# Why the sum of square formula for Wedderburn decomposition wrong for non-algebraically closed field

I was following the chapter 18 of Abstract Algebra by Dummit and Foote about representation of finite groups, in section 2 there are two results.

First, if G is finite group and F is a field whose character dose not divide |G|, then the group ring FG is isomorphic to a direct product of matrix ring over division ring $R_1\times R_2\times \cdots R_r$, where each $R_i$ is a ring of $n_j\times n_j$ matrices over division ring $\Delta_i$.

Second, if we assume $F=\mathbb{C}$, then all $\Delta_i=\mathbb{C}$. $G$ has r inequivalent irreducible complex representation, each has degree $n_i$, and we have the sum of square formula $|G|=\sum_{i=1}^rn_i^2$.

But the second result is wrong for non-algebraically closed field F. My guess is the division ring might not equal to F, is this the true reason?

I try to work on a simple example, representation of order 3 group $Z_3$ over $\mathbb{Q}$. We have a trivial representation with degree 1 and an irreducible representation with degree 2, where we map the generator of the group to a matrix A satisfying $A^2+A+1=0$. Here we have $3\ne 1^2+2^2$.

But what goes wrong here? I can't see how is the matrix ring correspond to the degree 2 representation looks like. Correct me if there are anything wrong in my statement.

## 1 Answer

My guess is the division ring might not equal to $F$, is this the true reason?

Yes. For example if $G$ is the group of order $3$, then $\mathbb R[G]$ is isomorphic to $\mathbb R\times \mathbb C$, and clearly the $\mathbb R$-dimension is $3$ and not $2$.

Of course, you could patch the formula to incorporate the degrees of extension from the base field to the division ring used in the matrix, it's just more complicated to state than the sum of squares result.