2 Reviewing a few known facts

For the reader’s sake, in this section we review a few known facts about valuation theory, decomposition theory, and (formally) $p$ -adic fields, but do not reproduce proofs.

(A) Generalities about valuations and their Hilbert decomposition theory

For an arbitrary field $K$ and an arbitrary valuation $v$ of $K$ , we denote usually by ${\mathcal{O}}_{v},\mathfrak{m}_{v}$ the valuation ring/ideal of $v$ , by $vK=K^{\times }/{\mathcal{O}}^{\times }$ the value group of $v$ , and by $Kv=:{\mathcal{O}}_{v}/\mathfrak{m}_{v}:=\unicode[STIX]{x1D705}(v)$ the residue field of $v$ . Further, $U_{v}^{1}:=1+\mathfrak{m}_{v}\subset U_{v}$ denote the groups of principal $v$ -units and $v$ -units, respectively. One has the following canonical exact sequences:

$$\begin{eqnarray}1\rightarrow \mathfrak{m}_{v}\rightarrow {\mathcal{O}}_{v}\rightarrow Kv\rightarrow 0\quad \text{and}\quad 1\rightarrow U_{v}^{1}\rightarrow {\mathcal{O}}_{v}^{\times }\rightarrow (Kv)^{\times }\rightarrow 1.\end{eqnarray}$$

The set of ideals of ${\mathcal{O}}_{v}$ is totally ordered with respect to inclusion. The subrings ${\mathcal{O}}_{1}\subseteq K$ with ${\mathcal{O}}_{v}\subseteq {\mathcal{O}}_{1}$ are precisely the localizations ${\mathcal{O}}_{1}:=({\mathcal{O}}_{v})_{\mathfrak{m}_{1}}$ with $\mathfrak{m}_{1}\in \operatorname{Spec}({\mathcal{O}}_{v})$ and, moreover, $\mathfrak{m}_{1}\subset {\mathcal{O}}_{v}$ , and $({\mathcal{O}}_{v})_{\mathfrak{m}_{1}}$ is a valuation ring with valuation ideal $\mathfrak{m}_{1}$ . Further, if $v_{1}$ is the corresponding valuation of $K$ , then ${\mathcal{O}}_{0}:={\mathcal{O}}_{v}/\mathfrak{m}_{1}$ is a valuation ring of $Kv_{1}$ with valuation ideal $\mathfrak{m}_{0}:=\mathfrak{m}_{v}/\mathfrak{m}_{1}$ , say of a valuation $v_{0}$ of $Kv_{1}$ . We say that $v_{1}$ is a coarsening of $v$ , and denote $v_{1}\leqslant v$ and $v_{0}:=v/v_{1}$ .

Conversely, if $v_{1}$ is a valuation of $K$ and $v_{0}$ is a valuation on the residue field $Kv_{1}$ , then the preimage of the valuation ring ${\mathcal{O}}_{v_{0}}\subseteq Kv_{1}$ under ${\mathcal{O}}_{v_{1}}\rightarrow Kv_{1}$ is a valuation ring ${\mathcal{O}}\subseteq {\mathcal{O}}_{v_{1}}$ having as valuation ideal the preimage $\mathfrak{m}\subset {\mathcal{O}}$ of $\mathfrak{m}_{v_{0}}$ . Hence, if $v$ is the valuation defined by ${\mathcal{O}}$ on $K$ , then $Kv=(Kv_{1})v_{0}$ and one has a canonical exact sequence of totally ordered groups:

$$\begin{eqnarray}0\rightarrow v_{0}(Kv_{1})\rightarrow vK\rightarrow v_{1}K\rightarrow 0.\end{eqnarray}$$

The relation between coarsening and decomposition theory is as follows. Let $\tilde{K}|K$ be a Galois extension and $\tilde{v}|v$ be a prolongation of $v$ to $\tilde{K}$ . Then the coarsenings $\tilde{v}_{1}$ of $\tilde{v}$ are in a canonical bijection with the coarsenings $v_{1}$ of $v$ via ${\mathcal{O}}_{\tilde{v}_{1}}\mapsto {\mathcal{O}}_{v_{1}}:={\mathcal{O}}_{\tilde{v}_{1}}\cap K$ ; thus, ${\mathcal{O}}_{\tilde{v}_{1}}={\mathcal{O}}_{\tilde{v}}\cdot {\mathcal{O}}_{v_{1}}$ . Let $\tilde{v}_{1}|v_{1}$ be given coarsenings of $\tilde{v}|v$ and $\tilde{K}\tilde{v}_{1}|Kv_{1}$ be the corresponding residue field extension. Then $\tilde{v}_{0}:=\tilde{v}/\tilde{v}_{1}$ is canonically a prolongation of $v_{0}:=v/v_{1}$ .

(B) Hilbert decomposition in elementary abelian extensions

Let $K$ be a field of characteristic prime to $p$ containing $\unicode[STIX]{x1D707}_{n}$ , where $n=p^{e}$ is a power of the prime number $p$ , and let $\tilde{K}=K[\sqrt[n]{K}]$ be the maximal $\mathbb{Z}/n$ elementary abelian extension of $K$ . Let $v$ be a valuation of $K$ , $\tilde{v}$ be some prolongation of $v$ to $\tilde{K}$ , and $V_{\tilde{v}}\subseteq T_{\tilde{v}}\subseteq Z_{\tilde{v}}$ be the ramification, the inertia, and the decomposition, groups of $\tilde{v}|v$ , respectively. We remark that because $\operatorname{Gal}(\tilde{K}|K)$ is commutative, the groups $V_{\tilde{v}}$ , $T_{\tilde{v}}$ , and $Z_{\tilde{v}}$ depend on $v$ only. Therefore, we will simply denote them by $V_{v}$ , $T_{v}$ , and $Z_{v}$ . Finally, we denote by $K^{Z}\subseteq K^{T}\subseteq K^{V}$ the corresponding fixed fields in $\tilde{K}$ . One has the following, see e.g. Pop [Reference PopPop10, §2] (where the case $n=p$ is dealt with; but the proof is similar for general $n=p^{e}$ and we will not reproduce the details here).

