Why is the $\mu_n$ representation rational?

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In their paper "On the irregularity of cyclic coverings of algebraic surfaces" by F. Catanese and C. Ciliberto, the authors consider the following situation.

Let $A = V/\Lambda$ be a $g$-dimensional complex abelian variety and let $\mathbb{G}_n$ be a cyclic group of order $n$ consisting of automorphisms of $A$. Then, of course, we have a group action of $\mathbb{G}_n$ on $A$. Denote by $\mu_n$ the group of $n$-th roots of unity, we get a representation $\rho \colon \mu_n \to GL(\Lambda \otimes \mathbb{C})$, so we can decompose $\Lambda \otimes \mathbb{C} = \bigoplus H_\chi$, where for a character $\chi \colon \mu_n \to \mathbb{C}^*$, the $\chi$-eigenspace is denoted $H_\chi$ (i.e. $H_\chi = \{ v \in \Lambda \otimes \mathbb{C}| \rho(g)(v) = \chi(g)\cdot v, \, \forall g \in \mu_n\}$).

So far so good, now for the part which I do not understand. The authors write:

Since the $\mu_n$-representation is rational, if $\chi$ and $\chi'$ have the same order then $\dim(H_\chi) = \dim(H_{\chi'})$.

First of all, I don't really know what a rational representation is. According to Serre's book "Linear Representation of Finite Groups" a rational representation of should be a representation over an algebraic closure of $\mathbb{Q}$ which is isomorphic to a representation over $\mathbb{Q}$ (of the same group, of course). Is that the right definition?

If so, my question is: Why is the $\mu_n$-representation rational and why does the conclusion above hold? I reckon there are some general theorems behind this...

I would also be satisfied with a good reference, where these topics are treated nicely.

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Automorphisms of $A$ are automorphisms of its universal covering space $V$ that preserve the lattice $\Lambda$. So the action of $\mathbf{G}_n$ on $\Lambda$ extended by scalar is defined over $\mathbf{Z}$, hence it is a rational representation.