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!
Best Answer
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 $n$.