finite composition length, Theorem 4.23 implies that $J(A) \neq 0$ and $J(A) \neq J^2(A)$. Pick maximal proper left sub-ideal of $J(A)$ containing $J^2(A)$: $J(A) \supset M \supseteq J^2(A)$ and an element $b \in J(A) \setminus M$. Consider an $A$-module homomorphism $$\varphi: A / J(A) \to J(A) / M: a + J(A) \mapsto ab + M.$$ It is well-defined since if $a \in J(A)$ then $ab \in J(A)^2 \subseteq M$. We claim that $\varphi = 0$. From Theorem 4.23 we know that the module $A/J(A)$ is semisimple, hence it can be written (up to isomorphism) as $S_1\oplus \dots \oplus S_{k}$ for some simple modules $S_j$. From the assumption all components in the composition series of $A$ are pairwise non-isomorphic. In particular, $S_j$ is not isomorphic to $J(A)/M$. Therefore, from Schur's lemma we know that $\varphi$ restricted to $S_j$ is $0$. Since this is true for every $j$, $\varphi = 0$. This, however, contradicts with $\varphi(1_{A} + J(A)) = a + M \neq 0$. $\square$