Galois Group of Cyclotomic Field Extension: A More Natural Approach

Lots of textbooks, when they consider the Galois group of Cyclotomic field extension, do not show how to observe the fact that $\mathrm{Gal}(\mathbb{Q}[{\zeta }_n]/\mathbb{Q})\subseteq U(\mathbb{Z}/n\mathbb{Z})$$\mathrm{Gal}(\mathbb{Q}[\zeta_n]/\mathbb{Q})\subseteq U(\mathbb{Z}/n\mathbb{Z})$. They just define the injective group homomorphism, which is not natural.

Lemma. $(\mathbb{Z}/n\mathbb{Z},+,\cdot )\cong {\mathrm{End}}_{\mathrm{Ab}}(\mathbb{Z}/n\mathbb{Z},+)$$(\mathbb{Z}/n\mathbb{Z},+,\cdot)\cong\mathrm{End}_{\mathrm{Ab}}(\mathbb{Z}/n\mathbb{Z},+)$.

Proof. Let $k\in \mathbb{Z}/n\mathbb{Z}$$k\in\mathbb{Z}/n\mathbb{Z}$ act on $\mathbb{Z}/n\mathbb{Z}$$\mathbb{Z}/n\mathbb{Z}$ via $a⟼ka$$a\longmapsto ka$, this is an injective ring homomorphism from

$$\begin{array}{}\text{(1)}& (\mathbb{Z}/n\mathbb{Z},+,\cdot )\to {\mathrm{End}}_{\mathrm{Ab}}(\mathbb{Z}/n\mathbb{Z},+)\end{array}$$

. To see it is surjective, let $\phi \in {\mathrm{End}}_{\mathrm{Ab}}(\mathbb{Z}/n\mathbb{Z},+)$$\varphi\in \mathrm{End}_{\mathrm{Ab}}(\mathbb{Z}/n\mathbb{Z},+)$, then $\phi (m)=m\phi (1)=\phi (1)m$$\varphi(m)=m\varphi(1)=\varphi(1)m$, $\phi (1)\in \mathbb{Z}/n\mathbb{Z}$$\varphi(1)\in\mathbb{Z}/n\mathbb{Z}$. $◻$$\square$

Corollary. ${\mathrm{Aut}}_{\mathrm{Grp}}(\mathbb{Z}/n\mathbb{Z},+)\cong U(\mathbb{Z}/n\mathbb{Z},+,\cdot )$$\mathrm{Aut}_{\mathrm{Grp}}(\mathbb{Z}/n\mathbb{Z},+)\cong U(\mathbb{Z}/n\mathbb{Z},+,\cdot)$

Notice that the roots of $x^n-1$$x^n-1$ form a multiplicative subgroup of $\mathbb{C}$$\mathbb{C}$ and it is cyclic, $⟨{\zeta }_n⟩\cong \mathbb{Z}/n\mathbb{Z}$$\langle \zeta_n\rangle\cong\mathbb{Z}/n\mathbb{Z}$. And the elements of Galois group send roots to roots, and preserve multiplication, hence $\mathrm{Gal}(\mathbb{Q}[{\zeta }_n]/\mathbb{Q})$$\mathrm{Gal}(\mathbb{Q}[\zeta_n]/\mathbb{Q})$ is a subgroup of the automorphism group of $⟨{\zeta }_n⟩$$\langle \zeta_n\rangle$, hence the Galois group is a subgroup of $U(\mathbb{Z}/n\mathbb{Z})$$U(\mathbb{Z}/n\mathbb{Z})$. To define the injection, we only need to know $\phi (1)$$\varphi(1)$, i.e. $\sigma ({\zeta }_n)$$\sigma(\zeta_n)$ just like what we do in the lemma.

$$\begin{array}{}\text{(2)}& \sigma ({\zeta }_n)={\zeta }_n^k,\sigma ⟼k\mathrm{mod}n\end{array}$$

Since $\mathbb{Q}[{\zeta }_n]\cong \mathbb{Q}[X]/({\mathrm{\Phi }}_n(X))$$\mathbb{Q}[\zeta_n]\cong\mathbb{Q}[X]/(\Phi_n(X))$, the degree of the extension is $\phi (n)$$\varphi(n)$, also the extension is Galois extension, hence

$$\begin{array}{}\text{(3)}& |\mathrm{Gal}(\mathbb{Q}[{\zeta }_n]/\mathbb{Q})|=[\mathbb{Q}[{\zeta }_n]:\mathbb{Q}]\end{array}$$

Hence

$$\begin{array}{}\text{(4)}& \mathrm{Gal}(\mathbb{Q}[{\zeta }_n]/\mathbb{Q})\cong U(\mathbb{Z}/n\mathbb{Z})\end{array}$$