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!

  • 3
    $\begingroup$ 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. $\endgroup$ – Dietrich Burde May 8 at 15:15

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