2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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, 2018 at 11:43
  • 1
    $\begingroup$ 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, 2018 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, 2018 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, 2018 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, 2018 at 15:00

2 Answers 2

Reset to default
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.