Galois Group of $x^p-1$ cyclic

Question

Prove: The Galois Group of $x^p-1$ over $\mathbb{Q}$ is cyclic when $p$ is prime.

I know that the Galois group is $\mathit{Gal}(\mathbb{Q}(\omega):\mathbb{Q})$ where $\omega$ is the $p^\mathrm{th}$ root of unity. I also know that if $h \in \mathit{Gal}(\mathbb{Q}(\omega):\mathbb{Q})$, then $h(\omega)=\omega^k$ for some $k$ where $1 \le k \le p-1$.

Any help is appreciated. Thanks!

Answer 1

Let $\mathbb{Z}_p^*$ be the set $\{1,2,\dots, p-1\}$ under multiplication mod 7. We know that $\mathbb{Z}_p^*$ is cyclic because $p$ is prime. Thus any group that is isomorphic to $\mathbb{Z}_p^*$ will also be cyclic. By our work in problem 8, it follows that $\mathit{Gal}(\mathbb{Q}(\omega):\mathbb{Q})=\{h_1, h_2, \dots, h_{p-1}\}$ where $h_i(\omega)=\omega^i$. Notice that all indices and powers are in $\mathbb{Z}_p^*$. Consider the function $\sigma:\mathit{Gal}(\mathbb{Q}(\omega):\mathbb{Q}) \to \mathbb{Z}_p^*$ defined $\sigma(h_j(\omega))=j$ for some $j$ where $1 \le j \le p-1$. First notice that the function is well defined. Suppose $h_m=h_n \in \mathit{Gal}(\mathbb{Q}(\omega):\mathbb{Q})$. This can only happen if $m \equiv n \mod 7$. Then $\sigma(h_n(\omega))=n\equiv m=\sigma(h_m(\omega))$. Therefore, $\sigma$ is well-defined. We must now check the $\sigma$ is a homomorphism. Let $h_i, h_j \in \mathit{Gal}(\mathbb{Q}(\omega):\mathbb{Q})$. Then $h_i(\omega)=\omega^i$ and $h_j(\omega)=\omega^j$. Observe that $\sigma(h_i(h_j(\omega))=\sigma(h_i(\omega^j))=\sigma((h_i(\omega))^j)=\sigma((\omega^i)^j)=\sigma(\omega^{ij})=\sigma((\omega^j)^i)=\sigma((h_j(\omega))^i)=\sigma(h_j(\omega^i))=\sigma(h_j(h_i(\omega))$. Thus, $\sigma(h_i(h_j(\omega))=\sigma(h_j(h_i(\omega))=ij$ and the function is a homomorphism. Lastly, we must check that $\sigma$ is a bijection. Suppose $\sigma(h_n(\omega))=\sigma(h_m(\omega))$ for some $h_n, h_m \in \mathit{Gal}(\mathbb{Q}(\omega):\mathbb{Q})$. Then $n=m$ which implies that $h_m=h_n$, thus $\sigma$ is injective. Now let $k \in \mathbb{Z}_p^*$. Then $1 \le k \le p-1$ which implies that there exists some $h_k \in \mathit{Gal}(\mathbb{Q}(\omega):\mathbb{Q})$. However, we can see that $\sigma(h_k(\omega))=k$ and therefore $\sigma$ is surjective. Thus, $\mathit{Gal}(\mathbb{Q}(\omega):\mathbb{Q}) \cong \mathbb{Z}_p^*$. Therefore, $\mathit{Gal}(\mathbb{Q}(\omega):\mathbb{Q})$ is a cyclic group. $\blacksquare$

Answer 2

HINT. An automorphism of $\mathbb{Q}(\zeta)$, where $\zeta^n=1$ and $\zeta$ is a primitive root of unity, is entirely determined by its action on $\zeta$. (Why?)

Show if $\sigma$ is in the Galois group, that $\sigma(\zeta)=\zeta^a$ for some $a$ and that $0 \leq a <n$ and $\text{gcd}(a,n)=1$ (Why?)

Now that you know what $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ looks like. Create a map to $(\mathbb{Z}/n\mathbb{Z})^\times$. What is the map given your work above?

Show that this map is a homomorphism.

Show that this map is injective.

Then you are done! Why? You have an injection between finite sets of the same cardinality so....

Note that this actually shows $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\times$ for $n$ not just prime and you obtain your result when $n=p$. The general result is really no harder and just takes a bit of careful work with integers mod $n$.