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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.

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  • $\begingroup$ This seems like a simple question, however the reason for my confusion is this: "Let k = C and let $A$ be a commutative algebra. Must A have exactly n simple A-modules up to isomorphism?". This was an exam question, and I believe the above is a counterexample. However the examiner's report states this: "Many candidates guessed correctly that the answer was no, but then tried to find examples with A semisimple". I believe my solution is a counterexample, but it is also semisimple. The examiner's comments seems to suggest no solutions are semisimple. $\endgroup$ – Daven May 11 '18 at 11:43
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    $\begingroup$ What is $n$ in your comment? And what are (non-trivial) examples where $A$ actually has exactly $n$ simple modules up to isomorphism? $\endgroup$ – Claudius May 11 '18 at 12:23
  • $\begingroup$ Good question. n is the dimension of $A$ over $\mathbb{C}$. Regarding your second question, I don't know. This is part of the reason that the question confuses me. $\endgroup$ – Daven May 11 '18 at 14:43
  • $\begingroup$ I suppose the composition series I have suggested does not actually have isomorphic composition factors (as modules), as the algebra $A$ acts on each one differently. $(1,0,0)*(1,0,0) \neq (1,0,0)*(0,1,0)$, but this equality should hold if they were isomorphic as modules as I suggested $\endgroup$ – Daven May 11 '18 at 14:59
  • $\begingroup$ So in fact $A$ itself is probably an example of a module having $n$ simple modules to isomorphism $\endgroup$ – Daven May 11 '18 at 15:00
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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.

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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.

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