Proof Consider the following element of ${\mathbb{L}} _p^{\text {ha}}$ :

$$ \begin{align*}\alpha=\sum_{k=1}^{\infty}p^{\frac{\lfloor\sqrt{2}\cdot p^k\rfloor}{p^k}}.\end{align*} $$

If $\alpha \in {\mathbb{C}} _p$ , then there exists a p-adic algebraic number $\beta \in \overline {{\mathbb{Q}} }_p$ that $v_p(\alpha -\beta )>2$ . This shows that the canonical expansion of $\beta $ in ${\mathbb{L}} _p^{\text {ha}}$ has the form

$$ \begin{align*}\beta=\sum_{k=1}^{\infty}p^{\frac{\lfloor\sqrt{2}\cdot p^k\rfloor}{p^k}}+\text{ terms with exponent greater than } 2.\end{align*} $$

Thus, $\operatorname {\mathrm {supp}}(\beta )$ has accumulation value $\sqrt {2}$ . However, this is impossible: Lampert showed in [Reference LampertLam86, Theorem 2] that the set

$$ \begin{align*}{\mathcal{A}}:=\left\{\alpha\in{\mathbb{L}}_p\middle\vert \{\text{accumulation value of }\operatorname{\mathrm{supp}}(\alpha)\}\subset {\mathbb{Q}}\right\}\end{align*} $$

is an algebraically closed field. Since the support of every p-adic rational number lies in ${\mathbb{Z}} \subset {\mathbb{Q}} $ , $\overline {{\mathbb{Q}} }_p$ is a subfield of ${\mathcal{A}} $ . On the other hand, $\beta $ does not belong to ${\mathcal{A}} $ . This contradiction shows that ${\mathbb{L}} _p^{\text {ha}}$ is not contained in ${\mathbb{C}} _p$ . In particular, ${\mathbb{L}} _p^{\text {ha}}$ contains $\overline {{\mathbb{Q}} }_p$ as a proper subfield.

To show that ${\mathbb{L}} _p^{\text {ha}}$ is not complete and does not contain ${\mathbb{C}} _p$ , we can consider the sequence $\left (\sum _{k=1}^n p^{k-1/k}\right )_{n\geq 1}$ in $\overline {{\mathbb{Q}} }_p\subseteq {\mathbb{L}} _p^{\text {ha}}$ , which clearly converges in ${\mathbb{C}} _p$ but has non-hyper-algebraic limit $\sum _{k=1}^\infty p^{k-1/k}$ in ${\mathbb{L}} _p$ : the p-power-free part of the denominators of elements of its support is unbounded.

To prove ${\mathbb{L}} _p^{\text {ha}}$ is not dense in ${\mathbb{L}} _p$ , we consider the element $\gamma =\sum _{k=1}^\infty p^{-\frac {1}{p^k-1}}$ in ${\mathbb{L}} _p$ . If it lies in the completion of ${\mathbb{L}} _p^{\text {ha}}$ , then there exists an element $\delta \in {\mathbb{L}} _p^{\text {ha}}$ that $v_p(\gamma -\delta )>~1$ . This leads to a contradiction if we consider the canonical expansion of $\delta $ in ${\mathbb{L}} _p^{\text {ha}}$

$$ \begin{align*}\delta=\sum_{k=1}^\infty p^{-\frac{1}{p^k-1}}+\text{ terms with exponent greater than 1}.\end{align*} $$

The p-power-free parts of the denominators of elements of $\operatorname {\mathrm {supp}}(\delta )$ are unbounded, suggesting that $\delta $ is not hyper-algebraic.

Proof

1. (1) For any positive integer N, the sets $\frac {1}{pN}{\mathbb{Z}} [1/p]$ and $\frac {1}{N}{\mathbb{Z}} [1/p]$ are identical.

2. (2) One has

   $$ \begin{align*}\{r_q\}_{q\in{\mathbb{Q}}}\subseteq {\mathbb{F}}_{p^{{\mathfrak{F}}_\alpha}}\cap{\mathbb{F}}_{p^s}={\mathbb{F}}_{p^{\gcd({\mathfrak{F}}_\alpha,s)}}.\end{align*} $$

   The result follows from the minimality of ${\mathfrak{F}} _\alpha $ .

