# Artin-Wedderburn theorem and square dimension

Let $A$ be a finite-dimensional simple algebra over $\mathbb{C}$ of dimension $n$. By Wedderburn's theorem, we have that $A$ is isomorphic to a matrix ring $M_r(\mathbb{C})$, which is of dimension $r^2$ over $\mathbb{C}$, which would imply that $n$ the dimension of $A$ over $\mathbb{C}$ is necessarily square.

This is false, however, as there exist irreducible representations of group algebras $\mathbb{C}G$ with non-square (and in fact square-free) dimension.

Where is my confusion here?

• $\mathbb{C}G$ is semisimple, not simple (unless $G$ is the trivial group). – darij grinberg Feb 28 '15 at 17:51
• Yes, but irreducible representations of $G$ are simple. – wedderburndimension Feb 28 '15 at 17:53
• Oh, I see what you mean. But irreducible representations of $G$ are $\mathbb CG$-modules, not $\mathbb C$-algebras. That's a different beast. There is no Artin-Wedderburn theorem for modules. – darij grinberg Feb 28 '15 at 17:56
• I see, thanks! The simple submodules of $\mathbb{C}G$ are generally not themselves simple rings. – wedderburndimension Feb 28 '15 at 18:06

The (unique) left simple modulo of the algebra $M_r(\Bbb{C})$ is the column space $\Bbb{C}^r$. In other words the simple component of dimension $r^2$ has a simple module of dimension $r$.
Observe that in case of a finite group $G$ of order $n$ and dimensions of simple representations $d_1,d_2,\ldots,d_k$ we have the equation $$n=d_1^2+d_2^2+\cdots+d_k^2.$$ It all fits magically together :-)