What are all possiblities for Galois group $G(L(\sqrt[5]{3})/L)$? If $L$ is a subfield of $\mathbb C$ (which implies it has characteristic 0 and $\mathbb Q$ as its prime subfield) , what can Galois group $G(L(\sqrt[5]{3})/L)$ be? I start to solve this problem by considering $\sqrt[5]{3}$ in $\mathbb C$. The polynomial of $\sqrt[5]{3}$ is $$f(x)=x^5-3$$ and all its roots are $\sqrt[5]{3}$,$\sqrt[5]{3} \zeta_5$,$\sqrt[5]{3} \zeta_5^2$,$\sqrt[5]{3} \zeta_5^3$, $\sqrt[5]{3} \zeta_5^4$. Then I think there are 2 cases for L, which are 
(1) $L$ does not contain $\zeta_5$. Then the Galois group is trivial since $\sqrt[5]{3}$ can only be sent to itself.
(2) $L$ does not contain $\zeta_5$. This is the part where I get confused. Since $G(L(\sqrt[5]{3})/L)$? fixes $\zeta_5$, there will still be only one automorphism. Then in conclusion $G(L(\sqrt[5]{3})/L)$ can only be trivial.
Am I right about this? I feel that this problem is not that easy and there should be more case for $L$.
 A: Firstly, you need to define what you mean by $\sqrt[5]{3}\in\mathbb{C}$. I will assume that $\sqrt[5]{3}$ is the unique real 5th root of 3 in $\mathbb{C}$. Let $S$ be a transcendence basis of $L$ over $\mathbb{Q}$, and let $K := \mathbb{Q}(S)$. Let $K'$ be the splitting field of $x^5-3$ over $K$. Then $K'/K$ is Galois with Galois group the semididirect product $C_5\rtimes C_4$, and $L(\sqrt[5]{3})$ is a subfield of the compositum $LK'$. Your job is to classify the intermediate extensions of $LK'/L$ of the form $L(\sqrt[5]{3})$. By the "Diamond theorem", there is a bijection between intermediate extensions of $LK'/L$ and those of $K'/K'\cap L$, obtained by "intersecting with $K'$". Thus, in terms of Galois theory, $L\cap K'$ (and hence $L$) corresponds to a subgroup $H_1$ of $C_5\rtimes C_4$, and $L(\sqrt[5]{3})\cap K'$ (and hence $L(\sqrt[5]{3})$) corresponds to a subgroup $H_2\le H_1$ of index at most 5 inside $H_1$. Note that $[L(\sqrt[5]{3}):L] = [H_1:H_2]\le 5$, but does not have to divide 5. I believe the following is a complete classification, but I'll let you work out the details and check to see if there are any cases I missed.


*

*If $L$ contains $\zeta_5$, then $H_1$ is contained in the unique cyclic normal subgroup $C_5\lhd C_5\rtimes C_4$ (which is the Sylow-5 subgroup of $C_5\rtimes C_4$), and $H_2 = 1$. If $L$ doesn't contain any 5th roots of 3, then $H_1 = C_5\lhd C_5\rtimes C_4$, and so the the extension is Galois with group isomorphic to the cyclic group $C_5$ of order 5, generated by the automorphism sending $\sqrt[5]{3}\mapsto \zeta_5\sqrt[5]{3}$.

*If $L$ contains $\zeta_5$, and contains a 5th root of 3, then it must contain all 5th roots of 3, so it contains $\sqrt[5]{3}$ and the Galois group is trivial (the extension is trivial). This corresponds to the case where $H_1 = H_2 = 1$.

*If $L$ doesn't contain $\zeta_5$, and doesn't contain any 5th roots of 3, then the extension is of degree 5, but not Galois. This corresponds to the case $H_1 = C_5\rtimes C_4$ or an index 2 subgroup of $C_5\rtimes C_4$.

*If $L$ doesn't contain $\zeta_5$ or $\zeta_5+\zeta_5^{-1}$, and contains a 5th root of 3, but doesn't contain $\sqrt[5]{3}$, then the extension is of degree 4 and we have $L(\sqrt[5]{3})\cong L(\zeta_5)$, which is Galois over $L$ with cyclic Galois group isomorphic to $(\mathbb{Z}/5\mathbb{Z})^\times\cong C_4$. This corresponds to the case where $H_1$ is precisely the Sylow-4 subgroup of $C_5\rtimes C_4$.

*If $L$ doesn't contain $\zeta_5$, but does contain $\zeta_5+\zeta_5^{-1}$ and contains a 5th root of 3 but not $\sqrt[5]{3}$, then the extension is quadratic and is isomorphic to $L(\zeta_5+\zeta_5^{-1})$. This corresponds to the case where $H_1$ is the unique nontrivial proper subgroup of the Sylow 4-subgroup.

*If $L$ doesn't contain $\zeta_5$ and contains $\sqrt[5]{3}$, then the extension is trivial.
