Let $\xi = \cos \frac{2\pi}{13}+i\sin\frac{2\pi}{13}$. Find degree of $\xi+\xi^5+\xi^8$ over $\mathbb Q$.
This question was given to me as a "quick question" on a exam, so I suppose it has an elegant solution but I can't see any. One idea would be to find the minimal polynomial of this element, but that seems too tedious. Computing sub-fields of $\mathbb Q (\xi)$ also seems like a lot of work.
The sum is in $\mathbf Q(\xi)$, whose Galois group $G$ over $\mathbf Q$ is $(\mathbf Z/13\mathbf Z)^\times$, a cyclic group of order $12$, so the sum has degree dividing $12$. To find the degree of the sum over $\mathbf Q$., we can determine the subgroup $H$ of $G$ fixing the sum. Then by the Galois correspondence, the degree of the sum is $[G:H] = |G|/|H|$. It would be a bad idea to think of trying to find the minimal polynomial over $\mathbf Q$.
We know a basis of $\mathbf Q(\xi)$ over $\mathbf Q$ is $\{1, \xi, \xi^2, \ldots, \xi^{11}\}$, or also (multiplying all basis elements by $\xi$) $\{\xi, \xi^2, \xi^3, \ldots, \xi^{12}\}$, so $\mathbf Q$-linear combinations of this set are equal only when coefficients of like powers of $\xi$ are equal.
A generator of $G$ is $2 \bmod 13$ since $2, 2^2, 2^3, 2^4 = 3, 2^6 = -1$ are all not $1 \bmod 13$, so the subgroups of $G$ are generated by $2, 4, 8, 3, -1, 1 \bmod 13$. Applying these elements of $G$ other than $1$ to $\xi + \xi^5 + \xi^8$ (as exponents on $13$th roots of unity), we get the $5$ sums $$ \xi^2 + \xi^{10} + \xi^3, \ \ \xi^4 + \xi^5 + \xi^8, \ \ \xi^8 + \xi + \xi^{12}, \ \ \xi^3 + \xi^2 + \xi^{11}, \ \ \xi^{12} + \xi^{8} + \xi^5, $$ and none of these equal $\xi + \xi^5 + \xi^8$ (because $\{\xi, \xi^2, \ldots, \xi^{12}\}$ is $\mathbf Q$-linearly independent), so the subgroup of $G$ fixing $\xi + \xi^5 + \xi^8$ is $\{1\}$. Thus $\xi + \xi^5 + \xi^8$ has degree $12$ over $\mathbf Q$.
Are you sure the number you were asked about was really $\xi + \xi^5 + \xi^8$? That is a weird sum because it has no nice properties. The subgroup $\langle 5 \bmod 13\rangle$ is $\{1, 5, 12, 8\}$, so I suspect the question may have been intended to be about $\xi + \xi^5 + \xi^8 + \xi^{12}$. Then the exponents on $\xi$ in the sum would be running over a subgroup of $G$ with order $4$, which makes $\xi + \xi^5 + \xi^8 + \xi^{12}$ fixed by $\langle 5 \bmod 13\rangle$ but not by anything more (the only larger subgroup is all of $G$), so $\xi + \xi^5 + \xi^8 + \xi^{12}$ has degree $12/4 = 3$ over $\mathbf Q$. (Using a computer algebra system, a cubic polynomial with root $\xi + \xi^5 + \xi^8 + \xi^{12}$ is $x^3 + x^2 - 4x + 1$, which is irreducible since it's irreducible mod $2$, but you do not need to know such things to figure out the sum has degree $3$ over $\mathbf Q$ using Galois theory.)