3. (3) By the first assertion, we may assume that N is coprime to p. Suppose the contrary that $N=d\cdot {\mathfrak{T}} _\alpha +r$ with $d\in {\mathbb{Z}} _{\geq 1}$ and $r\in \{1,\ldots ,{\mathfrak{T}} _\alpha -1\}$ . Take $q\in \operatorname {\mathrm {supp}}(\alpha )$ . Then the inclusion $q\in \frac {1}{{\mathfrak{T}} _\alpha }{\mathbb{Z}} [1/p]\cap \frac {1}{N}{\mathbb{Z}} [1/p]$ allows us to write

   $$ \begin{align*}q=\frac{a_1\cdot p^{v_1}}{{\mathfrak{T}}_\alpha}=\frac{a_2\cdot p^{v_2}}{N},\end{align*} $$

   where $a_1,a_2,v_1,v_2\in {\mathbb{Z}} $ with $a_1,a_2$ coprime to p. By comparing the p-adic valuation, we get $v_1=v_2$ . Since

   $$ \begin{align*}a_2\cdot p^{v_2}=(d\cdot {\mathfrak{T}}_\alpha+r)\cdot q=d\cdot a_1\cdot p^{v_1}+r\cdot q=d\cdot a_1\cdot p^{v_2}+r\cdot q,\end{align*} $$

   we obtain that $q=\frac {a_2-d\cdot a_1}{r}\cdot p^{v_2}\in \frac {1}{r}{\mathbb{Z}} [1/p]$ , which contradicts the minimality of ${\mathfrak{T}} _\alpha $ .

Proof The first and the second assertions follow from the definition of addition and multiplication on ${\mathbb{L}} _p$ . In particular, if ${\mathbb{Q}} _p(\alpha )$ is unramified over ${\mathbb{Q}} _p$ , then ${\mathbb{Q}} _p(\alpha )=\text {Frac}W({\mathbb{F}} _{p^{{\mathfrak{f}} _\alpha }})$ . As a result, every element in ${\mathbb{Q}} _p(\alpha )$ has the form $\sum _{k>\!\!\!>-\infty }[\zeta _k]p^k$ , where $\zeta _k\in {\mathbb{F}} _{p^{{\mathfrak{f}} _\alpha }}$ for all k. This shows that ${\mathfrak{T}} _\alpha =1$ and ${\mathfrak{F}} _\alpha ={\mathfrak{f}} _\alpha $ .

For the third assertion, the result is trivial when $\vert \operatorname {\mathrm {supp}}(\alpha )\vert =1$ . Now we suppose $\vert \operatorname {\mathrm {supp}}(\alpha )\vert \geq 2$ and write $\alpha =[\zeta ]p^{v_p(\alpha )}-A$ for some $\zeta \in \overline {{\mathbb{F}} }_p$ with $v_p(A)>v_p(\alpha )$ . Then $\zeta \in {\mathbb{F}} _{p^{{\mathfrak{F}} _\alpha }}$ , ${\mathfrak{T}} _A\mid {\mathfrak{T}} _\alpha $ and ${\mathfrak{F}} _A\mid {\mathfrak{F}} _\alpha $ . The result follows from the expansion

$$ \begin{align*}\alpha^{-1}=[\zeta^{-1}]p^{-v_p(\alpha)}\sum_{k=0}^\infty \left([\zeta^{-1}]p^{-v_p(\alpha)}\cdot A\right)^k,\end{align*} $$

where $v_p\left ([\zeta ^{-1}]p^{-v_p(\alpha )}\cdot A\right )>0,\ {\mathfrak{T}} _{[\zeta ^{-1}]p^{-v_p(\alpha )}\cdot A}\mid {\mathfrak{T}} _\alpha \text { and }{\mathfrak{F}} _{[\zeta ^{-1}]p^{-v_p(\alpha )}\cdot A}\mid {\mathfrak{F}} _\alpha $ .

Proof As we have mentioned, the action of ${\mathcal{G}} _{{{\mathbb{F}} }_p}$ on ${\mathbb{L}} _p$ sends p-adic algebraic numbers to their conjugates under the action of ${\mathcal{G}} _{{\mathbb{Q}} _p}$ . As a result, one has

(4.1) $$ \begin{align} \left\lvert\{g(\alpha)\colon g\in{\mathcal{G}}_{{\mathbb{F}}_p}\}\right\rvert\leq [{\mathbb{Q}}_p(\alpha)\colon{\mathbb{Q}}_p] \end{align} $$

for any $\alpha \in \overline {{\mathbb{Q}} }_p$ .

Suppose there exists $\alpha \in \overline {{\mathbb{Q}} }_p$ that ${\mathfrak{F}} _\alpha>[{\mathbb{Q}} _p(\alpha )\colon {\mathbb{Q}} _p]$ . Thus, there exists a rational number $q_0\in \operatorname {\mathrm {supp}}(\alpha )$ such that the minimal finite field containing $C_{q_0}(\alpha )$ is ${\mathbb{F}} _{p^r}$ with $r>[{\mathbb{Q}} _p(\alpha )\colon {\mathbb{Q}} _p]$ . Consider the set

$$ \begin{align*}\{g(C_{q_0}(\alpha))\colon g\in{\mathcal{G}}_{{\mathbb{F}}_p}\}=\left\{C_{q_0}(\alpha),C_{q_0}(\alpha)^p,\ldots,C_{q_0}(\alpha)^{p^n},\ldots\right\}.\end{align*} $$

