I am attempting the following question:
For each of the following algebras A, determine the number of non-isomorphic A-modules, and describe each simple A-module by giving a vector space basis and by defining the action of some algebra generators on this basis.
i) $A=\mathbb{C}C_3\times\mathbb{C}C_4$
ii) $A=\mathbb{C}[C_3\times C_4]$
My attempt:
$i)$ By Maschke's theorem, $\mathbb{C}C_3$ is semisimple. So by Artin-Wedderbern, there exist $n_1,\cdots,n_r$ such that $\mathbb{C}C_3\cong M_{n_1}\times\cdots\times M_{n_{r}}$. But $\mathbb{C}C_3$ is finite dimensional and commutative so every simple $\mathbb{C}C_3$ module is one-dimensional. So we must have that $n_1=\cdots=n_r=1$. But $n_1+\cdots+n_r=\text{dim}(\mathbb{C}C_3)=3$ so $r=3$ and there are exactly three (one-dimensional) simple $\mathbb{C}C_3$ modules up to isomorphism. Similarly, there are exactly four (one-dimensional) simple $\mathbb{C}C_4$ modules up to isomorphism. I think that this will mean that there are $3\cdot 4=12$ simple modules up to isomorphism, but I am struggling to justify it.
I am unsure how to continue with this question- how can I describe the simple modules? As they are all one-dimensional, surely the vector space basis will just be $\{v\}$ for some vector $v$?
ii) I think the same logic largely applies here. $C_3\times C_4$ is semisimple, the group ring is finite-dimensional and commutative, so we obtain that there are 12 one-dimensional simple modules up to isomorphism. Again, I'm not sure how to continue.
Thanks in advance for your help
Best Answer
If $M$ is a finite dimensional irreducible $A$-module and $m \in M$ is any non-zero element, then since $M = A\cdot m$ by irreducibility, we get a surjective morphism of left $A$-modules $A \twoheadrightarrow M$; if its kernel is the left ideal $I \subset A$ then we get an isomorphism $A/I \cong M$ and (as in both of your cases $A$ is semisimple, by Maschke's theorem) if $J \subseteq A$ is a complement of $I$'s (i.e. $A = I \oplus J$ is a decomposition of $A$ into left $A$-submodules) we get $J \cong M$. In other words, searching for irreducible $A$ modules is the same as searching for minimal left ideals in $A$ (the word minimal pops up because $M$ is irreducible).
This little argument - as well as analogues of it - is pretty frequently used in representation theory to transform the analysis of representations of a ring $A$ to that of its inner structure.
In your case this is pretty simple because both of your rings are commutative and their ideals are easily described:
$\mathbb{C}C_3 \times \mathbb{C}C_4 \cong \mathbb{C}[T]/(T^3 - 1) \times \mathbb{C}[T]/(T^4 - 1)$. In this case the minimal ideals are $3 + 4 = 7$ (for example, $(T-1)\times(0)$ and $(0)\times (T+1)$ are two of them). Note that each of these ideals determines a representation on which one of the two factors acts as zero.
$\mathbb{C}C_{12} \cong \mathbb{C}[T]/(T^{12} - 1)$ which has 12 minimal ideals, on which $T$ acts as one of the 12th roots of unity.
Hope this was helpful :)