# A question on the group algebra $k[G]$ of a finite abelian group

Let $G$ be a finite abelian group and $k$ a field in which the group order is invertible. Then the group algebra $k[G]$ is a semisimple ring by Maschke's theorem. As $k[G]$ is also abelian, the Artin-Wedderburn theorem implies, that $$k[G]=K_1\oplus\ldots\oplus K_n$$ for fields $K_i$.

I have two questions and I have to admit that they are both a little awkward:

1. Is $K_i$ an algebraic field extension of $k$? I am sure that this is true but I can't find an argument.
2. If (1) is true and $k$ is algebraically closed, I can deduce, that each irreducible representation of $G$ is one-dimensional. However, this is quite easy to prove directly and the structure of $k[G]$ (obtained by non-trivial theorems) seems to contain more information. Can I deduce something about the number of irreducible representations? What can I say for $k$ not algebraically closed?

1. Yes, because it's a finite extension of $k$.
2. If $k$ is algebraically closed, then $k[G] \cong k^{|G|}$ and $G$ has $|G|$ one-dimensional irreps. The story is more interesting for $k$ not algebraically closed; consider, for example, $k = \mathbb{R}$ and $G = C_n$. See Frobenius-Schur indicator for some details. An easy observation is that in general $G$ has at most $|G|$ irreps, but they aren't necessarily one-dimensional over $k$. For $G = C_n$ we have $k[G] \cong k[x]/(x^n - 1)$ and you can be quite explicit about how this splits up depending on which $n^{th}$ roots of unity $k$ has via the Chinese remainder theorem.