The cardinality of this set is the minimal positive integer d that $C_{q_0}(\alpha )=C_{q_0}(\alpha )^{p^d}$ , which is the same as the minimal positive integer d that $C_{q_0}(\alpha )\in {\mathbb{F}} _{p^d}$ . This shows that $r=d$ , that is, $\left \lvert \{g(C_{q_0}(\alpha ))\colon g\in {\mathcal{G}} _{{\mathbb{F}} _p}\}\right \rvert =r$ . Since the following map is surjective:

$$ \begin{align*}\{g(\alpha)\colon g\in{\mathcal{G}}_{{\mathbb{F}}_p}\}\longrightarrow \{g(C_{q_0}(\alpha))\colon g\in{\mathcal{G}}_{{\mathbb{F}}_p}\},\ g(\alpha)\longmapsto C_{q_0}(g(\alpha))=g(C_{q_0}(\alpha)),\end{align*} $$

we know that $\left \lvert \{g(\alpha )\colon g\in {\mathcal{G}} _{{\mathbb{F}} _p}\}\right \rvert \geq r$ , which contradicts to (4.1).

Proof

1. (1) By restricting the morphisms in $\text {Hom}_{\mathbf {Ab}}\left (\langle A/{\mathbb{Z}} \rangle ,\overline {{\mathbb{F}} }_p^\times \right )$ to $A/{\mathbb{Z}} $ , we obtain an injection

   $$ \begin{align*}\iota\colon\text{Hom}_{\mathbf{Ab}}\left(\langle A/{\mathbb{Z}}\rangle,\overline{{\mathbb{F}}}_p^\times\right)\longrightarrow\text{Hom}_{\mathbf{Set}}\left(A/{\mathbb{Z}},\overline{{\mathbb{F}}}_p^\times\right).\end{align*} $$

   we are left to show that the image of this map is exactly $\text {Hom}^{\text {adm}}_{\mathbf {Set}}\left (A/{\mathbb{Z}} ,\overline {{\mathbb{F}} }_p^\times \right )$ .

   For any $f\in \text {Hom}^{\text {adm}}_{\mathbf {Set}}\left (A/{\mathbb{Z}} ,\overline {{\mathbb{F}} }_p^\times \right )$ , any extension $\widetilde {f}\in \text {Hom}_{\mathbf {Ab}}\left ({\mathbb{Q}} /{\mathbb{Z}} ,\overline {{\mathbb{F}} }_p^\times \right )$ of f has image f by the injection $\iota $ . This implies that $\text {Hom}^{\text {adm}}_{\mathbf {Set}}\left (A/{\mathbb{Z}} ,\overline {{\mathbb{F}} }_p^\times \right )$ is contained in the image of $\iota $ .

   For any $h=\iota (a)\in \text {Hom}_{\mathbf {Set}}\left (A/{\mathbb{Z}} ,\overline {{\mathbb{F}} }_p^\times \right )$ with some $a\in \text {Hom}_{\mathbf {Ab}}\left (\langle A/{\mathbb{Z}} \rangle ,\overline {{\mathbb{F}} }_p^\times \right )$ , a extends uniquely to a group homomorphism $\widetilde {a}\in \text {Hom}_{\mathbf {Ab}}\left ({\mathbb{Q}} /{\mathbb{Z}} ,\overline {{\mathbb{F}} }_p^\times \right )$ . This is because $\overline {{\mathbb{F}} }_p^\times $ is divisible, and consequently injective in $\mathbf {Ab}$ . Since $\widetilde {a}\mid _{A/{\mathbb{Z}} }=a\mid _{A/{\mathbb{Z}} }=\iota (a)=h$ , we know that h is admissible.

2. (2) The following proof is given by Lahtonen (cf. [Reference LahtonenLah24]). Let $N=\vert \text {Hom}_{\mathbf {Ab}}\left (\langle A/{\mathbb{Z}} \rangle ,\overline {{\mathbb{F}} }_p^\times \right )\vert $ . Suppose there exists a rational number $q\in {\mathbb{Q}} $ that $q+{\mathbb{Z}} \in \langle A/{\mathbb{Z}} \rangle $ and $q\notin \frac {1}{N}{\mathbb{Z}} [1/p]$ . We write $q=\frac {u}{p^r\cdot v}$ , where $u,v\in {\mathbb{Z}} _{\geq 1}$ , $r\in {\mathbb{Z}} _{\geq 0}$ with $\gcd (u,v)=\gcd (p,v)=1$ . Since $q\notin \frac {1}{N}{\mathbb{Z}} [1/p]$ , one knows that v does not divide N.

