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.