(C) Formally $p$ -adic fields and $p$ -adic valuations

We recall a few basic facts about $p$ -adic valuations and (formally) $p$ -adically closed fields; see Ax and Kochen [Reference Ax and KochenAK66] and Prestel and Roquette [Reference Prestel and RoquettePR85] for more details.

1. (1) A valuation $v$ of a field $k$ is called (formally) $p$ -adic if its residue field $kv$ is a finite field, say $\mathbb{F}_{q}$ with $q=p^{f_{v}}$ elements, and the value group $vk$ has a minimal positive element $1_{v}$ such that $v(p)=e_{v}\cdot 1_{v}$ for some natural number $e_{v}>0$ . The number $d_{v}:=e_{v}f_{v}$ is called the $p$ -adic rank (or degree) of the $p$ -adic valuation $v$ . Note that a field $k$ carrying a $p$ -adic valuation $v$ must necessarily have $\operatorname{char}(k)=0$ , as $v(p)\neq \infty$ , and $\operatorname{char}(kv)=p$ .

2. (2) Let $v$ be a $p$ -adic valuation of $k$ with valuation ring ${\mathcal{O}}_{v}$ . Then ${\mathcal{O}}_{1}:={\mathcal{O}}_{v}[1/p]$ is the valuation ring of the unique maximal proper coarsening $v_{1}$ of $v$ , which is called the canonical coarsening of  $v$ . Note that setting $k_{0}:=kv_{1}$ , and $v_{0}:=v/v_{1}$ the corresponding valuation on $k_{0}$ , we have: $v_{0}$ is a $p$ -adic valuation of $k_{0}$ with $e_{v_{0}}=e_{v}$ and $f_{v_{0}}=f_{v}$ ; hence, $d_{v_{0}}=d_{v}$ and, moreover, $v_{0}$ is a discrete valuation of $k_{0}$ . In particular, the following hold.

   1. (a) $v$ has rank one if and only if $v_{1}$ is the trivial valuation if and only if $v=v_{0}$ .

   2. (b) Giving a $p$ -adic valuation $v$ of a field $k$ of $p$ -adic rank $d_{v}=e_{v}f_{v}$ is equivalent to giving a place $\mathfrak{p}$ of $k$ with values in a finite extension $\boldsymbol{k}_{0}$ of $\mathbb{Q}_{p}$ such that the residue field $k_{0}:=k\mathfrak{p}$ of $\mathfrak{p}$ is dense in $\boldsymbol{k}_{0}$ , and $\boldsymbol{k}_{0}|\mathbb{Q}_{p}$ has ramification index $e_{v}$ and residual degree $f_{v}$ .

   3. (c) If $v_{i}<v$ is a strict coarsening of $v$ , then $v_{i}\leqslant v_{1}$ , and the quotient valuation $v/v_{i}$ on the residue field $kv_{i}$ is a $p$ -adic valuation with $e_{v/v_{i}}=e_{v}$ and $f_{v/v_{i}}=f_{v}$ ; thus, $d_{v/v_{i}}=d_{v}$ . (Actually, $\unicode[STIX]{x1D705}(v_{i}/v_{1})\cong kv_{1}$ and $\unicode[STIX]{x1D705}(v_{i}/v)\cong kv$ canonically.)

3. (3) Let $v$ be a $p$ -adic valuation of $k$ , $l|k$ a finite field extension, and denote by $w|v$ the prolongations of $v$ to $l$ . Then the following hold.

   1. (a) All prolongations $w|v$ are $p$ -adic valuations. Further, the fundamental equality holds for the finite extension $l|k$ , i.e., $[l:k]=\sum _{w|v}e(w|v)f(w|v)$ , where $e(w|v)$ and $f(w|v)$ are the ramification index and the residual degree, respectively, of $w|v$ .

   2. (b) For each $w|v$ , let $w_{1}$ be the canonical coarsening of $w$ , and $w_{0}=w/w_{1}$ be the canonical quotient on the residue field $lw_{1}$ . Then by general decomposition theory of valuations one has $e(w|v)=e(w_{1}|v_{1})e(w_{0}|v_{0})$ and $f(w|v)=f(w_{0}|v_{0})$ . Further, $e_{w}=e(w_{0}|v_{0})e_{v}$ and $f_{w}=f(w|v)f_{v}$ ; thus, $d_{w}=e(w_{0}|v_{0})f(w|v)d_{v}$ .

   3. (c) In particular, if $l|k$ is Galois, and $w^{Z}$ is the restriction of $w$ to the decomposition field $l^{Z}$ of $w$ , then $e(w|w^{Z})=e(w|v)$ and $f(w|w^{Z})=f(w|v)$ ; thus, $w^{Z}$ is a $p$ -adic valuation having $p$ -adic rank equal to the one of $v$ . Further, the same is true for infinite Galois extensions  $l|k$ .

4. (4) A field $k$ is called (formally) $p$ -adically closed if $k$ carries a $p$ -adic valuation $v$ such that for every finite extension $\tilde{k}|k$ , one has: if $v$ has a prolongation $\tilde{v}$ to $\tilde{k}$ with $d_{\tilde{v}}=d_{v}$ , then $\tilde{k}=k$ . One has the following characterization of the $p$ -adically closed fields: for a field $k$ endowed with a $p$ -adic valuation $v$ , and its canonical coarsening $v_{1}$ , the following are equivalent.

   1. (i) $k$ is $p$ -adically closed with respect to $v$ .

   2. (ii) $v$ is Henselian and $v_{1}k$ is divisible (maybe trivial).

   3. (iii) $v_{1}$ is Henselian, $v_{1}k$ is divisible (maybe trivial), and the residue field $k_{0}:=kv_{1}$ is relatively algebraically closed in its $v_{0}=v/v_{1}$ completion $\boldsymbol{k}_{0}$ (itself a finite extension of  $\mathbb{Q}_{p}$ ).

   Further, the $p$ -adic valuation of a $p$ -adically closed field is definable and unique.