   Notice that the element $z^\prime := \frac {u}{v}+{\mathbb{Z}} $ has order v in $\langle A/{\mathbb{Z}} \rangle \subseteq {\mathbb{Q}} /{\mathbb{Z}} $ . Fix a v-th primitive root $\zeta _v$ of unity in $\overline {{\mathbb{F}} }_p^\times $ , then the map $z^\prime \longmapsto \zeta _v$ induces a morphism d in $\text {Hom}_{\mathbf {Ab}}\left (\langle z^\prime \rangle ,\overline {{\mathbb{F}} }_p^\times \right )$ with order v. Since $\overline {{\mathbb{F}} }_p^\times $ is injective in $\mathbf {Ab}$ , d extends to a morphism $\widetilde {d}\in \text {Hom}_{\mathbf {Ab}}\left (\langle A/{\mathbb{Z}} \rangle ,\overline {{\mathbb{F}} }_p^\times \right )$ . The order of $\widetilde {d}$ in $\text {Hom}_{\mathbf {Ab}}\left (\langle A/{\mathbb{Z}} \rangle ,\overline {{\mathbb{F}} }_p^\times \right )$ , which divides N by Lagrange’s theorem, is a multiple of v. This contradicts the assertion that v does not divide N. Thus, $\langle A/{\mathbb{Z}} \rangle \subset \frac {1}{N}{\mathbb{Z}} [1/p]/{\mathbb{Z}} $ , which allows us to conclude the proof.

Proof we can set A in Lemma 4.3 (2) to be $\operatorname {\mathrm {supp}}(\alpha )$ , and we obtain $\operatorname {\mathrm {supp}}(\alpha )\subseteq \frac {1}{N}{\mathbb{Z}} [1/p]$ , where

$$ \begin{align*}N=\left\lvert\text{Hom}_{\mathbf{Ab}}\left(\langle \operatorname{\mathrm{supp}}(\alpha)/{\mathbb{Z}}\rangle,\overline{{\mathbb{F}}}_p^\times\right)\right\rvert.\end{align*} $$

By Lemma 4.3 (1), we have $N=\left \lvert \text {Hom}^{\text {adm}}_{\mathbf {Set}}\left (\operatorname {\mathrm {supp}}(\alpha )/{\mathbb{Z}} ,\overline {{\mathbb{F}} }_p^\times \right )\right \rvert $ . Thus, ${\mathfrak{T}} _\alpha \leq N\leq [{\mathbb{Q}} _p(\alpha )\colon {\mathbb{Q}} _p]$ , as promised.

Proof we apply the transfinite Newton algorithm on the equation $T^p-\alpha =0$ to get a root $\beta $ . Set $\beta =\sum _{\omega }[c_\omega ]\cdot p^{k_\omega }$ , where the ordinal $\omega $ runs through the well-ordered set $\operatorname {\mathrm {supp}}(\beta )$ . Recall that for any ordinal $\omega $ , let $\beta _\omega =\sum _{\rho <\omega }[c_\rho ]\cdot p^{k_\rho }$ and

$$ \begin{align*}\Phi_\omega(T)=(T+\beta_\omega)^p-\alpha=T^p+\sum_{k=1}^{p-1}\binom{p}{k}\beta_\omega^k\cdot T^{p-k}+\beta_\omega^p-\alpha.\end{align*} $$

The first step is easy: since $\beta _0=0$ and $\Phi _0(T)=T^p-\alpha $ , the Newton polygon $\text {Newt}\left (\Phi _0\right )$ consists of a single horizontal segment with residue polynomial given by

$$ \begin{align*}\text{Res}_{\Phi_0}(T)=T^p-C_0(\alpha)\in{\mathbb{F}}_{p^{{\mathfrak{F}}_\alpha}}[T],\end{align*} $$

which splits in ${\mathbb{F}} _{p^{{\mathfrak{F}} _\alpha }}$ . This shows that $\beta _1\in {\mathbb{L}} _p^{\text {ha}}({\mathfrak{T}} _\alpha ,{\mathfrak{F}} _\alpha )$ and $v_p(\beta _1)=0$ .

For any $\omega \geq 1$ , since $v_p(\beta _\omega )=v_p(\beta _1)=0$ , we know that $v_p(\binom {p}{k}\beta _\omega ^k)=1$ for all $k=1,2,\ldots ,p-1$ . This implies that $\text {Newt}\left (\Phi _\omega \right )$ is determined by the point $(p,v_p(\beta _\omega ^p-\alpha ))$ for every $\omega \geq 1$ .

Since $k_\omega \in {\mathbb{Q}} $ increases monotonically with respect to the ordinal $\omega $ , we set $\omega _0$ to be the minimal ordinal $\rho $ that satisfies $k_\rho \geq \frac {1}{p-1}$ .

1. (1) Suppose $\omega <\omega _0$ and $\beta _\rho \in {\mathbb{L}} _p^{\text {ha}}({\mathfrak{T}} _\alpha ,{\mathfrak{F}} _\alpha )$ for every $\rho \leq \omega $ . Then $\text {Newt}\left (\Phi _\omega \right )$ consists of a single segment with slope $k_\omega =s_{\max }^{\Phi _\omega }=\frac {1}{p}v_p(\beta _\omega ^p-\alpha )<\frac {1}{p-1}$ .

   Figure 1

   $\text {Newt}\left (\Phi _{\omega }\right )$ , $1\leq \omega <\omega _0$ .

