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I am studying Galois theory and have come across many questions regarding finding the splitting field of $x^n-a$ over $\mathbb{Q}$.

The specific question I was looking at was

Let $L$ be the splitting field of $x^6-2$ over $\mathbb{Q}$. Determine $[L:\mathbb{Q}]$ and describe the Galois group.

In this case I know that $L=\mathbb{Q}(\sqrt[6]{2}, i\sqrt{3})$. I think the degree would be $12$ since the minimal polynomial of $\sqrt[6]{2}$ is $x^6-2$ and the minimal polynomial of $i\sqrt{3}$ is $x^2+3$ and the former is nonreal, so we can multiply the field degrees. I don't know how I would find the Galois group. And are there any methods that would work more generally?

user346096
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There is no General Method of finding the galois Group its requires alot of computation even with a pc.Though on small Splitting fields it can be done.To find the galois group you must ue facts like that an automorphism is defined by its values on the basis of the extension.And say the degree of the extention is $m$ then try to find all the automorphism that can exist .using facts that for every pair of roots for a irreducible polynomial that its splittfield is a subfield of your extension there is an automorphism $\in [L:Q]$ such that $h(a)=a'$ .Then you also know that the order of the galois group is at much $d_1d_2...d_m$ where $d_i$ is the order of a minimum polynomial of a subfield of $[L:Q]$ So you work out all the cases by hand about what can an element inside the gaois group can be.It might be helpfull

How to determine the Galois group of irreducible polynomials of degree $3,4,5$ see also

How to find the Galois group of a polynomial?

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Jam
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  • What about methods for determining the degree of the field extension $[L:\mathbb{Q}]$? – user346096 Jul 30 '16 at 20:37
  • Degree of the minimum polynomial For algebraic extensions .A field $K$ over a field $F$ is in particular a vector space over $F$, and $[K:F]$ is its dimension. For $F(\alpha)$ it's true that this dimension is the degree of the minimal polynomial of $\alpha$ over $F$, because $K$ then has a basis over $F$ given by $1,\alpha,...,\alpha^n$ – Jam Jul 31 '16 at 18:51