5. (5) Finally, for every field $k$ endowed with a $p$ -adic valuation $v$ , there exist $p$ -adic closures $\widehat{k},\widehat{v}$ such that $d_{\widehat{v}}=d_{v}$ . Moreover, the space of the $k$ -isomorphy classes of $p$ -adic closures of $k,v$ has a concrete description as follows: let $v_{1}$ be the canonical coarsening of $v$ , and $\boldsymbol{k}_{0}|\mathbb{Q}_{p}$ the completion of the residue field of $k_{0}=kv_{1}$ with respect to the discrete valuation $v_{0}=v/v_{1}$ . Recalling the canonical exact sequence $1\rightarrow I_{v_{1}}\xrightarrow[{}]{}D_{v}\xrightarrow[{}]{\text{pr}}G_{\boldsymbol{k}_{0}}\rightarrow 1$ , one has that the space of the isomorphy classes of $p$ -adic closures of $k,v$ is in bijection with the space of conjugacy classes of sections of $\text{pr}$ and thus with $\text{H}_{\text{cont}}^{1}(G_{\boldsymbol{k}_{0}},I_{v_{1}})$ .

6. (6) In the above notation, the following hold.

   1. (a) Let $k,v$ be a $p$ -adically closed field. Then $k_{0}=kv_{1}$ is $p$ -adically closed (with respect to $v_{0}$ ), and $k^{\text{abs}}$ is actually the relative algebraic closure of $\mathbb{Q}$ in $k_{0}$ . Further, $\overline{k}=k\overline{\mathbb{Q}}$ .

   2. (b) The elementary equivalence class of a $p$ -adically closed field $k$ is determined by both the absolute subfield $k^{\text{abs}}:=k\cap \overline{\mathbb{Q}}=k_{0}\cap \overline{\mathbb{Q}}$ of $k$ and the completion $\boldsymbol{k}_{0}$ of $k_{0}=kv_{1}$ with respect to $v_{0}$ (which equals the completion of $k^{\text{abs}}$ with respect to $v_{0}$ as well).

   3. (c) If $N$ is $p$ -adically closed with respect to the $p$ -adic valuation $w$ , and $k\subseteq N$ is a subfield which is relatively closed in $N$ , then $k$ is $p$ -adically closed with respect to $v:=w|_{k}$ , $v$ and $w$ have equal $p$ -adic ranks, and $N$ and $k$ are elementary equivalent.

   4. (d) If $N|k$ is an extension of $p$ -adically closed fields of the same rank, the following hold.

      • ∙ $\tilde{k}|k\mapsto N\tilde{k}$ defines a bijection from the set of algebraic extensions $\tilde{k}|k$ of $k$ onto the set of algebraic extensions of $N$ .

      • ∙ The canonical projection $G_{N}\rightarrow G_{k}$ is an isomorphism.

   5. (e) In particular, if $L|l$ is an extension of $p$ -adically closed fields of the same rank, in the notation from the Introduction, the following canonical projections are isomorphisms:

      

(D) Valuations and rigid elements

We recall the result of Arason et al. [Reference Arason, Elman and JacobAEJ87, Theorem 2.16]; see also Koenigsmann [Reference KoenigsmannKoe95], Ware [Reference WareWar81], Efrat [Reference EfratEfr99], and especially Topaz [Reference TopazTop15] for much more about this. The point is that one can recover valuations of a field $K$ from particular subgroups $T\subset K^{\times }$ as follows: let $T\subset K^{\times }$ be a subgroup with $-1\in T$ . We say that $x\in K^{\times }\backslash T$ is $T$ -rigid if $1+x\in T\cup xT$ ; and, by abuse of language, we say that $K$ is $T$ -rigid if all $x\in K^{\times }\backslash T$ are $T$ -rigid.

3 Proof of Theorem B

To (1): Let $\widehat{K},\widehat{w}$ be a $p$ -adic closure of $K,w$ . Then $\widehat{w}$ prolongs $w$ and has $p$ -adic rank $d_{\widehat{w}}=d_{w}$ and thus equal to $d_{v}$ by the fact that $d_{w}=d_{v}$ . Therefore, since $k$ is $p$ -adically closed, $k$ must be relatively algebraically closed in $\widehat{K}$ . We conclude by using (†) from § 2(C)(6)(e), with $l:=k$ and $L:=\widehat{K}$ , and taking into account that the isomorphism $\operatorname{Gal}(\widehat{K}^{\prime \prime }|\widehat{K})\rightarrow \operatorname{Gal}(k^{\prime \prime }|k)$ factors through $\operatorname{Gal}(K^{\prime \prime }|K)\rightarrow \operatorname{Gal}(k^{\prime \prime }|k)$ and thus gives rise to a liftable section of $\operatorname{Gal}(K^{\prime }|K)\rightarrow \operatorname{Gal}(k^{\prime }|k)$ .

To (2): The proof of assertion (2) is divided into three main steps, whereas the hypothesis $p>2$ is used only in Step 2. This might be relevant when trying to address the case  $p=2$ .

Step 1. By Kummer theory, $\operatorname{pr}_{K}^{\prime }:\operatorname{Gal}(K^{\prime }|K)\rightarrow \operatorname{Gal}(k^{\prime }|k)$ is Pontrjagin dual to the canonical embedding $k^{\times }/p\rightarrow K^{\times }/p$ . Second, given a liftable section $s^{\prime }:\operatorname{Gal}(k^{\prime }|k)\rightarrow \operatorname{Gal}(K^{\prime }|K)$ of $\operatorname{pr}_{K}^{\prime }$ , it follows by Kummer theory that the Pontrjagin dual of $s^{\prime }:\operatorname{Gal}(k^{\prime }|k)\rightarrow \operatorname{Gal}(K^{\prime }|K)$ is a surjective projection $K^{\times }/p\rightarrow k^{\times }/p$ , whose kernel $\unicode[STIX]{x1D6F4}/p\subset K^{\times }/p$ is a complement of $k^{\times }/p\subset K^{\times }/p$ . That means that $s^{\prime }$ gives rise canonically to a presentation of $K^{\times }/p$ as a direct sum