   

   Since $\beta _\omega ^p-\alpha \in {\mathbb{L}} _p^{\text {ha}}({\mathfrak{T}} _\alpha ,{\mathfrak{F}} _\alpha )$ by Corollary 3.7, we know that

   $$\begin{align*}v_p(\beta_\omega^p-\alpha)\in \operatorname{\mathrm{supp}}(\beta_\omega^p-\alpha)\subseteq \frac{1}{{\mathfrak{T}}_\alpha}{\mathbb{Z}}[1/p].\end{align*}$$

   This implies that $k_\omega =\frac {1}{p}v_p(\beta _\omega ^p-\alpha )$ also belongs to $\frac {1}{{\mathfrak{T}} _\alpha }{\mathbb{Z}} [1/p]$ . The residue polynomial of $\Phi _\omega (T)$ is given by

   $$ \begin{align*}\text{Res}_{\Phi_\omega}(T)=T^p+C_{v_p(\beta_\omega^p-\alpha)}(\beta_\omega^p-\alpha),\end{align*} $$

   where $C_{v_p(\beta _\omega ^p-\alpha )}(\beta _\omega ^p-\alpha )\in {\mathbb{F}} _{p^{{\mathfrak{F}} _\alpha }}$ . Thus, any root of this residue polynomial lies in ${\mathbb{F}} _{p^{{\mathfrak{F}} _\alpha }}$ . This shows that $\beta _{\omega +1}\in {\mathbb{L}} _p^{\text {ha}}({\mathfrak{T}} _\alpha ,{\mathfrak{F}} _\alpha )$ . Since the case of limit ordinals is self-indicating, we can show by transfinite induction that $\beta _{\omega }\in {\mathbb{L}} _p^{\text {ha}}({\mathfrak{T}} _\alpha ,{\mathfrak{F}} _\alpha )$ for all $\omega \leq \omega _0$ .

2. (2) Now we deal with $\omega =\omega _0+1$ .

   1. (a) If $k_{\omega _0}=s_{\max }^{\Phi _{\omega _0}}=\frac {1}{p-1}$ , then $\text {Newt}\left (\Phi _{\omega _0}\right )$ consists of a single segment with slope equals to

      $$ \begin{align*}k_{\omega_0}=\frac{1}{p-1}=\frac{1}{p}v_p(\beta_{\omega_0}^p-\alpha)\in \frac{1}{{\mathfrak{T}}_\alpha}{\mathbb{Z}}[1/p].\end{align*} $$

      Since this segment contains the point (Figure 2) $(p-1,1)$ , one knows that

      $$ \begin{align*}\text{Res}_{\Phi_{\omega_0}}(T)=T^p+C_0(\beta_{\omega_0})^{p-1}T+C_{v_p(\beta_{\omega_0}^p-\alpha)}(\beta_{\omega_0}^p-\alpha)\in{\mathbb{F}}_{p^{{\mathfrak{T}}_\alpha}}[T],\end{align*} $$

      whose root lies in ${\mathbb{F}} _{p^{p\cdot {\mathfrak{F}} _\alpha }}$ . In this case, one has $\beta _{\omega _0+1}\in {\mathbb{L}} _p^{\text {ha}}({\mathfrak{T}} _{\alpha },p\cdot {\mathfrak{F}} _\alpha )$ .

   2. (b) If $k_{\omega _0}=s_{\max }^{\Phi _{\omega _0}}>\frac {1}{p-1}$ , then $\text {Newt}\left (\Phi _{\omega _0}\right )$ consists of two segments, where the vertexes of the segment with maximal slope is given by $(p-1,1)$ and $(p,v_p(\beta _{\omega _0}^p-\alpha ))$ . Thus,

      $$ \begin{align*}k_{\omega_0}=\frac{v_p(\beta_{\omega_0}^p-\alpha)-1}{p-(p-1)}\in \frac{1}{{\mathfrak{T}}_\alpha}{\mathbb{Z}}[1/p]\end{align*} $$

      and one has

      $$ \begin{align*}\text{Res}_{\Phi_{\omega_0}}(T)=C_0(\beta_{\omega_0})^{p-1}T+C_{v_p(\beta_{\omega_0}^p-\alpha)}(\beta_{\omega_0}^p-\alpha),\end{align*} $$

      whose root lies in ${\mathbb{F}} _{p^{{\mathfrak{F}} _\alpha }}$ . In this case, one has $\beta _{\omega _0+1}\in {\mathbb{L}} _p^{\text {ha}}({\mathfrak{T}} _{\alpha },{\mathfrak{F}} _\alpha )$ .

   Figure 2

   $\text {Newt}\left (\Phi _{\omega _0}\right )$ , if $k_{\omega _0}=\frac {1}{p-1}$ .

   
   Figure 3

   $\text {Newt}\left (\Phi _{\omega _0}\right )$ , if $k_{\omega _0}>\frac {1}{p-1}$ .

