Exercise 6.2.4

Proof.

(a)

${\zeta }_n$ is a root of $x^n-1\in \mathbb{Q}[x]$. Write ${𝕌}_n$ the set of $n$th roots of unity in $ℂ$ : $${𝕌}_n=\{{\zeta }_n^k,0\le k\le n-1\}$$

and $|{𝕌}_n|=n$.

As $x^n-1=\underset{\zeta \in {𝕌}_n}{\prod <!--\; FUNCTION\; APPLICATION\; -->}(x-\zeta )$, $x^n-1$ is separable, and the splitting field of $x^n-1$ over $\mathbb{Q}$ is $\mathbb{Q}({\zeta }_n,\dots ,{\zeta }_n^{n-1})=\mathbb{Q}({\zeta }_n)$

Conclusion: $\mathbb{Q}({\zeta }_n)$ is the splitting field of the separable polynomial $x^n-1\in \mathbb{Q}[x]$ over $\mathbb{Q}$.

(b)

Let $\sigma \in \mathrm{Gal}(\mathbb{Q}({\zeta }_n):\mathbb{Q})$.

As ${\zeta }_n$ is a root of $x^n-1\in \mathbb{Q}[x]$, by Proposition 6.1.4(a), $\sigma ({\zeta }_n)$ is a root of $x^n-1$, thus $\sigma ({\zeta }_n)\in {𝕌}_n$, so

$$\sigma ({\zeta }_n)={\zeta }_n^i,i\in \mathbb{N}.$$

(c)

Note that ${\zeta }_n=e^{2\mathit{i\pi }∕n}$ is an element of order $n$ in the group ${𝕌}_n$. Indeed, for all $k\in \mathbb{Z}$,

$${\zeta }_n^k=1⟺e^{2\mathit{i\pi k}∕n}=1⟺k∕n\in \mathbb{Z}⟺n\mid k.$$

$\sigma$ being a field isomorphism, $\sigma ({\zeta }_n)\in {𝕌}_n$ is also of order $n$. Indeed, for all $k\in \mathbb{Z}$,

$$\sigma {({\zeta }_n)}^k=1⟺\sigma ({\zeta }_n^k)=1⟺{\zeta }_n^k=1⟺n\mid k.$$

If the order of an element $\zeta$ is $|\zeta |=n$, then for all integer $j$, the order of ${\zeta }^j$ in ${𝕌}_n$ is

$$|{\zeta }^j|=\frac{n}{n\wedge j}.$$

Indeed for all $k\in \mathbb{Z}$,

${({\zeta }^j)}^k=1⟺n\mid \mathit{jk}⟺\frac{n}{n\wedge j}\mid \frac{j}{n\wedge j}k⟺\frac{n}{n\wedge j}\mid k$ (since $\frac{n}{n\wedge j}\wedge \frac{j}{n\wedge j}=1$).

If we apply this result to ${\zeta }_n^i=\sigma ({\zeta }_n)$, we obtain

$$\frac{n}{n\wedge i}=|{\zeta }_n^i|=|\sigma ({\zeta }_n|=n,$$

thus

$$n\wedge i=1.$$

(d)

Let

$\phi :\{\begin{array}{ccc}\hfill \mathrm{Gal}(\mathbb{Q}({\zeta }_n)∕\mathbb{Q})\hfill & \hfill \to \hfill & \hfill {(\mathbb{Z}∕\mathit{n\mathbb{Z}})}^{\text{∗}}\hfill \\ \hfill \sigma \hfill & \hfill \mapsto \hfill & \hfill [i]:\sigma ({\zeta }_n)={\zeta }_n^i\hfill \end{array}$

Note that $\phi$ is well defined, since ${\zeta }_n^i={\zeta }_n^j$ implies $i\equiv j(\mathit{mod}n)$ and so $[i]=[j]$.

We show that $\phi$ is a group homomorphism.

If $\sigma ,\tau \in \mathrm{Gal}(\mathbb{Q}({\zeta }_n)∕\mathbb{Q})$, and $\phi (\sigma )=[i],\phi (\tau )=[j]$, then $\sigma ({\zeta }_n)={\zeta }_n^i,\tau ({\zeta }_n)={\zeta }_n^j,$ thus

$$(\sigma \circ \tau )({\zeta }_n)=\sigma ({({\zeta }_n)}^j)={(\sigma ({\zeta }_n))}^j={({\zeta }_n^i)}^j={\zeta }_n^{\mathit{ij}},$$

therefore

$$\phi (\sigma \circ \tau )=[\mathit{ij}]=[i][j]=\phi (\sigma )\phi (\tau ).$$

$\phi$ is injective :

If $\phi (\sigma )=[1]$, then $\sigma ({\zeta }_n)={\zeta }_n$. Since $\sigma \in \mathrm{Gal}(\mathbb{Q}({\zeta }_n)∕\mathbb{Q})$, $\sigma$ is uniquely determined by the image of ${\zeta }_n$, thus $\sigma =1_{\mathbb{Q}({\zeta }_n)}$. The kernel of $\phi$ is trivial, thus $\phi$ is injective.

Conclusion: there exist an injective group homomorphism

$$\phi :\mathrm{Gal}(\mathbb{Q}({\zeta }_n)∕\mathbb{Q})\to {(\mathbb{Z}∕\mathit{n\mathbb{Z}})}^{\text{∗}}.$$

(e)

As $\mathbb{Q}({\zeta }_n)$ is the splitting field of a separable polynomial over $\mathbb{Q}$, $$|\mathrm{Gal}(\mathbb{Q}({\zeta }_n)∕\mathbb{Q})|=[\mathbb{Q}({\zeta }_n):\mathbb{Q}].$$

If we suppose that $[\mathbb{Q}({\zeta }_n):\mathbb{Q}]=\varphi (n)$, $\phi$ is an injection between two set of same cardinality, thus $\phi$ is a bijection, and so $\phi$ is a group isomorphism. Conversely, if $\phi$ is a group isomorphism, then $[\mathbb{Q}({\zeta }_n):\mathbb{Q}]=|\mathrm{Gal}(\mathbb{Q}({\zeta }_n)∕\mathbb{Q})|=\varphi (n)$

Conclusion: $[\mathbb{Q}({\zeta }_n):\mathbb{Q}]=\varphi (n)$ if and only if $\mathrm{Gal}(\mathbb{Q}({\zeta }_n)∕\mathbb{Q})\simeq {(\mathbb{Z}∕\mathit{n\mathbb{Z}})}^{\text{∗}}$.

(f)

If $p$ is prime, we know that $f=1+x+\cdots +x^{p-1}$ is irreducible over $\mathbb{Q}$, so $f$ is the minimal polynomial of ${\zeta }_p$ over $\mathbb{Q}$. This implies that $[\mathbb{Q}({\zeta }_p):\mathbb{Q}]=p-1=\varphi (p)$.

By part (e), we know then that $\mathrm{Gal}(\mathbb{Q}({\zeta }_p)∕\mathbb{Q})\simeq {(\mathbb{Z}∕\mathit{p\mathbb{Z}})}^{\text{∗}}$ (and so this group is cyclic with order $p-1$).