I'm starting to learn representation theory, and thinking about a very basic example. I was wondering if anyone can confirm/clarify my thinking here.
Let $G$ the cyclic group of order 3. Then let's consider it's regular representation over $\mathbb{C}$ -- the free vector space V over $\mathbb{C}$ with basis given by $ \{e_x | x \in G\}$, and homomorphism $\rho: G \rightarrow Aut(V)$ where an element $g \in G$ acts by $g \cdot \sum_{x \in G}a_{x}e_x = \sum_{x \in G} a_{x}e_{gx}$.
Now, over $\mathbb{C}$ we should be able to decompose this representation into three 1-dimensional representations, is that correct? Let $r$ denote the generator of the group and $\omega$ a primitive third root of unity. I imagine that I can take the 1-dimensional $\mathbb{C}$-span of the vector $e_1 + e_r + e_{r^2}$, and similarly the spans of vectors $\omega e_1 + \omega^2 e_r + e_{r^2}$, and $\omega^2 e_1 + \omega e_r + e_{r^2}$. These 1-dimensional $\mathbb{C}$-vector subspaces of $V$ are fixed by $G$. Does this make sense?
I was wondering now how things change over $\mathbb{Q}$? I'm clearly not allowed to consider the two 1-dimensional subspaces involving $\omega$ above. I still have the subspace defined by the span of $e_1 + e_r + e_{r^2}$ over $\mathbb{Q}$, on which $G$ acts trivially, and therefore, by Maschke's theorem I should still be able to write down a two dimensional complement of this (something like the span $<e_1-e_r,e_r-e_{r^2}>$ over $\mathbb{Q}$). Does this make sense? Is it the case that over $\mathbb{Q}$ the irreducible representations of the cyclic group of order $3$ are of degree 1 and degree 2?