Element of degree 5 in $\mathbb{Q}(\zeta_{25})$ I want to find an element of degree 5 over $\mathbb{Q}$, in $\mathbb{Q}(\zeta_{25})$. I searched, but only found elements of degree 1, 4, 10, and 20. Is there such an element or no?
Here $\zeta_{25} = e^{2\pi i/25}$.
 A: The Galois group is multiplicative group of integers mod 25 which has $\phi(25)=25-5=20$ elements. This group is cyclic. A generator for this group is $2$ (that is, powers of 2 modulo 25 provide all numbers less than 25 coprime to 25).
This means the automorphism  $\sigma$ of the cyclotomic field that generates the Galois group is the map given by $\zeta\mapsto \zeta^2$. (here $\zeta $denotes a primitive 25th root of unity)
To get a subfield of degree 5, take the subgroup generated by$\sigma^5$ and apply those automorphisms to $\zeta$ and sum them.
Now $\sigma^5:\zeta\mapsto\zeta^{2^5}=\zeta^{32}=\zeta^7$.
For the group generated by this element we get $\zeta^7,\zeta^{49}=\zeta^{24},\zeta^{24\times7}=\zeta^{18}, \zeta^{18\times7}=\zeta^{126}=\zeta$.
Summing these four elements: $\zeta^7+\zeta^{24}+\zeta^{18}+\zeta$ is an element of degree $5$.
EDIT: In response to user1952009's comment below I indicate how to show this number is not rational.
Suffices to show, e.g., it is not fixed by the the automorphism $\sigma^3:\zeta\mapsto \zeta^8$. Assume otherwise leading to an equality:
$$\zeta +\zeta^7+\zeta^{18}+\zeta^{24}=\zeta^8+\zeta^{6}+\zeta^{19}+\zeta^{17}  $$
As the $\zeta$ has degree 20, to keep all powers within  $20$ in the above equation, let us multiply by $\zeta$ which gives us
$$\zeta^2+\zeta^8+\zeta^{19}+1=\zeta^9+\zeta^7+\zeta^{20}+\zeta^{18}$$
Rearrange the above to get a polynomial expression of degree $<20$ giving $g(\zeta)=0$. This contradicts the fact that $\zeta$ has   minimal polynomial $X^{20}+x^{15}+X^{10}+X^5+1=0$
