A group algebra isomorphism

Question

Is there a prime $p$ at which the group algebra $\Bbb R[\Bbb F_p^n]$ isomorphic to $\Bbb R^{p^n}$? Is so how to define the isomorphism?

Answer 1

Note that we only have to do the case where $n = 1$, as $\mathbb{R}[\mathbb{F}_p^n] \cong \mathbb{R}[\mathbb{F}_p]^{\otimes n}$. If $\mathbb{R}^p \cong \mathbb{R}[\mathbb{F}_p]$, then we obtain a decomposition $1 = \sum_{i = 1}^p e_i$ into primitive idempotents such that $\mathbb{R}[\mathbb{F}_p] e_i$ is one dimensional for all $1 \leq i \leq p$ and these modules are pairwise non-isomorphic.

Note that for $p > 2$, every group homomorphism $\mathbb{F}_p \to \mathbb{R}^*$ is trivial as $\mathbb{R}$ only has $1$ as a $p$th roots of unity. In this case, there exists only one one-dimensional module (up to isomorphism), hence $\mathbb{R}[\mathbb{F}_p] \not\cong \mathbb{R}^p$.

For $p = 2$, we have $\mathbb{F}_p \cong C_2$, the group of order two. Let $C_2 = \{1,z\}$, then there are exactly two group homomorphisms $\chi_i : C_2 \to \mathbb{R}^*$ ($i = 1,2$), where $\chi_1$ is the trivial homomomorphism and $\chi_2(z) = -1$. From these homomorphisms we get idempotents $e_{\chi_i} = \frac{1}{2} (\chi_i(1) 1 + \chi_i(z) z)$ such that $1 = e_1 + e_2$ and $\mathbb{R}[C_2]e_i$ is one-dimensional. This shows that $\mathbb{R}[\mathbb{F}_2] \cong \mathbb{R}^2$.