# 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.

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 $2$.