$\mathbb{C}^3$ as a pointwise alegbra has only one simple module: true or false? Let $A = \mathbb{C}^3$ be the algebra with pointwise addition and multiplication. That is, $(a_1, a_2, a_3)*(b_1, b_2, b_3) = (a_1b_1,a_2b_2,a_3b_3)$, $(a_1, a_2, a_3)+(b_1, b_2, b_3) = (a_1+b_1,a_2+b_2,a_3+b_3)$.
Am I correct in saying that $A$ has only one simple module up to isomorphism? My reasoning is this: every simple module appears as a composition factor in the composition series. We have a composition series for $A$: $\{0\} \subset A_1 \subset A_2 \subset A$, where $A_1 = \{(a_1, 0, 0), a_1 \in \mathbb{C}\}$, $A_2 = \{(a_1, a_2, 0), a_1, a_2 \in \mathbb{C}\}$. The composition factors are then isomorphic to $\mathbb{C}$, and so $A$ only has one simple module up to isomorphism.
 A: You can view your algebra as a quiver algebra with three vertices and no arrows. Then each vertex gives a distinct isomorphism class of a simple module, and these are all. 
To obtain this, note that a module over your algebra is nothing else than a vector space with three orthogonal idempotent linear transformations $f_1,f_2$ and $f_3$ such that $f_1+f_2+f_3=1$. Elementary linear algebra now shows that $V$ is a direct sum $V_1\oplus V_2\oplus V_3$ where $V_i$ is the image of $f_i$, and $V_i$ is $f_i$-invariant and annihilated by the other two operators. 
This shows that the simple modules are of the form $\mathbb C$ where only one $e_i$ acts as the identity and the others act trivially.
A: The composition series I have suggested does not actually have isomorphic composition factors (as modules), as the algebra A acts on each one differently. (1,0,0)∗(1,0,0)≠(1,0,0)∗(0,1,0), but this equality should hold if they were isomorphic as modules as I suggested.
Hence the composition series produced gives 3 different simple modules.