For every $k$ -subfield $K_{\unicode[STIX]{x1D6FC}}\subset K$ which is relatively algebraically closed in  $K$ , one has a commutative diagram of surjective projections

and $s^{\prime }$ gives rise canonically to a liftable section $s_{\unicode[STIX]{x1D6FC}}^{\prime }$ of $\operatorname{pr}_{\unicode[STIX]{x1D6FC}}^{\prime }:\operatorname{Gal}(K_{\unicode[STIX]{x1D6FC}}^{\prime }|K_{\unicode[STIX]{x1D6FC}})\rightarrow \operatorname{Gal}(k^{\prime }|k)$ , etc. In particular, one has corresponding canonical presentations as direct sums

defined by the sections $s_{\unicode[STIX]{x1D6FC}}^{\prime }:\operatorname{Gal}(k^{\prime }|k)\rightarrow \operatorname{Gal}(K_{\unicode[STIX]{x1D6FC}}^{\prime }|K_{\unicode[STIX]{x1D6FC}})$ induced by $s^{\prime }:\operatorname{Gal}(k^{\prime }|k)\rightarrow \operatorname{Gal}(K^{\prime }|K)$ .

Indeed, let $D_{\unicode[STIX]{x1D6FC}}:=\operatorname{im}(s_{\unicode[STIX]{x1D6FC}}^{\prime })\subset \operatorname{Gal}(K_{\unicode[STIX]{x1D6FC}}^{\prime }|K_{\unicode[STIX]{x1D6FC}})$ . Then, by the definition of $s_{\unicode[STIX]{x1D6FC}}^{\prime }$ , it follows that $D_{\unicode[STIX]{x1D6FC}}$ is the image of $D=\operatorname{im}(s^{\prime })$ under the canonical projection $\operatorname{Gal}(K^{\prime }|K)\rightarrow \operatorname{Gal}(K_{\unicode[STIX]{x1D6FC}}^{\prime }|K_{\unicode[STIX]{x1D6FC}})$ . In other words, by Pontrjagin duality, the projection $K_{\unicode[STIX]{x1D6FC}}^{\times }/p\rightarrow k^{\times }/p$ factors through the inclusion $K_{\unicode[STIX]{x1D6FC}}^{\times }/p{\hookrightarrow}K^{\times }/p$ . Hence, $\unicode[STIX]{x1D6F4}_{\unicode[STIX]{x1D6FC}}/p$ is mapped into $\unicode[STIX]{x1D6F4}/p$ under $K_{\unicode[STIX]{x1D6FC}}^{\times }/p{\hookrightarrow}K^{\times }/p$ , which proves the claim.

Now let $T/p:=\unicode[STIX]{x1D6F4}/p\cdot {\mathcal{O}}_{v}^{\times }/p$ and $T\subset K^{\times }$ be the corresponding subgroup (thus containing the $p$ th powers in $K^{\times }$ ). Then, for every $k$ -subfield $K_{\unicode[STIX]{x1D6FC}}\subset K$ which is relatively algebraically closed in $K$ , by the remarks above one has that $T_{\unicode[STIX]{x1D6FC}}:=T\cap K_{\unicode[STIX]{x1D6FC}}^{\times }\subset K_{\unicode[STIX]{x1D6FC}}^{\times }$ satisfies $T_{\unicode[STIX]{x1D6FC}}/p=\unicode[STIX]{x1D6F4}_{\unicode[STIX]{x1D6FC}}/p\cdot {\mathcal{O}}_{v}^{\times }/p$ .

Finally, let $(K_{\unicode[STIX]{x1D6FC}})_{\unicode[STIX]{x1D6FC}}$ be the family of all the $k$ -subfields $K_{\unicode[STIX]{x1D6FC}}\subset K$ which are relatively algebraically closed in $K$ and satisfy $\text{tr}.\text{deg}(K_{\unicode[STIX]{x1D6FC}}|k)=1$ . Then, by Pop [Reference PopPop10, Theorem B], for every subfield $K_{\unicode[STIX]{x1D6FC}}$ , there exists a unique $p$ -adic valuation $w_{\unicode[STIX]{x1D6FC}}$ of $K_{\unicode[STIX]{x1D6FC}}$ prolonging the $p$ -adic valuation $v$ of $k$ to $K_{\unicode[STIX]{x1D6FC}}$ and having the same $p$ -adic rank as  $v$ . Our final aim is to show that there exists a (unique) $p$ -adic valuation $w$ of $K$ such that $w_{\unicode[STIX]{x1D6FC}}$ is the restriction of $w$ to $K_{\unicode[STIX]{x1D6FC}}$ for each $K_{\unicode[STIX]{x1D6FC}}$ .

Proof. We first show that ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}\subset T_{\unicode[STIX]{x1D6FC}}$ . Indeed, let $v$ be the $p$ -adic valuation of $k$ , and further consider: first, the canonical coarsening $v_{1}$ of $v$ and the canonical $p$ -adic valuation $v_{0}:=v/v_{1}$ on the residue field $k_{0}:=kv_{1}$ of $v_{1}$ . Second, consider the $p$ -adic valuation $w_{\unicode[STIX]{x1D6FC}}$ of $K_{\unicode[STIX]{x1D6FC}}$ , and let $w_{\unicode[STIX]{x1D6FC}1}$ and $w_{\unicode[STIX]{x1D6FC}0}:=w_{\unicode[STIX]{x1D6FC}}/w_{\unicode[STIX]{x1D6FC}1}$ and $K_{\unicode[STIX]{x1D6FC}0}$ be correspondingly defined. Notice that $w_{\unicode[STIX]{x1D6FC}}|_{k}=v$ implies that $w_{\unicode[STIX]{x1D6FC}1}|_{k}=v_{1}$ and $w_{\unicode[STIX]{x1D6FC}0}|_{k_{0}}=v_{0}$ . The following hold.

1. (a) First, by Fact 2, it follows that $\sqrt[p]{1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}}}$ is contained in the decomposition field of $w_{\unicode[STIX]{x1D6FC}}$ over $K$ , which is actually the fixed field of $Z_{w_{\unicode[STIX]{x1D6FC}}}$ in $K_{\unicode[STIX]{x1D6FC}}^{\prime }$ . Second, the fixed field of $\operatorname{im}(s^{\prime })$ in $K_{\unicode[STIX]{x1D6FC}}^{\prime }$ is, by the mere definitions, generated as a field extension of $K$ by $\sqrt[p]{\unicode[STIX]{x1D6F4}_{\unicode[STIX]{x1D6FC}}^{^{\prime }}}$ . Thus, since $\operatorname{im}(s^{\prime })\subset Z_{w_{\unicode[STIX]{x1D6FC}}}$ , it follows by Kummer theory that $1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}}\subset \unicode[STIX]{x1D6F4}_{\unicode[STIX]{x1D6FC}}$ .

