TITLE: Realizability of irreducible representations of dihedral groups
QUESTION [12 upvotes]: Consider a dihedral group of degree n and order 2n. Its two-dimensional irreducible representations can be realized over the field $\mathbb{Q}(\cos(2\pi/n),\sin(2\pi/n))$, with the usual action by rotations and reflections. Also, any splitting field of characteristic zero for this group must contain $\mathbb{Q}(\cos(2\pi/n))$, because this is the field generated by the characters of the irreducible representations. (Here, splitting field for a finite group means a field over which all the irreducible representations are realized).
When n is a multiple of 4, these two fields are the same; otherwise, they are not. My question: when n is not a multiple of 4, what conditions would ensure that the smaller subfield $\mathbb{Q}(\cos(2\pi/n))$ is a splitting field for the dihedral group? I think we can restrict attention to n odd.
For instance, when $n = 3$, the dihedral group of degree 3, order 6, has $\cos(2\pi/3) = -1/2$, $\sin(2\pi/3) = \sqrt{3}/2$. So, any splitting field must contain $\mathbb{Q}(1/2) = \mathbb{Q}$. Also, $\mathbb{Q}(1/2,\sqrt{3}/2) = \mathbb{Q}(\sqrt{3})$ is a splitting field since it contains the usual representation given by rotations and reflections.
However, $\mathbb{Q}$ is also a splitting field. To see this, we think of the group as the symmetric group of degree three and take the standard representation. So in this case, we see that $\mathbb{Q}(\cos(2\pi/n))$ works as splitting field even though it is smaller than the field $\mathbb{Q}(\cos(2\pi/n),\sin(2\pi/n))$.
NOTE: It is not true in general that if the characters all take values in a field, the representation can be realized over that field. The standard counterexample is the quaternion group, whose characters all take rational values but whose irreducible representations are realized only when we go to $\mathbb{Q}(i)$. However, some weaker variant of the result may be true for the groups that we are interested in here, namely, the dihedral groups.

REPLY [4 votes]: Since, as Pete Clark pointed out, it is one way to answer the original question, 
it may be worth explaining with Schur index theory, rather than invariant theory, why a finite 
irreducible complex linear group $G$ in dimension $n$ which contains even a single pseudoreflection is conjugate within ${\rm GL}(n,\mathbb{C})$ to a subgroup of ${\rm GL}(n,F)$, where $F$ is the
field generated by the traces of the elements of $G$. Let $\chi$ be the character of the given representation of $G$. Let $x$ be a (genuine, ie non-scalar) pseudo-reflection of $G$. Then
${\rm Res}^{G}_{\langle x \rangle}(\chi) = (n-1)1 + \lambda$ for a non-trivial character $\lambda$
of $\langle x \rangle$. 
Recall that the Schur index $m_{F}(\chi)$ is the smallest positive multiplicity with which $\chi$
occurs as a constituent of a character afforded by a representation of $G$ over $F$. A key property of $m_{F}(\chi)$, from the general theory, is that $m_{F}(\chi)$ even divides the multiplicity
with which $\chi$ occurs in the character afforded by a representation of $G$ realised over $F$.
Now certainly the regular representation of $G$ can be realized over $F$, and $\chi$ occurs with multiplicity $n = \chi(1)$ in the regular character. Hence $m_{F}(\chi)$ divides $n$. But by Frobenius reciprocity, $\chi$ occurs with multiplicity $n-1$ in the permutation representation 
of $G$ induced from the trivial representation of $\langle x \rangle$, and this representation is certainly also realised over the field $F$. Hence $m_{F}(\chi)$ also divides $n-1$, so we conclude
that $m_{F}(\chi) = 1$, that is, the given representation may be realised over $F$.