# How to Determine Generators and Relations of Galois group of $X^7-2$

I try to understand how one may describe the Galois group of polynomials of the form $$X^n-a\in\mathbb{Q}[X]$$, in terms of generators and relations, where $$X^n-a$$ ($$n>1$$) is assumed to be irreducible. In Dummit and Foote they did this for $$X^8-2$$ and my professor did it for $$X^5-3$$.

I'm not sure I really understand the process though, I tried to mimic it for $$X^7-2$$.

The splitting field of $$X^7-2$$ is given by $$\mathbb{Q}(\sqrt[7]{2},\zeta_7)=\mathbb{Q}(\alpha,\zeta)$$ ($$\zeta_p$$ is the $$p^{\operatorname{th}}$$ root of unity). Since $$\alpha$$ and $$\zeta$$ generates $$\mathbb{Q}(\alpha,\zeta)/\mathbb{Q}$$, an automorphism of $$\mathbb{Q}(\alpha,\zeta)$$ is determined by its action on $$\alpha$$ and $$\zeta$$. An automorphism must map a root of an irreducible polynomial in $$\mathbb{Q}[X]$$ to another root of the same polynomial. This means that an automorphism of $$\mathbb{Q}(\alpha,\zeta)$$ must be of the form $$\zeta\mapsto \zeta^k,\quad 1\leq k\leq 6$$ $$\alpha\mapsto \zeta^j\alpha,\quad 0\leq j\leq 6.$$

Since $$[\mathbb{Q}(\alpha,\zeta):\mathbb{Q}]=7\cdot 6=42$$ and the extension is Galois, all of the above maps must be automorphisms. Now, define $$\sigma$$ and $$\tau$$ as follows $$\sigma(\zeta)=\zeta^3,\quad \tau(\zeta)=\zeta$$ $$\sigma(\alpha)=\alpha,\quad \tau(\alpha)=\alpha\zeta.$$ One may check that $$\sigma$$ and $$\tau$$ has order $$6$$ respectively $$7$$.

Let $$N=\langle\tau\rangle$$. Then $$N$$ is a subgroup of $$G$$ order $$7$$. Note that $$N$$ is normal since $$\sigma\circ\tau\circ\sigma^{-1}(\zeta)=\sigma\circ\sigma^{-1}(\zeta)=\zeta$$ and $$\sigma\circ\tau\circ\sigma^{-1}(\alpha)=\sigma\circ\tau(\alpha)=\sigma(\alpha\zeta)=\sigma(\alpha)\sigma(\zeta)=\alpha\zeta^3$$. This shows that $$\sigma\circ\tau\circ\sigma^{-1}=\tau^3.$$ This gives us (for some reason) that $$\operatorname{Gal}(\mathbb{Q}(\alpha,\zeta)/\mathbb{Q})=\langle\sigma,\tau|\sigma^6=\tau^7=1\quad \sigma\circ\tau\circ\sigma^{-1}=\tau^3\rangle.$$

Is the above procedure correct? What I did is almost identical to what my professor did. But I did choose the map $$\zeta\mapsto\zeta^3$$ instead of $$\zeta\mapsto\zeta^2$$ (which my professor did choose for the polynomial $$X^5-3$$), since I thought the goal is to find an automorphism that generates all $$\zeta^k$$.

Why is it enough to compute the conjugate to determine the Galois group? I don't really understand what's so special about $$N$$ being normal. How can I be sure that the relations describes the whole Galois group?

I would be really happy if someone could explain this to me. Thanks for taking your time!

• The Galois group is $C_7\rtimes U(C_7)\cong C_7\rtimes C_6$, see this question. There the Galois group of $X^n-a$ is explained, with detailed references. Then it is clear what generators and relations are. – Dietrich Burde May 8 at 15:15