2. (b) Since, by the mere definition, one has $\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}1}}\subset \mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}}$ and $p$ is invertible in ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}1}}$ , it follows that $1+\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}1}}\subset 1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}}$ . Thus, one has finally $1+\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}1}}\subset \unicode[STIX]{x1D6F4}_{\unicode[STIX]{x1D6FC}}$ as well.

3. (c) Since $w_{\unicode[STIX]{x1D6FC}}$ and $v$ have the same $p$ -adic rank, it follows by the discussion in § 2(C)(5) that $w_{\unicode[STIX]{x1D6FC}0}$ and $v_{0}$ are discrete $p$ -adic valuations of the same $p$ -adic rank and hence $k_{0}$ is dense in $K_{\unicode[STIX]{x1D6FC}0}$ . Therefore, since $w_{\unicode[STIX]{x1D6FC}0}|v_{0}$ are discrete valuations, and $k_{0}$ is dense in $K_{\unicode[STIX]{x1D6FC}0}$ under $k_{0}{\hookrightarrow}K_{\unicode[STIX]{x1D6FC}0}$ , one has that ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}0}}^{\times }={\mathcal{O}}_{v_{0}}^{\times }\cdot (1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}0}})$ and $K_{\unicode[STIX]{x1D6FC}0}^{\times }=k_{0}^{\times }\cdot (1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}0}})$ as well.

4. (d) Since $K_{\unicode[STIX]{x1D6FC}0}^{\times }={\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}1}}^{\times }/(1+\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}1}})$ , $k_{0}^{\times }={\mathcal{O}}_{v_{1}}^{\times }/(1+\mathfrak{m}_{v_{1}})$ , and $1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}0}}=(1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}})/(1+\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}1}})$ , from the equality $K_{\unicode[STIX]{x1D6FC}0}^{\times }=k_{0}^{\times }\cdot (1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}0}})$ above, it follows that ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}1}}^{\times }={\mathcal{O}}_{v_{1}}^{\times }\cdot (1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}})$ .

5. (e) Similarly, the equalities ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}0}}^{\times }={\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }/(1+\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}1}})$ and ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}0}}^{\times }={\mathcal{O}}_{v_{0}}^{\times }\cdot (1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}0}})$ imply that ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }={\mathcal{O}}_{v}^{\times }\cdot (1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}})$ .

Hence, since ${\mathcal{O}}_{v}^{\times },1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}}\subset T_{\unicode[STIX]{x1D6FC}}$ , one finally has ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }={\mathcal{O}}_{v}^{\times }\cdot (1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}})\subset T_{\unicode[STIX]{x1D6FC}}$ , as claimed.

We next show that $K_{\unicode[STIX]{x1D6FC}}$ is $T_{\unicode[STIX]{x1D6FC}}$ -rigid. To do so, we first notice that by the discussion above, for any fixed element $\unicode[STIX]{x1D70B}\in {\mathcal{O}}_{v}$ of minimal positive value $1_{v}\in vk$ , the following holds: let $x\in {\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}1}}^{\times }$ be an arbitrary $w_{\unicode[STIX]{x1D6FC}1}$ -unit. Then there exist $m\in \mathbb{Z}$ , $\unicode[STIX]{x1D716}\in {\mathcal{O}}_{v}^{\times }$ , and $x_{1}\in 1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}}$ such that

Now let $x\in K_{\unicode[STIX]{x1D6FC}}^{\times }\backslash T_{\unicode[STIX]{x1D6FC}}$ be given. Then one has the following possibilities.

1. (1) $w_{\unicode[STIX]{x1D6FC}1}(x)>0$ . Then $1+x$ is a principal $w_{\unicode[STIX]{x1D6FC}1}$ -unit and, therefore, $1+x\in \unicode[STIX]{x1D6F4}_{\unicode[STIX]{x1D6FC}}$ by assertion (b) above. Since $\unicode[STIX]{x1D6F4}_{\unicode[STIX]{x1D6FC}}\subset T_{\unicode[STIX]{x1D6FC}}$ , we conclude that $1+x\in T_{\unicode[STIX]{x1D6FC}}$ .

2. (2) $w_{\unicode[STIX]{x1D6FC}1}(x)<0$ . Then $1+x=x(1+x^{-1})$ . Since $w_{\unicode[STIX]{x1D6FC}1}(x^{-1})>0$ , by the discussion above, it follows that $1+x^{-1}\in T_{\unicode[STIX]{x1D6FC}}$ . Therefore, one finally has that $1+x\in xT_{\unicode[STIX]{x1D6FC}}$ .

3. (3) $w_{\unicode[STIX]{x1D6FC}1}(x)=0$ or, equivalently, $x\in {\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}1}}^{\times }$ . Let $x=\unicode[STIX]{x1D70B}^{m}\unicode[STIX]{x1D716}x_{1}$ be as given in (♯) above. One has:

$(\unicode[STIX]{x1D6FC})$

if $m>0$ , then $x\in \unicode[STIX]{x1D70B}^{m}\cdot {\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }$ and thus $1+x\in {\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }$ as well. Hence, by the relation (♯) above, $1+x=\unicode[STIX]{x1D702}_{1}\cdot \unicode[STIX]{x1D702}_{0}$ for some $\unicode[STIX]{x1D702}_{1}\in 1+p^{2}\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}}\subset \unicode[STIX]{x1D6F4}_{\unicode[STIX]{x1D6FC}}$ , $\unicode[STIX]{x1D702}_{0}\in {\mathcal{O}}_{v}^{\times }$ . Thus, finally, $1+x\in T_{\unicode[STIX]{x1D6FC}}$ ;

$(\unicode[STIX]{x1D6FD})$

if $m<0$ , then $1+x=x(1+x^{-1})$ , and $x^{-1}$ has value $-m>0$ . But then, by the first case above, $1+x^{-1}\in T_{\unicode[STIX]{x1D6FC}}$ . Hence, $1+x=x(1+x^{-1})\in xT_{\unicode[STIX]{x1D6FC}}$ and thus $1+x\in xT_{\unicode[STIX]{x1D6FC}}$ ;

$(\unicode[STIX]{x1D6FE})$

if $m=0$ , then $x\in {\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }\subset T_{\unicode[STIX]{x1D6FC}}$ and thus $x\not \in K_{\unicode[STIX]{x1D6FC}}^{\times }\backslash T_{\unicode[STIX]{x1D6FC}}$ .

