2. Preliminaries

One of the basic tools needed in the proof of the generalization of Geyer’s result is a special case of the renowned Pontryagin–van Kampen theorem. Here, and in the rest of this note, $l$ stands for a prime number, $\mathbb{Z}_l$ is the ring of $l$ -adic numbers, viewed as a profinite abelian group or as a principal ideal domain. We also write ${\hat{\mathbb{Z}}}\;:\!=\;\prod _l\mathbb{Z}_l$ for the Prüfer group [Reference Fried and Jarden3, p. 12]. Thus, $\mathbb{Z}_l$ is the free pro- $l$ cyclic group and $\hat{\mathbb{Z}}$ is the free procyclic group.

The proof of Proposition 2.1 uses a special case of the Pontryagin–van Kampen duality theorem saying that every locally compact abelian topological group $A$ is canonically isomorphic to its double dual group $A^{**}$ , where $A^*={\mathrm{Hom}}(A,\mathbb{R}/\mathbb{Z})$ . The proof of that special case needed in our proposition, dealing only with abelian profinite groups, appears in [Reference Ribes and Zalesskii10, Section 2.9]. It is much simpler than the proof of the general theorem [Reference Hewitt and Ross5, p. 376, Thm. 24.2].

We denote the algebraic closure of a field $K$ by $\tilde{K}$ and its separable algebraic closure by $K_{\mathrm{sep}}$ . We write $\textrm{Gal}(K)$ for the absolute Galois group $\textrm{Gal}(K_{\mathrm{sep}}/K)$ of $K$ . If $A$ is a closed subgroup of $\textrm{Gal}(K)$ , then $K_{\mathrm{sep}}(A)$ denotes the fixed field of $A$ in $K_{\mathrm{sep}}$ .

Proof. Let $R=K_{\mathrm{sep}}(A)$ . Then, a theorem of Artin says that $\mathrm{char}(K)=0$ , $K_{\mathrm{sep}}={\tilde{K}}$ , and ${\tilde{K}}=R(\sqrt{-1})$ [Reference Lang7, p. 299, Cor. 9.3]. Let $\tau$ be the unique element of order $2$ of $\textrm{Gal}(R)$ defined by $\tau (\sqrt{-1})=-\sqrt{-1}$ .

By [Reference Lang7, p. 452, Prop. 2.4], $R$ is real closed. Let $\lt$ be the ordering of $K$ induced by the unique ordering of $R$ . If $R'$ is a real closed field extension of $K$ in $\tilde{K}$ whose ordering extends $\lt$ , then by [Reference Lang7, p. 455, Thm. 2.9], there exists a unique $K$ -isomorphism $R\to R'$ .

Let $\sigma$ be an element of the centralizer $C_{\textrm{Gal}(K)}(A)$ of $A$ in $\textrm{Gal}(K)$ . Then, $\sigma R$ is a real closure of $(K,\lt )$ and $\textrm{Gal}(\sigma R)\cong \mathbb{Z}/2\mathbb{Z}$ . Also, $\tau (\sigma R)=\tau \sigma R=\sigma \tau R=\sigma R$ . By the preceding paragraph applied to $\sigma R$ rather than to $R$ , the restriction of $\tau$ to $\sigma R$ is the identity map. In other words, $\tau \in \textrm{Gal}(\sigma R)$ . Since ${\mathrm{ord}}(\tau )=2$ , the element $\tau$ generates $\textrm{Gal}(\sigma R)$ , so $R=\sigma R$ . The uniqueness of the $K$ -isomorphism of $R$ into $R$ implies that $\sigma \in \textrm{Gal}(R)=A$ , as desired.

Proof. If $A$ has a non-unit element $\alpha$ of a finite order, then by Lemma 2.2, $\langle \alpha \rangle \cong \mathbb{Z}/2\mathbb{Z}$ and $\langle \alpha \rangle$ is its own centralizer in $\textrm{Gal}(K)$ . Since $A$ is abelian, $A$ is contained in that centralizer. Therefore, $A=\langle \alpha \rangle$ .

Otherwise, $A$ is torsion-free. Hence, by Proposition 2.1, $A$ has the desired structure.

Given a profinite group $G$ and a prime number $l$ , we write $\mathrm{cd}_l(G)$ for the $\textit{l}$ th cohomology dimension of $\textbf{G}$ [Reference Ribes9, p. 196, Def. 1.1]. Also, we write $\zeta _n$ for a primitive root of unity of order $n$ .

Proof of (a). Let $O$ be the ring of integers of $E$ , $\bar{E}$ the residue field of $E$ , $\pi$ a prime element of $O$ , $U$ the group of invertible elements of $O$ , and $U^{(1)}=1+\pi O$ the subgroup of $1$ -units of $O$ . Reduction modulo $\pi O$ yields the following short exact sequence

\begin{align*} \textbf{1}\buildrel \over \longrightarrow U^{(1)}\buildrel \over \longrightarrow U\buildrel \over \longrightarrow{\bar{E}}^\times \buildrel \over \longrightarrow \textbf{1}, \end{align*}

where $\textbf{1}$ is the trivial group. By [Reference Serre11, p. 213, Chap. XIV, Prop. 10], $U^{(1)}$ is isomorphic to a direct product of a finite abelian group with a free abelian group. Since ${\bar{E}}^\times$ is also finite, the torsion group of $U$ is finite. That group is the group of roots of unity in $E$ .

