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1) How can I compute the galois group $\text{Gal}(K[\sqrt[n] X]/K)$ where $K=\mathbb C(X)$ ?

I would say that the minimal polynomial of $\sqrt[n]X$ on $K$ is $t^n-X\in K[t]$ which is separable, therefore $|\text{Gal}(K[\sqrt[n] X]/K)|=n$. Let define $X_1,...,X_n$ all the solutions. I want to define $\sigma_k: X_i\mapsto X_i^k$ and conclude that

$$\text{Gal}(K[\sqrt[n] X]/K)=\{\sigma _0,...,\sigma _{n-1}\}\cong \mathbb Z/n\mathbb Z$$

What do you think ?

2) How can I compute the galois group $\text{Gal}(\mathbb F_7[\alpha]/\mathbb F_7)$ where $\alpha$ is a root of $t^6-1$ ?

I don't know what to do here.

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  • $\begingroup$ The problem is that $K(\sqrt[n]{X})$ is not usually a Galois extension of $K$, so its Galois group is not usually well-defined. For example, $\mathbb{Q}(\sqrt[3]{2})$ is not a Galois extension of $\mathbb{Q}$. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 29, 2014 at 18:53
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    $\begingroup$ @Jyrki: oh, yes, I did miss that. But in any case the OP's solution did not mention this, and the proposed Galois automorphisms are wrong. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 30, 2014 at 4:21

1 Answer 1

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Hints:

  1. This is a Kummer type root extension and thus Galois. You should observe that the (other) roots of $t^n-X$ are of the form $X_k=\zeta^k \root n\of X$, where $\zeta=e^{2\pi i/n}\in\Bbb{C}\subset K$ is a primitive root of unity of order $n$, and $k=0,1,2,\ldots,n-1$. Show that the Galois group is generated by the automorphism $\sigma$ that is fully determined by $\sigma(X_0)=X_1$.
  2. What does Little Fermat tell you about the six non-zero elements of $\Bbb{F}_7$?
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  • $\begingroup$ 1) could you explain ? 2) I don't understand. $\endgroup$
    – idm
    Commented Nov 29, 2014 at 23:11
  • $\begingroup$ 1) Which numbers occur as automorphic images of $\root3\of2$? It's the same here (except that this time we are dealing with the nth root instead of a cubic root). 2) If $a\in\Bbb{F}_7$ what can you say about $a^6$? $\endgroup$
    – Jyrki Lahtonen
    Commented Nov 29, 2014 at 23:28
  • $\begingroup$ I mean, you probably can show that $Gal(\Bbb{Q}(\root3\of2,\omega)/\Bbb{Q}(\omega)$ is isomorphic to $C_3$? There $\omega=(-1+i\sqrt3)/2$ is the primitive third root of unity. Part 1 is reusing the same ideas. $\endgroup$
    – Jyrki Lahtonen
    Commented Nov 30, 2014 at 0:00

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