Q:

Finding Galois Group

Let $\mathbb Q\subset\mathbb Q(\sqrt{3}+\sqrt[3]{5})$. I want to find the Galois group of the given field extension. 

It would be easy for me if I could find a basis of the given field extension but how does one find a basis of that field extension ? Are there any better ways of finding the Galois group in this case ?

A:

As user lhf points out in the comments, $\mathbb{Q}(\sqrt3 + \sqrt[3]{5})$ is most probably equal to $\mathbb{Q}(\sqrt3, \sqrt[3]{5})$. (Why?)
Now $M:= \mathbb{Q}(\sqrt3, \sqrt[3]{5})/\mathbb{Q}$ is not normal, because $[M:\mathbb{Q}] = 6$, thus the minimal polynomial $X^3-5$ of $\sqrt[3]{5}$ over $\mathbb{Q}$ is also irreducible over $\mathbb{Q}(\sqrt3)$ and is not decomposable into linear factors over $M$.
$X^3-5=(X-\sqrt[3]{5}) \cdot (X-\zeta\sqrt[3]{5}) \cdot (X-\zeta^2\sqrt[3]{5})$ where $\zeta = \exp(\frac{2\pi i}{3}) = -\frac{1}{2} + i\frac{\sqrt3}{2}$.
Now $L:= M(i) = \mathbb{Q}(\sqrt3, \sqrt[3]{5},i)$ is a normal closure of $M$ over $\mathbb{Q}$ and $L/\mathbb{Q}$ is a Galois extension.
$[L:\mathbb{Q}] = 12 = \#Gal(L/\mathbb{Q})$ Now consider $$f(X) = (X^2-3)\cdot(X^3-5)\cdot(X^2+1) \in \mathbb{Q}[X]$$
A $\sigma \in Gal(L/\mathbb{Q})$ is uniquely defined by the images $\sigma(\sqrt3)$, $\sigma(\sqrt[3]{5})$ and $\sigma(i)$. If you keep in mind that there are 12 distinct $\mathbb{Q}$-automorphisms and you know how $\sigma$ maps roots of an irreducible factor of a polynomial, it should be very easy to find the Galois group.