For the $T$ -rigidity of $K$ , let $x\in K\backslash T$ be given. If $x\in k$ , then $x\in k\backslash {\mathcal{O}}_{v}^{\times }$ (by the definition of  $T$ ). An easy case by case analysis, namely $v(x)>0$ or $v(x)<0$ , shows that $1+x\in {\mathcal{O}}_{v}^{\times }\cup x{\mathcal{O}}_{v}^{\times }$ , etc. Finally, if $x\not \in k$ , then letting $K_{\unicode[STIX]{x1D6FC}}\subset K$ be the relative algebraic closure of $k(x)$ in $K$ , one has: since $x\in K\backslash T$ , one must have $x\in K_{\unicode[STIX]{x1D6FC}}\backslash T_{\unicode[STIX]{x1D6FC}}$ . Thus, by the discussion above, it follows that $1+x\in T_{\unicode[STIX]{x1D6FC}}\cup xT_{\unicode[STIX]{x1D6FC}}$ and, therefore, $1+x\in T\cup xT$ , etc.

This concludes the proof of Lemma 4. ◻

Step 2. Using Lemma 4 above and applying the Arason–Elman–Jacob theorem 3, we get: there exists a valuation $w$ on $K$ such that $|{\mathcal{O}}_{w}^{\times }/({\mathcal{O}}_{w}\cap T)|\leqslant 2$ and $1+\mathfrak{m}_{w}\subset T$ . Hence, letting ${\mathcal{O}}_{w}^{\times }T\subset K^{\times }$ be the subgroup generated by $T$ and ${\mathcal{O}}_{w}$ , one has ${\mathcal{O}}_{w}/({\mathcal{O}}_{w}\cap T)=({\mathcal{O}}_{w}^{\times }T)/T$ and thus $|({\mathcal{O}}_{w}^{\times }T)/T|\leqslant 2$ . We claim that ${\mathcal{O}}_{w}^{\times }\subset T$ . Indeed, first, one has $k^{\times }={\mathcal{O}}_{v}^{\times }\cdot \unicode[STIX]{x1D70B}^{\mathbb{Z}}$ as direct sum and hence $(k^{\times }/p)/({\mathcal{O}}_{v}^{\times }/p)=\unicode[STIX]{x1D70B}^{\mathbb{Z}/p}$ . Second, by definitions, one has that $K^{\times }/p=\unicode[STIX]{x1D6F4}/p\cdot k^{\times }/p$ and $T/p=\unicode[STIX]{x1D6F4}/p\cdot {\mathcal{O}}_{v}^{\times }/p$ , both of which being direct sums. Thus, finally one gets that

$$\begin{eqnarray}K^{\times }/p=\unicode[STIX]{x1D6F4}/p\cdot k^{\times }/p=\unicode[STIX]{x1D6F4}/p\cdot {\mathcal{O}}_{v}^{\times }/p\cdot \unicode[STIX]{x1D70B}^{\mathbb{Z}/p}=T/p\cdot \unicode[STIX]{x1D70B}^{\mathbb{Z}/p},\end{eqnarray}$$

where the dot denotes direct sums; in particular, one has $|K^{\times }/T|=|(K^{\times }/p)/(T/p)|=p$ . Hence, considering the canonical inclusions of groups $T\subseteq {\mathcal{O}}_{w}^{\times }T\subseteq K^{\times }$ , we get

$$\begin{eqnarray}p=|K^{\times }/T|=|K^{\times }/({\mathcal{O}}_{w}^{\times }T)|\cdot |({\mathcal{O}}_{w}^{\times }T)/T|.\end{eqnarray}$$

Since $|({\mathcal{O}}_{w}^{\times }T)/T|\leqslant 2$ and $2<p$ , it follows that $|({\mathcal{O}}_{w}^{\times }T)/T|=1$ is the only possibility; hence, $T={\mathcal{O}}_{w}^{\times }T$ , and, finally, ${\mathcal{O}}_{w}^{\times }\subseteq T$ . Hence, we conclude that $|K^{\times }/{\mathcal{O}}_{w}^{\times }|\geqslant p$ and therefore we have the following result.

$\bullet$ The valuation $w$ is a non-trivial valuation of $K$ .

Step 3. Recalling that ${\mathcal{O}}_{w}^{\times }\subset T$ , one has that the canonical projection $K^{\times }/{\mathcal{O}}_{w}^{\times }\rightarrow K^{\times }/T$ is surjective. Therefore, if $b\in K$ is a generator of $K^{\times }/T$ , e.g., $b=\unicode[STIX]{x1D70B}\in k_{0}$ has $v_{0}(\unicode[STIX]{x1D70B})=1$ , then $b$ is not a $w$ -unit and $w(b)$ is not divisible by  $p$ in $wK=K^{\times }/{\mathcal{O}}_{w}^{\times }$ and hence $wK$ is not divisible by  $p$ .

For every subfield $K_{\unicode[STIX]{x1D6FC}}\subset K$ as in the proof of Lemma 4, let $v_{\unicode[STIX]{x1D6FC}}:=w|_{K_{\unicode[STIX]{x1D6FC}}}$ be the restriction of $w$ to  $K_{\unicode[STIX]{x1D6FC}}$ . Then ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}={\mathcal{O}}_{w}\cap K_{\unicode[STIX]{x1D6FC}}$ and, therefore, ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}^{\times }$ is contained in $T_{\unicode[STIX]{x1D6FC}}=T\cap K_{\unicode[STIX]{x1D6FC}}$ .

