Why $Gal(\mathbb Q(\zeta _n)/\mathbb Q)\hookrightarrow (\mathbb Z/n\mathbb Z)^\times $

Question

Let $\zeta _n=e^{\frac{2i\pi}{n}}$ an let $\mu_n=\{1,\zeta _n,\zeta _n^2,...,\zeta _n^{n-1}\}$.

1) Show that $\mathbb Q(\mu_n)=\mathbb Q(\zeta _n)$ and that $\mathbb Q(\zeta _n)/\mathbb Q$ is Galoisienne.

2) Show that $$Gal(\mathbb Q(\zeta _n)/\mathbb Q)\hookrightarrow (\mathbb Z/n\mathbb Z)^\times$$ where $(\mathbb Z/n\mathbb Z)^\times $ is the multiplicative group of $\mathbb Z/n\mathbb Z$.

My work

1) Since $\zeta _n\in\mathbb Q(\mu_n)$ and that if $k\in \mu_n$, then $k=\zeta _n^t\in\mathbb Q(\zeta _n)$ for a certain $t$, we also have $\mathbb Q(\mu_n)\subset \mathbb Q(\zeta _n)$. Therefore $$\mathbb Q(\zeta _n)=\mathbb Q(\mu_n).$$

Q1) Is it correct ?

Moreover, $\mathbb Q(\mu_n)$ is by definition the splitting field of $X_n-1$, therefore $X^n-1$ split over $\mathbb Q(\zeta_n)$ and since all extension of $\mathbb Q$ is separable, we get that $\mathbb Q(\zeta _n)/\mathbb Q$ is galoisienne.

Q2) Is it correct ?

2) If $\sigma \in Gal(\mathbb Q(\zeta _n)/\mathbb Q)$, then $\sigma $ is determined by $\sigma (\zeta _n)=\zeta _n^k$. Then let denote $\sigma _k(\zeta _n)=\zeta _n^k$. Since $\sigma $ is a field morphism, it's injectif, and thus $$\sigma _k(\zeta _n)=1\iff \zeta _n^k=1\iff k\mid n$$

Q3) then how I can conclude ? I'm not sure about what I'm doing.

Answer 1

Q1) Everything is fine.

Q2) You're right.

Q3) Let define the following map:

$$\varphi:\left\{\begin{array}{ccc} (\mathbb{Z}/n\mathbb{Z})^\times & \rightarrow & \textrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\\ [d] & \mapsto & \sigma:\zeta_n\mapsto{\zeta_n}^d\end{array}\right..$$ $\varphi$ is a group isomorphism. One has: $$\textrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^\times.$$

Going further:

Definition. Let $\zeta$ be a root of unity, then $\mathbb{Q}(\zeta)/\mathbb{Q}$ is called a cyclotomic extension.

One has the:

Proposition. $\mathbb{Q}(\zeta)/\mathbb{Q}$ is a galois extension. Let $n$ be the order of $\zeta$, $\textrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^\times$.

Proof. $\zeta$ is a $n$th primitive root of the unity, then $\zeta$ is a root of $\Phi_n$ the $n$th cyclotomic polynomial and $\zeta$ is a generator of all the roots of $\Phi_n$. Hence, $\mathbb{Q}(\zeta)$ is the splitting field of $\Phi_n$ (separable over $\mathbb{Q}$, since $\Phi_n$ is irreducible over $\mathbb{Q}$ and $\textrm{char}(\mathbb{Q})=0$) and $\mathbb{Q}(\zeta)/\mathbb{Q}$ is a galois extension. The following map: $$\varphi:\left\{\begin{array}{ccc} (\mathbb{Z}/n\mathbb{Z})^\times & \rightarrow & \textrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})\\ [d] & \mapsto & \sigma:\zeta_n\mapsto{\zeta_n}^d\end{array}\right.$$ is a group isomorphism. $\Box$

Answer 2

The most canonical way to see the isomorphism in question is to write its inverse as $$\sigma: \mathbb Z/n\mathbb Z\to Gal(\mathbb Q(\mu _n)/\mathbb Q): \bar k\mapsto \sigma_{\bar k}$$ where $\sigma_{\bar k}$ is the unique field automorphism of $\mathbb Q(\mu _n)$ mapping $\omega\in \mu _n$ to $\sigma_{\bar k}(\omega)=\omega^k$
But why does such an automorphism exist?
Answer: because the cyclotomic polynomial $\Phi_n(X)$ of level $n$ is irreducible (Gauß,1808) so that given any two of its roots, which form the set $\mu_n$, there is an automorphism of $\mathbb Q(\mu _n)$ sending the first root to the second root (Kronecker's theorem).
It is then clear that $ \sigma$ is an isomorphism of groups. This presentation has the advantage that no choice of an $n$-th root of unity $\zeta_n$ need be made.
However let me emphasize that whatever presentation is chosen the irreducibility of the cyclotomic polynomial and Kronecker's theorem are involved and theses are really serious mathematical results, triumphs of 19th century algebra/number theory.