Automorphism of $\mathbb Q (\sqrt[5]{2}, \zeta_5)$ 
Show that if $\sigma \in \mathrm{Aut}(\mathbb Q (\sqrt[5]{2}, \zeta_5)/\mathbb Q)$, then $\sigma (\zeta_5) = \zeta^i_5$ for some $i=1,2,3,4$ and $\sigma (\sqrt[5]{2}) = \zeta^j_5 \sqrt[5]{2}$ for some $j=1,2,3,4,5$.

Help please.
 A: Hint:


*

*$(\zeta_5)^4+(\zeta_5)^3+(\zeta_5)^2+(\zeta_5)+1=0$

*$(\sqrt[5]{2})^5-2=0$

*For any $\sigma$, we have $\sigma(a)=a$ for all $a\in\mathbb{Q}$ (by definition, elements of $\mathrm{Aut}(L/K)$ are automorphisms of $L$ that fix $K$).
A: Hints:


*

*You know that $\zeta_5^5=1$. Therefore
$$
1=\sigma(1)=\sigma(\zeta_5^5)=\sigma(\zeta_5)^5,
$$
so $\sigma(\zeta_5)$ has to be a fifth root of unity. It cannot be equal to $1$, for $\sigma$ is injective, and $\sigma(1)=1$. Can you see that this leaves the four listed alternatives for $\sigma(\zeta_5)$.

*In the same way $\sigma(\root 5\of 2)$ has to be a fifth root of $2$. Can you see, why this leaves the five listed alternatives for  $\sigma(\root 5\of 2)$.

*To get all 20 automorphisms (that you asked about in a comment to Zev's answer) you need to be able to argue in favor of the following claims:

*

*Why is an automorphism $\sigma$ fully specified, when we know $\sigma(\zeta_5)$ and $\sigma(\root 5\of 2)$?

*Why are we free to use all the $4\cdot5=20$ pairs of choices for $\sigma(\zeta_5)$ and $\sigma(\root 5\of 2)$?



The first item in the last bullet is straightforward field theory. For its part the second item is most conveniently handled by referring to suitable theorems of Galois theory, and is perhaps a bit trickier (although standard).
A: Hint with different approach:
Note I am using this field extension definition: an extension of a field K is a triple (i, K, L), where L is another field, and i is a monomorphism of K into L. In other words, K is embedded in L.
This becomes routine with the extension of monomorphisms theorem. I'm leaving out some gritty details. Perhaps you should fill these in.
Let i be the inclusion map of $\mathbb Q $ in $\mathbb Q (\sqrt[5]{2})(\zeta_5)$, that is $i:$ $\mathbb Q $ $\hookrightarrow$ $\mathbb Q (\sqrt[5]{2})(\zeta_5)$
Then the minimal polynomial of $\sqrt[5]{2}$ over $\mathbb Q $ is $x^5 - 2$. So i($x^5 - 2$) = $x^5 - 2$ and $x^5 - 2$ has roots $(-1)^{2/5}$$\sqrt[5]{2}$, $-(-1)^{3/5}$$\sqrt[5]{2}$, $(-1)^{4/5}$$\sqrt[5]{2}$ in $\mathbb Q (\sqrt[5]{2})(\zeta_5)$.
Hence there is a monomorphism $j_k$ extending i such that $j_k$: $\mathbb Q (\sqrt[5]{2})$ $\hookrightarrow$ $\mathbb Q (\sqrt[5]{2})(\zeta_5)$ 
Given by:
$j_k$: $\sqrt[5]{2}$ $\mapsto$ $\zeta_5^k$$\sqrt[5]{2}$ for k = 0, 1, 2, 3, 4.
The minimal polynomial of $\zeta_5$ over $\mathbb Q (\sqrt[5]{2})$ is $x^4 + x^3 + x^2 + x + 1$. So $j_k$($x^4 + x^3 + x^2 + x + 1$) = $x^4 + x^3 + x^2 + x + 1$ and has roots $-(-1)^{1/5}$, $(-1)^{2/5}$, $-(-1)^{3/5}$, $(-1)^{4/5}$.
Hence there is a monomorphism (which is actually an automorphism) $\sigma_{k,l}$ extending $j_k$ such that:
$\sigma_{k,l}$: $\mathbb Q (\sqrt[5]{2})(\zeta_5)$ $\hookrightarrow$ $\mathbb Q (\sqrt[5]{2})(\zeta_5)$.
Given by:
$\sigma_{k,l}$: $\sqrt[5]{2}$ $\mapsto$ $\zeta_5^k$$\sqrt[5]{2}$ for k = 0, 1, 2, 3, 4.
$\sigma_{k,l}$: $\zeta_5$ $\mapsto$ $\zeta_5^l$ for l = 1, 2, 3, 4.
