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What is an example of a finite dimensional algebra $A$ over $\mathbb{C}$ with a simple module of dimension $2013$?

I don't know if the general case holds here (there exists $A$ and a simple module of dimension $n$) or if there is a specific example for $2013$.

My first idea was a group algebra, something like $\mathbb{C}C_{2013k}$, but this is a commutative algebra so all simple modules will be one-dimensional.

Bhaskar Vashishth
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Andrew
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1 Answers1

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HINT:

You can make it a group algebra: $S_n$ has an irreducible representation of dimension $n-1$.

orangeskid
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  • Am I right in the thinking that the span of ${(1,0,\ldots,0 ,-1),(0,1,\ldots ,0,-1),\ldots ,(0,0,\ldots,1,-1)}\subset\mathbb{C}^{2014}$ is a simple module? – Andrew Jun 06 '15 at 23:00
  • @Andrew: Yes, you are right, it is a simple $S_{2013}$ module. – orangeskid Jun 06 '15 at 23:25
  • How do I prove that it is a simple module? – Andrew Jun 06 '15 at 23:31
  • @Andrew: Simple means : "the transforms of every non-zero vector in the module span the whole module". The module consists of all the $n$-dimensional vector with the sum of components $0$. Take one such vector $v \ne 0$. Then not all of its components are equal. Say $v_i \ne v_j$. Try to find combination of $v$ and a two component flip of $v$ that span $e_i - e_j$, then you are almost done. – orangeskid Jun 07 '15 at 06:24