Proof. By the first part of the proof of Lemma 4, we have that ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }\subset T_{\unicode[STIX]{x1D6FC}}$ . Since ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}^{\times }\subseteq T_{\unicode[STIX]{x1D6FC}}$ as well, it follows that the element-wise product ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }{\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}^{\times }$ is contained in $T_{\unicode[STIX]{x1D6FC}}$ . Since $T_{\unicode[STIX]{x1D6FC}}$ is a proper subgroup of $K_{\unicode[STIX]{x1D6FC}}^{\times }$ , it follows that ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }{\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}^{\times }\neq K^{\times }$ as well. The following is well-known valuation theoretical nonsense: let $\mathfrak{n}$ be the largest common ideal of ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}$ and ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}$ . Then ${\mathcal{O}}:={\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}{\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}$ equals both the localization of ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}$ at $\mathfrak{n}$ and the localization of ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}$ at $\mathfrak{n}$ . Further, ${\mathcal{O}}$ is the smallest valuation ring of $K$ which contains both ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}$ and ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}$ ; or, equivalently, ${\mathcal{O}}$ is the valuation ring of the finest common coarsening of $v_{\unicode[STIX]{x1D6FC}}$ and $w_{\unicode[STIX]{x1D6FC}}$ . We now claim that one has

$$\begin{eqnarray}{\mathcal{O}}^{\times }={\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}^{\times }{\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }.\end{eqnarray}$$

Indeed, let $v_{\unicode[STIX]{x1D6FC}}^{1}$ and $w_{\unicode[STIX]{x1D6FC}}^{1}$ be the valuations of $\unicode[STIX]{x1D705}(\mathfrak{n}):={\mathcal{O}}/\mathfrak{n}$ defined by ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}/\mathfrak{n}$ and ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}/\mathfrak{n}$ , respectively. Then $v_{\unicode[STIX]{x1D6FC}}^{1}$ and $w_{\unicode[STIX]{x1D6FC}}^{1}$ are independent, and one has exact sequences

$$\begin{eqnarray}1\rightarrow (1+\mathfrak{n})\rightarrow {\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}^{\times }\rightarrow {\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}^{1}}^{\times }\rightarrow 1\quad and\quad 1\rightarrow (1+\mathfrak{n})\rightarrow {\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }\rightarrow {\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}^{1}}^{\times }\rightarrow 1.\end{eqnarray}$$

Since $v_{\unicode[STIX]{x1D6FC}}^{1}$ and $w_{\unicode[STIX]{x1D6FC}}^{1}$ are independent valuations of $\unicode[STIX]{x1D705}(\mathfrak{n})$ , one has that ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}^{1}}^{\times }{\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}^{1}}^{\times }=\unicode[STIX]{x1D705}(\mathfrak{n})^{\times }$ and therefore

$$\begin{eqnarray}({\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}^{\times }{\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times })/(1+\mathfrak{n})=\unicode[STIX]{x1D705}(\mathfrak{n})^{\times }.\end{eqnarray}$$

On the other hand, one also has ${\mathcal{O}}^{\times }/(1+\mathfrak{n})=\unicode[STIX]{x1D705}(\mathfrak{n})^{\times }$ . Further, $1+\mathfrak{n}$ is contained in both ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}^{\times }$ and ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }$ and hence we conclude that ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}^{\times }{\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }={\mathcal{O}}^{\times }$ , as claimed.

By contradiction, suppose that ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}\neq {\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}$ . Recall that the valuation ring ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}$ has finite residue field and hence ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}$ is minimal among the valuation rings of $K_{\unicode[STIX]{x1D6FC}}$ and, in particular, ${\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}$ cannot be contained in ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}$ . Therefore, in the above notation, one has that ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}\subset {\mathcal{O}}$ strictly or, equivalently, $\mathfrak{n}\subset \mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}}$ is a strict inclusion. On the other hand, if $b\in k$ is any element of minimal positive value  $1_{v}$ , then $\mathfrak{m}_{w_{\unicode[STIX]{x1D6FC}}}=b{\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}$ and, therefore, $b\not \in \mathfrak{n}$ . Thus, we have

$$\begin{eqnarray}b\in {\mathcal{O}}^{\times }={\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}^{\times }{\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}^{\times }\subseteq T_{\unicode[STIX]{x1D6FC}},\end{eqnarray}$$

contradicting the fact that $w(b)$ generates $wK/w(T)\cong \mathbb{Z}/p$ . Thus, we conclude that one must have ${\mathcal{O}}_{w_{\unicode[STIX]{x1D6FC}}}={\mathcal{O}}_{v_{\unicode[STIX]{x1D6FC}}}$ , and Lemma 5 is proved.◻

We next claim that $w$ is a $p$ -adic valuation of $K$ having $p$ -adic rank $d_{w}=d_{v}$ . Indeed, for $t\in {\mathcal{O}}_{w}$ , let $K_{\unicode[STIX]{x1D6FC}}\subset K$ be the relative algebraic closure of $k(t)$ in $K$ . Then $K_{\unicode[STIX]{x1D6FC}}|k$ has transcendence degree ${\leqslant}1$ and, therefore, $w|_{K_{\unicode[STIX]{x1D6FC}}}=w_{\unicode[STIX]{x1D6FC}}$ is the $p$ -adic valuation $w_{\unicode[STIX]{x1D6FC}}$ by Lemma 5. In particular, if $b\in k$ is such that $v(b)=1_{v}$ is the minimal positive element of $v(k^{\times })$ , it follows that $w_{\unicode[STIX]{x1D6FC}}(b)$ is the minimal positive element of $w_{\unicode[STIX]{x1D6FC}}K_{\unicode[STIX]{x1D6FC}}$ under $vk{\hookrightarrow}w_{\unicode[STIX]{x1D6FC}}K_{\unicode[STIX]{x1D6FC}}$ and, further, $kv=K_{\unicode[STIX]{x1D6FC}}w_{\unicode[STIX]{x1D6FC}}$ is the finite field of cardinality $f_{v}=f_{w_{\unicode[STIX]{x1D6FC}}}$ . One has the following.