   

3. (3) For the case of $\omega>\omega _0$ , we have (Figure 3) $k_\omega>\frac {1}{p-1}$ . With the same calculation as above, one can prove by transfinite induction that for any ordinal $\omega \geq \omega _0+1$ , $\beta _\omega \in {\mathbb{L}} _p^{\text {ha}}({\mathfrak{T}} _\alpha ,{\mathfrak{F}} _{\beta _{\omega _0+1}})$ .

The result follows.

Proof Fix a $2(p-1)$ -th primitive root of unity $\tilde {\zeta }_{2(p-1)}$ . Let

$$ \begin{align*}{\mathcal{W}}:=\left\{\tilde{\zeta}_{2(p-1)}^{2k+1}\colon k\in{\mathbb{N}}_{<p-1}\right\}.\end{align*} $$

By choosing $\zeta _{2(p-1)}$ in the expansion of the $p^2$ -th primitive root of unity given by Example 2.5 (see also [Reference Wang and YuanWY21, Theorem 3.3]) in ${\mathcal{W}} $ , we get $p-1$ different $p^2$ -th primitive roots of unity $r_0,r_1,\ldots ,r_{p-2}$ , satisfying $[C_{\frac {1}{p(p-1)}}(r_k)]=\tilde {\zeta }_{2(p-1)}^{2k+1}$ and $[C_{\frac {1}{p-1}}(r_k)]=0$ for every $k\in {\mathbb{N}} _{<p-1}$ .

Similarly, for every $c\in \{0\}\cup {\mathcal{W}} $ , there exists a p-th root of unity (not necessarily primitive) $\xi _c$ that $v_p\left (\xi _c-1-c\cdot p^{\frac {1}{p-1}}\right )> \frac {1}{p-1}$ . Thus, for any $k\in {\mathbb{N}} _{<p-1}$ and $c\in \{0\}\cup {\mathcal{W}} $ , $r_k\cdot \xi _c$ is a $p^2$ -th primitive root of unity, satisfying $[C_{\frac {1}{p(p-1)}}(r_k\cdot \xi _c)]=\tilde {\zeta }_{2(p-1)}^{2k+1}$ and $[C_{\frac {1}{p-1}}(r_k\cdot \xi _c)]=c$ . This enumerates all $p(p-1)\, p^2$ -th primitive roots of unity. The result follows.

Proof of Proposition 4.6

The case of $n=1$ follows immediately from [Reference Wang and YuanWY21, Proposition 3.4].

Let $\zeta _{p^2}$ be any $p^2$ -th primitive root of unity. By Lemma 4.8, there exists another $p^2$ -th primitive root of unity $\zeta _{p^2}^\prime $ and a p-th root of unity $\xi _c$ (not necessarily primitive) that $\zeta _{p^2}^p=\zeta _{p^2}^\prime \cdot \xi _c$ and $C_{\frac {1}{p-1}}(\zeta _{p^2}^\prime )=0$ . By applying Lemma 4.7, we have

$$ \begin{align*}\zeta_{p^2}^\prime\in{\mathbb{L}}_p^{\text{ha}}({\mathfrak{T}}_{(\zeta_{p^2}^\prime)^p},{\mathfrak{F}}_{(\zeta_{p^2}^\prime)^p})={\mathbb{L}}_p^{\text{ha}}(p-1,2).\end{align*} $$

Since $\xi _c\in {\mathbb{L}} _p^{\text {ha}}(p-1,2)$ , we know that $\zeta _{p^2}\in {\mathbb{L}} _p^{\text {ha}}(p-1,2)$ . On the other hand, by [Reference Wang and YuanWY21, Theorem 3.3], one has ${\mathfrak{T}} _{\zeta _{p^2}}\geq p-1$ and ${\mathfrak{F}} _{\zeta _{p^2}}\geq 2$ . This implies that ${\mathfrak{T}} _{\zeta _{p^2}}=p-1$ and ${\mathfrak{F}} _{\zeta _{p^2}}=2$ .

when $n\geq 3$ , we can set $\alpha =\left (\zeta _{p^n}\right )^p$ in Lemma 4.7 inductively to get the result. One should notice that when $n\geq 3$ , we no longer know if the analog of Lemma 4.8 holds for $\zeta _{p^n}$ . Thus, the hyper-inertia index is multiplied by p when n increases by $1$ .

Proof It suffices to note that any r-th root of unity lies in $W({\mathbb{F}} _{p^{\text {ord}_rp}})$ .

