Exercise 9.2.11

Proof.

(a)

Let $f^{\prime }=\mathit{fq}$, and $e^{\prime }=n∕f^{\prime }$. Then $p-1=\mathit{ef}=e^{\prime }{f}^{\prime }$, and $e=e^{\prime }q$.

By section 9.2,

$L_f$ is the fixed field of ${\tilde{H}}_f=⟨\sigma ⟩$, where $\sigma ({\zeta }_p)={\zeta }_p^{g^e}$.

${\tilde{H}}_f$ is the set of automorphisms $\xi$ such that ${\zeta }_p\mapsto \xi ({\zeta }_p)={\zeta }_p^i,i\in H_f=\{1,g^e,g^{2e},\cdots ,g^{(f-1)e}\}$.

This result applied to $f^{\prime }$ gives:

$L_{\mathit{fq}}$ is the fixed field of ${\tilde{H}}_{\mathit{fq}}=⟨\tau ⟩$, where $\tau ({\zeta }_p)={\zeta }_p^{g^{e^{\prime }}}={\zeta }_p^{g^{e∕q}}$.

By the Galois correspondence, $\mathrm{Gal}(\mathbb{Q}({\zeta }_p)∕L_{\mathit{fq}})={\tilde{H}}_{\mathit{fq}}=⟨\tau ⟩$.

As $\mathbb{Q}\subset L_f$ is a Galois extension, $\tau L_f=L_f$ (Theorem 7.2.5).

If ${\sigma }^{\prime }:L_f\to L_f$ is the restriction of $\tau$ to $L_f$, then ${\sigma }^{\prime }\in \mathrm{Gal}(L_f∕L_{\mathit{fq}})$.

The restriction mapping $\psi :\mathrm{Gal}(\mathbb{Q}({\zeta }_p)∕L_{\mathit{fq}})\to \mathrm{Gal}(L_f∕L_{\mathit{fq}})$ is a surjective mapping by the proof of Theorem 7.2.7, so every element of $\mathrm{Gal}(L_f∕L_{\mathit{fq}})$ is of the form $\psi ({\tau }^k)={\sigma {}^{\prime }}^k,k\in \mathbb{Z}$, therefore

$$\mathrm{Gal}(L_f∕L_{\mathit{fq}})=⟨{\sigma }^{\prime }⟩.$$

Since $|\mathrm{Gal}(L_f∕L_{\mathit{fq}})|=q$ (Proposition 9.2.1), the order of ${\sigma }^{\prime }$ is $q$.

Note: as $\tau ({\zeta }_p)={\zeta }_p^{g^{e∕q}},\tau ((f,\lambda ))=(f,g^{e∕q}\lambda )$, for every period $(f,\lambda )$ (Lemma 9.2.4(d)), and $(f,\lambda )\in L_f$, so

$${\sigma }^{\prime }((f,\lambda ))=(f,g^{e∕q}\lambda ).$$

(b)

Since $q\mid p-1,p\wedge q=1$, therefore ${\Phi }_q(x)=\frac{x^q-1}{x-1}$ is irreducible over $\mathbb{Q}({\zeta }_p)$ by Exercise 9.1.16. Hence ${\Phi }_q$ is a fortiori irreducible over the subfields $L_f,L_{\mathit{fq}}$ of $\mathbb{Q}({\zeta }_p)$. Consequently

$$[L_f(\omega ):L_f]=[L_{\mathit{fq}}(\omega ):L_{\mathit{fq}}]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_q)=q-1.$$

$L_f(\omega )$ is the splitting field of ${\Phi }_q$ over $L_f$, $L_f\subset L_f(\omega )$ is thus a Galois extension, and similarly $L_{\mathit{fq}}\subset L_{\mathit{fq}}(\omega )$ is Galois.

By Exercises 8.3.2 and 8.2.7, $L_f(\omega )$ is a Galois extension of $L_{\mathit{fq}}$, a fortiori of $L_{\mathit{fq}}(\omega )$.

Let

$$\phi :\{\begin{array}{ccc}\hfill \mathrm{Gal}(L_f(\omega )∕L_{\mathit{fq}}(\omega ))\hfill & \hfill \to \hfill & \hfill \mathrm{Gal}(L_f∕L_{\mathit{fq}})\hfill \\ \hfill \sigma \hfill & \hfill \mapsto \hfill & \hfill \sigma {\text{|}}_{L_f}\hfill \end{array}$$

This mapping is well defined since $L_f$ is a normal extension of $L_{\mathit{fq}}$, so $\sigma L_f=L_f$, and $\sigma$ fixes the elements of $L_{\mathit{fq}}(\omega )$, a fortiori the elements of $L_{\mathit{fq}}$.

$\phi$ is a group homomorphism, and $\phi$ is injective:

if $\sigma \in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $\sigma (\omega )=\omega$, and $\sigma$ is the identity on $L_f$, thus $\sigma$ is the identity on $L_f(\omega )$, so $\sigma =e$, therefore $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{e\}$.

Moreover, $[L_f:L_{\mathit{fq}}]=q$ and $[L_f(\omega ):L_f]=[L_{\mathit{fq}}(\omega ):L_{\mathit{fq}}]=q-1$, therefore, by the Tower Theorem, $[L_f(\omega ):L_{\mathit{fq}}(\omega )]=q$. Hence $|\mathrm{Gal}(L_f(\omega )∕L_{\mathit{fq}}(\omega ))|=|\mathrm{Gal}(L_f,L_{\mathit{fq}})|=q$, so $\phi$ is a group isomorphism.

(c)

Let $\sigma ={\phi }^{-1}({\sigma }^{\prime })$. Then $\sigma$ is a generator of $\mathrm{Gal}(L_f(\omega )∕L_{\mathit{fq}}(\omega ))$, and $\phi (\sigma )={\sigma }^{\prime }$.

As $\sigma {\text{|}}_{L_f}={\sigma }^{\prime }$, by the note in part (a),

$$\sigma ((f,\lambda ))={\sigma }^{\prime }((f,\lambda ))=(f,g^{e∕q}\lambda ).$$

(d)

$H_f=⟨g^e⟩$, and $H_{\mathit{fq}}=⟨g^{e∕q}⟩$.

We show first that $g^{k(e∕q)}\notin H_f$ if $1\le k\le q-1$. If not, there would exist an integer $j$ such that $g^{k(e∕q)}=g^{\mathit{je}}$. As the order of $g$ is $p-1=\mathit{ef}$, $\mathit{ef}\mid k\frac{e}{q}-\mathit{je}$, so $\mathit{\lambda efq}=\mathit{ke}-\mathit{jeq},\lambda \in \mathbb{Z}$, therefore $\mathit{\lambda fq}=k-\mathit{jq}$, and so $q\mid k$. It is impossible since $1\le k\le q-1$.

If $0\le i<j\le q-1$, by the preceding result, ${(g^{i(e∕q)})}^{-1}{g}^{j(e∕q)}=g^{(j-i)(e∕q)}\notin H_f$, therefore $g^{i(e∕q)}{H}_f\ne g^{j(e∕q)}{H}_f$.

The $q$ left cosets $H_f,g^{e∕q}{H}_f,g^{2e∕q}{H}_f,\cdots ,g^{(q-1)e∕q}{H}_f$ are so distinct. Since $(H_{\mathit{fq}}:H_f)=q$, the set of left cosets is reduced to these $q$ cosets, which give a partition of $H_{\mathit{fq}}$:

$$H_{\mathit{fq}}=H_f\cup g^{e∕q}{H}_f\cup g^{2e∕q}{H}_f\cup \cdots \cup g^{(q-1)e∕q}{H}_f.$$