1. (a) $w(b)$ is the minimal positive element of $w(K^{\times })$ . Indeed, for $t\in \mathfrak{m}_{w}$ , in the above notation one has $w(t)=w_{\unicode[STIX]{x1D6FC}}(t)\geqslant w_{\unicode[STIX]{x1D6FC}}(b)=w(b)$ .

2. (b) $kv=Kw$ and thus $f_{v}=f_{w_{\unicode[STIX]{x1D6FC}}}$ . Indeed, if $t\in {\mathcal{O}}_{w}$ , then in the above notation the residue $\overline{t}\in Kw$ satisfies $\overline{t}\in K_{\unicode[STIX]{x1D6FC}}w_{\unicode[STIX]{x1D6FC}}=kv$ .

Therefore, $w$ is a $p$ -adic valuation of rank $d_{w}=d_{v}$ , which is unique, by the uniqueness of $w_{\unicode[STIX]{x1D6FC}}=w|_{K_{\unicode[STIX]{x1D6FC}}}$ for every subfield $K_{\unicode[STIX]{x1D6FC}}$ . This concludes the proof of Theorem B.

4 Proof of the other announced results

(A) Proof of Theorem A

The following stronger assertion holds (from which Theorem A immediately follows).

(B) Proof of Theorem B⁰

The proof is almost identical with the one of Theorem B⁰ from Pop [Reference PopPop10]. The proof of assertion (1) is identical with the proof of assertion (1) of Theorem B; thus, we omit it. Concerning the proof of assertion (2), let $s_{L}^{\prime }:\operatorname{Gal}(k^{\prime }|l)\rightarrow \operatorname{Gal}(K^{\prime }|L)$ be a given liftable section of $\operatorname{pr}_{L}^{\prime }:\operatorname{Gal}(K^{\prime }|L)\rightarrow \operatorname{Gal}(k^{\prime }|l)$ . Then considering the restriction

$$\begin{eqnarray}s^{\prime }:=\operatorname{pr}_{L}^{\prime }|_{\operatorname{ Gal}(k^{\prime }|k)}:\operatorname{Gal}(k^{\prime }|k)\rightarrow \operatorname{Gal}(K^{\prime }|K),\end{eqnarray}$$

it follows by mere definitions that $s^{\prime }$ is a liftable section of $\operatorname{pr}_{K}^{\prime }:\operatorname{Gal}(K^{\prime }|K)\rightarrow \operatorname{Gal}(k^{\prime }|k)$ . Hence, by Theorem B, there exists a unique $p$ -adic valuation $w^{1}$ of $K$ which prolongs the $p$ -adic valuation $v_{k}$ of $k$ to $K$ and has $d_{w^{1}}=d_{v_{k}}$ , and $s^{\prime }=s_{w^{1}}$ in the usual way.

Let $w=w^{1}|_{L}$ be the restriction of $w^{1}$ to  $L$ . Then $w$ prolongs the valuation $v$ of $l$ to $L$ . We claim that $w^{1}$ is the unique prolongation of $w$ to  $K$ . Indeed, let $w^{2}:=w^{1}\circ \unicode[STIX]{x1D70E}_{0}$ , with $\unicode[STIX]{x1D70E}_{0}\in \operatorname{Gal}(k|l)$ , be a further prolongation of $w$ to $K$ . Then, if $(w^{i})^{\prime }$ is a prolongation of $w^{i}$ to $K^{\prime }$ , $i=1,2$ , and $\unicode[STIX]{x1D70E}\in \operatorname{im}(s_{L}^{\prime })$ is a preimage of $\unicode[STIX]{x1D70E}_{0}$ , then $(w^{2})^{\prime }:=(w^{1})^{\prime }\circ \unicode[STIX]{x1D70E}$ is a prolongation of $w^{2}$ to $K^{\prime }$ . Therefore, if $Z_{w^{1}}\subset \operatorname{Gal}(K^{\prime }|K)$ is the decomposition group above $w^{1}$ , then $Z_{w^{2}}:=\unicode[STIX]{x1D70E}Z_{w^{1}}\unicode[STIX]{x1D70E}^{-1}$ is the decomposition group above $w^{2}$ . On the other hand, $\operatorname{im}(s^{\prime })\subseteq Z_{w^{1}}$ by Theorem B. Since $\operatorname{Gal}(k^{\prime }|k)$ is a normal subgroup of $\operatorname{Gal}(k^{\prime }|l)$ , it follows that $\operatorname{im}(s^{\prime })$ is normal in $\operatorname{im}(s_{L}^{\prime })$ . Hence, if $\unicode[STIX]{x1D70E}\in \operatorname{im}(s_{L}^{\prime })$ , it follows that $\unicode[STIX]{x1D70E}(\operatorname{im}(s^{\prime }))\unicode[STIX]{x1D70E}^{-1}=\operatorname{im}(s^{\prime })$ and, therefore, one has

$$\begin{eqnarray}Z_{w^{1}}\supseteq \operatorname{im}(s^{\prime })=\unicode[STIX]{x1D70E}(\operatorname{im}(s^{\prime }))\unicode[STIX]{x1D70E}^{-1}\subseteq \unicode[STIX]{x1D70E}Z_{w^{1}}\unicode[STIX]{x1D70E}^{-1}=Z_{w^{2}}.\end{eqnarray}$$

Hence, $\operatorname{im}(s^{\prime })\subset Z_{w^{1}},Z_{w^{2}}$ ; thus, by the uniqueness assertion of Theorem B, we must have $w^{1}=w^{2}$ . Equivalently, if $\unicode[STIX]{x1D70E}\in \operatorname{im}(s_{L}^{\prime })$ , then $\unicode[STIX]{x1D70E}Z_{w^{1}}\unicode[STIX]{x1D70E}^{-1}=Z_{w^{1}}$ and therefore $\unicode[STIX]{x1D70E}\in Z_{w^{1}}$ . Finally, we conclude that $d_{w}=d_{v}$ , as claimed, and this concludes the proof of Theorem B⁰ .

(C) Proof of Theorem A⁰

The following stronger assertion holds (from which Theorem A⁰ follows immediately).