Proof By [Reference GuillotGui18, Lemma 4.11] and its proof, there exists $s\geq 1$ that

$$\begin{align*}\langle p^s\rangle\times U_{{\mathbb{Q}}_p}^{(\mathbf{f}_K)}\subseteq {\mathcal{N}}_{K/{\mathbb{Q}}_p}K^\times.\end{align*}$$

It follows that $K\subseteq {\mathbb{Q}} _p\left (\zeta _{(p^s-1)p^{\textbf {f}_K}}\right )$ by the proof of [Reference GuillotGui18, Theorem 13.27]. On the other hand, we have $K\subseteq {\mathbb{Q}} _p\left (\zeta _{(p^n-1)p^{v_p(n)+2}}\right )$ by [Reference Koenigsmann and StockKS22, Theorem 3.1]. Since

$$ \begin{align*}{\mathbb{Q}}_p\left(\zeta_{(p^s-1)p^{\mathbf{f}_K}}\right)\cap{\mathbb{Q}}_p\left(\zeta_{(p^n-1)p^{v_p(n)+2}}\right)\subseteq {\mathbb{Q}}_p(\zeta_m),\end{align*} $$

we have $K\subseteq {\mathbb{Q}} _p(\zeta _m)$ .

Proof of Theorem 4.10

Let $m=\left (p^{n}-1\right )p^{{\textbf {f}}_{{\mathbb{Q}} _p(\alpha )}}$ . By Lemma 4.11, we know that $\alpha \in {\mathbb{Q}} _p(\zeta _m)$ .

Note $\text {ord}_{p^{n}-1}p=n$ . By Corollary 4.9, we know that

$$ \begin{align*}{\mathfrak{T}}_{\zeta_m}=\begin{cases} 1, & \text{ if }\mathbf{f}_{{\mathbb{Q}}_p(\alpha)}=0; \\ p-1, & \text{ if }\mathbf{f}_{{\mathbb{Q}}_p(\alpha)}\geq 1, \end{cases}\end{align*} $$

and

$$ \begin{align*}{\mathfrak{F}}_{\zeta_m}\begin{cases} \quad=\quad n, & \text{ if }\mathbf{f}_{{\mathbb{Q}}_p(\alpha)}=0; \\ \quad=\quad \text{lcm}(2,n), & \text{ if }\mathbf{f}_{{\mathbb{Q}}_p(\alpha)}=1,2; \\ \text{divides } \text{lcm}\left(2\cdot p^{ \mathbf{f}_{{\mathbb{Q}}_p(\alpha)}-1},n\right), & \text{ if }\mathbf{f}_{{\mathbb{Q}}_p(\alpha)}\geq 3. \end{cases}\end{align*} $$

Since $\alpha \in {\mathbb{Q}} _p(\zeta _m)\subseteq {\mathbb{L}} _p^{\text {ha}}({\mathfrak{T}} _{\zeta _m},{\mathfrak{F}} _{\zeta _m})$ , the result follows.

Proof Let ${\mathcal{O}} _K$ be the ring of integer of $K:= {\mathbb{Q}} _p(\alpha )$ with a uniformizer $\pi _K$ . Suppose $\pi _K^{{\mathfrak{e}} _\alpha }=p\cdot u$ , where u is a unit in ${\mathcal{O}} _K^\times $ .

Note that the polynomial $T^{{\mathfrak{e}} _\alpha }-\overline {u}\in {\mathbb{F}} _{p^{{\mathfrak{f}} _\alpha }}[T]$ has simple roots by the condition $\gcd ({\mathfrak{e}} _\alpha ,p)=1$ . Hensel lemma implies that there is a ${\mathfrak{e}} _\alpha $ -th root v of u in ${\mathcal{O}} _K^\times $ . If we set $\pi _K^\prime := \pi _K\cdot v^{-1}$ , then this element is also a uniformizer of K. Since $\pi _K^\prime $ is a ${\mathfrak{e}} _\alpha $ -th root of p, we have $\pi _K^\prime =p^{1/{\mathfrak{e}} _\alpha }\cdot [\zeta _e]$ for some ${\mathfrak{e}} _\alpha $ -th root $\zeta _e$ of unity in $\overline {{\mathbb{F}} }_p$ .

Proof of Theorem 4.12

If $\text {supp}(\alpha )\subseteq \frac {1}{{\mathfrak{T}} _\alpha }{\mathbb{Z}} $ , we can write $\alpha =\sum _{k>\!\!\!>-\infty }^{+\infty }[r_k]\cdot p^{\frac {k}{{\mathfrak{T}} _\alpha }}$ , where $r_k\in {\mathbb{F}} _{p^{{\mathfrak{F}} _\alpha }}$ for all k. Thus, $\alpha $ lies in ${\mathbb{Q}} _{p^{{\mathfrak{F}} _\alpha }}\left (p^{\frac {1}{{\mathfrak{T}} _\alpha }}\right )$ , where ${\mathbb{Q}} _{p^{{\mathfrak{F}} _\alpha }}:= W({\mathbb{F}} _{p^{{\mathfrak{F}} _\alpha }})\left [\frac {1}{p}\right ]$ is the unique unramified extension of ${\mathbb{Q}} _p$ with residue field ${\mathbb{F}} _{p^{{\mathfrak{F}} _\alpha }}$ . Since ${\mathfrak{T}} _\alpha $ is coprime to p (cf. Lemma 3.5), the field ${\mathbb{Q}} _{p^{{\mathfrak{F}} _\alpha }}\left (p^{\frac {1}{{\mathfrak{T}} _\alpha }}\right )$ is tamely ramified over ${\mathbb{Q}} _p$ , implying that ${\mathbb{Q}} _p(\alpha )$ is also tamely ramified over ${\mathbb{Q}} _p$ .

