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I am trying to understand the rudiments of galois theory but I have a hard time answering the following question: Suppose we have $f(X)=x^5-2$ over $\mathbb{Q}$. What is $Gal(E_f: \mathbb{Q})$?

After seeing similar questions here I am more confused: Isn't it true that since $f$ is irreducible, every possible permutation of its roots can be extended linearly to a endomorphism $E_f \rightarrow E_f$ while holding $\mathbb{Q}$ constant? If that is so then shouldn't the Galois Group be $S_5$?

I understand that this isn't the case, but can't really understand why...

Nick A.
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    Your reasoning shows that the Galois group of the splitting field of a polynomial of degree $n$, is a subgroup of $S_n$. You don't necessarily get all permutations as elements of the Galois group. – Watson May 29 '18 at 11:31
  • The Galois group is not $S_5$ because it has order $20$, not $120$. – lhf May 29 '18 at 11:32
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    See https://math.stackexchange.com/questions/1072830/galois-group-of-the-splitting-field-of-the-polynomial-x5-2-over-mathbb-q, https://math.stackexchange.com/questions/204552/computing-the-galois-group-of-polynomials-xn-a-in-mathbbqx, https://mathoverflow.net/questions/143739/galois-group-of-xn-2 – lhf May 29 '18 at 11:32
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    I think the problem is that I have understood the "swapping roots" theorem wrongly. Indeed I can find an endomorphism $σ$ that sents any root I want to any other root I want, but there might be constraints on $σ$. For example, I cannot have an $σ$ sending $2^{\frac{1}{5}}$ and $2^{\frac{1}{5}}ζ_5$ to themselves but do a cyclic permutation to the others. – Nick A. May 29 '18 at 11:33
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    This is exactly the problem: Indeed, for any pair of roots you find an automorphism sending the first root to the second, but you do not have much control what will happen to the other roots. You cannot even say what the second root will be sent to, in particular there is in general no way to extend a transposition of roots to an automorphisms, which would be sufficient to prove that the Galois group is $S_5$. – asdq May 29 '18 at 11:52

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First note that the splitting field of $f$ is given by $L = \mathbb{Q}(\sqrt[5]{2}, \zeta_5)$, where $\zeta_5$ is the primitive root of unity.

Now we know that any automorphism on $L$ is uniquely determined by the action on the adjoined elements $\sqrt[5]{2}$ and $\zeta_{5}$. Additionally we know that if $\alpha_1$ and $\alpha_2$ are roots of the same irreducible polynomial over $\mathbb{Q}$ then the identity automorphism on $\mathbb{Q}$ can be extended to a automorphism $\sigma$ of the splitting field s.t. $\sigma(\alpha_1) = \alpha_2$. In particular in our case we can have automorphism of $L$ s.t. $\sigma(\sqrt[5]{2}) = \zeta_5^2\sqrt[5]{2}$. Now we can do the same for $\zeta_5$ and for example get an automorphism $\tau$ on $L$, s.t. $\tau(\zeta_5) = \zeta_5^3$ and it is identity on $\mathbb{Q}$.

However note that this doesn't give you every possible permutation of roots, as it's not possible to permute the first three roots, while to keep the other two fixed. So to summarize there is an element of the Galois group of $f$ (assuming $f$ is irreducible) s.t. that it send any root to any other root of $f$, but it's not necessarily keeping the other roots fixed or permuting them in arbitrary manner.

Stefan4024
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