Conversely, if ${\mathbb{Q}} _p(\alpha )/{\mathbb{Q}} _p$ is tamely ramified, then we have

$$ \begin{align*}{\mathbb{Q}}_p(\alpha)={\mathbb{Q}}_{p^{{\mathfrak{f}}_\alpha}}\left(p^{1/{\mathfrak{e}}_\alpha}\cdot[\zeta_e]\right)\end{align*} $$

for some ${\mathfrak{e}} _\alpha $ -th root $\zeta _e\in \overline {{\mathbb{F}} }_p$ of unity by Lemma 4.13. Let

$$ \begin{align*}\alpha=\sum_{k=0}^{{\mathfrak{e}}_\alpha-1}c_k\cdot\left(p^{1/{\mathfrak{e}}_\alpha}\cdot[\zeta_e]\right)^k\end{align*} $$

with $c_k\in {\mathbb{Q}} _{p^{{\mathfrak{f}} _\alpha }}$ for $k=0,\ldots ,{\mathfrak{e}} _\alpha -1$ . If we set $c_k=\sum _{i>-\infty }\left [c_i^{(k)}\right ]p^i\in {\mathbb{Q}} _{p^{{\mathfrak{f}} _\alpha }}$ with $c_i^{(k)}\in {\mathbb{F}} _{p^{{\mathfrak{f}} _\alpha }}$ , then

(4.2) $$ \begin{align} \alpha=\sum_{k=0}^{{\mathfrak{e}}_\alpha-1}\sum_{i>-\infty}\left[c_i^{(k)}\cdot\zeta_e^k\right] p^{i+k/{\mathfrak{e}}_\alpha}. \end{align} $$

This shows that $\operatorname {\mathrm {supp}}(\alpha )\subseteq \frac {1}{{\mathfrak{e}} _\alpha }{\mathbb{Z}} $ . Thus,

$$ \begin{align*}\operatorname{\mathrm{supp}}(\alpha)\subseteq \frac{1}{{\mathfrak{e}}_\alpha}{\mathbb{Z}}\cap \frac{1}{{\mathfrak{T}}_\alpha}{\mathbb{Z}}[1/p]\subseteq {\mathbb{Z}}_{(p)}\cap\frac{1}{{\mathfrak{T}}_\alpha}{\mathbb{Z}}[1/p]=\frac{1}{{\mathfrak{T}}_\alpha}{\mathbb{Z}}.\end{align*} $$

To prove the second assertion, notice that the inclusion $\alpha \in {\mathbb{Q}} _{p^{{\mathfrak{F}} _\alpha }}\left (p^{\frac {1}{{\mathfrak{T}} _\alpha }}\right )$ implies ${\mathfrak{e}} _\alpha \mid {\mathfrak{T}} _\alpha $ and ${\mathfrak{f}} _\alpha \mid {\mathfrak{F}} _\alpha $ . On the other hand, if any coefficient $c_i^{(k)}\cdot \zeta _e^k$ in (4.2) is nonzero, then it is a $\text {lcm}({\mathfrak{e}} _\alpha ,p^{{\mathfrak{f}} _\alpha }-1)$ -th root of unity, that is, $c_i^{(k)}\cdot \zeta _e^k\in {\mathbb{F}} _{p^c}$ . As a result, one conclude by Lemma 3.5 that $\alpha \in {\mathbb{L}} _p^{\text {ha}}({\mathfrak{e}} _\alpha ,c)$ .

Proof Let $e_0:= \frac {{\mathfrak{e}} _\alpha }{\gcd ({\mathfrak{e}} _\alpha ,p^{{\mathfrak{f}} _\alpha }-1)}$ , then $\text {lcm}({\mathfrak{e}} _\alpha ,p^{{\mathfrak{f}} _\alpha }-1)=e_0\cdot (p^{{\mathfrak{f}} _\alpha }-1)$ , with $e_0$ a factor of ${\mathfrak{e}} _\alpha $ that coprime to $p^{{\mathfrak{f}} _\alpha }-1$ and p. The Chinese remainder theorem implies that

$$ \begin{align*}c=\text{lcm}(\text{ord}_{e_0}p,\text{ord}_{p^{{\mathfrak{f}}_\alpha}-1}p)=\text{lcm}(\text{ord}_{e_0}p,{\mathfrak{f}}_\alpha).\end{align*} $$

Since $e_0$ is a factor of ${\mathfrak{e}} _\alpha $ , we have $\text {ord}_{e_0}p$ divides $\text {ord}_{{\mathfrak{e}} _\alpha }p$ . The result follows from Euler’s theorem.