Proof of (b). By (a), $E$ has only finitely many roots of unity of order $l^j$ with $j\ge 1$ . Thus, there exists a non-negative integer $j$ with $\zeta _{l^j}\in E$ and $\zeta _{l^{j+1}}\notin E$ . By [Reference Lang7, p. 297, Thm. 9.1], $[E(\zeta _{l^{j+1}})\;:\;E(\zeta _{l^j})]=l$ . Apply the same argument to the field $E_1\;:\!=\;E(\zeta _{l^{j+1}})$ to find an integer $j_2\gt j_1\;:\!=\;j$ such that $\zeta _{l^{j_2}}\in E_1$ and $\zeta _{l^{j_2+1}}\notin E_1$ , so $[E_2\;:\;E_1]=l$ with $E_2\;:\!=\;E(\zeta _{l^{j_1+1}},\zeta _{l^{j_2+1}})$ . Continue to find a sequence $j_1\lt j_2\lt j_3\lt \dots$ and fields $E\subset E_1\subset E_2\subset E_3\subset \cdots$ such that $\zeta _{l^{j_{n+1}}}\in E_n\;:\!=\;E(\zeta _{l^{j_i+1}})_{i=1}^n$ and $\zeta _{l^{j_{n+1}+1}}\notin E_n$ , so $[E_{n+1}\;:\;E_n]=l$ , for each $n\ge 1$ . Hence, $l^\infty |[E(\zeta _{l^j})_{j\ge 1}\;:\;E]$ .

Proof of (c). The claim follows from (b) and [Reference Ribes9, p. 291, Cor. 7.4(i),(ii)].

Note that the citation in the proof of (c) relies on local class field theory.

3. Geyer’s theorem

We generalize Geyer’s theorem which asserts that every closed abelian subgroup of $\textrm{Gal}(\mathbb{Q})$ is procyclic [Reference Geyer4, p. 357, Satz 2.3].

Proof. Let $G$ be a closed pro- $p$ subgroup of $\textrm{Gal}(F)$ . By [Reference Ribes9, p. 256, Thm. 3.3], $\mathrm{cd}(G)\le 1$ . On the other hand, $\mathbb{Z}_p$ is a free pro- $p$ group of rank $1$ . Hence, by [Reference Ribes9, p. 217, Cor. 3.2], $\mathrm{cd}(\mathbb{Z}_p)=1$ . It follows from [Reference Ribes9, p. 221, Prop. 4.4] that $\mathrm{cd}(\mathbb{Z}_p\times \mathbb{Z}_p)=\mathrm{cd}(\mathbb{Z}_p)+\mathrm{cd}(\mathbb{Z}_p)=2$ . Therefore, $G\not \cong \mathbb{Z}_p\times \mathbb{Z}_p$ , as claimed.

Proof. We distinguish between two cases:

Case A: $K$ is a number field. We assume without loss that $K=\mathbb{Q}$ . By assumption, $\zeta _{l^2}\in M \newcommand{\hefreshD }{\mathop{\raise 1.5pt\hbox{${\smallsetminus }$}}} \newcommand{\hefreshS }{\mathop{\raise 0.85pt\hbox{$\scriptstyle \smallsetminus $}}} \mathchoice{\hefreshD }{\hefreshD }{\hefreshS }{\hefreshS }\mathbb{R}$ . Thus, $M$ cannot be embedded into $\mathbb{R}$ , that is $M$ is totally imaginary. Hence by [Reference Ribes9, p. 302, Thm. 8.8(a)], $\mathrm{cd}_l(\textrm{Gal}(M))\ne \infty$ .

Now we consider a prime number $p$ , a valuation $v$ of $M$ lying over $p$ , and the completion ${\hat M}_v$ of $M$ at $v$ . Then, $\zeta _{l^i}\in M\subseteq{\hat M}_v$ for each $i$ . Hence, by Lemma 2.4(b), $l^\infty |[{\hat M}_v\;:\;\mathbb{Q}_p]$ . Therefore, by [Reference Ribes9, p. 302, Thm. 8.8(b)], $\mathrm{cd}_l(\textrm{Gal}(M))\le 1$ .

Finally, by [Reference Ribes9, p. 217, Cor. 3.2 and p. 221, Prop. 4.4] and [Reference Ribes9, p. 217, Cor. 3.2],

\begin{align*} \mathrm{cd}_l(\mathbb{Z}_l\times \mathbb{Z}_l) =\mathrm{cd}_l(\mathbb{Z}_l)+\mathrm{cd}_l(\mathbb{Z}_l) =1+1=2. \end{align*}

Hence, $\textrm{Gal}(M)\not \cong \mathbb{Z}_l\times \mathbb{Z}_l$ , as claimed.

Case B: $K$ is a finite separable extension of $\mathbb{F}_p(t)$ with $t$ transcendental over $\mathbb{F}_p$ . We assume without loss that $K=\mathbb{F}_p(t)$ . By assumption, $M$ contains the field $L\;:\!=\;\mathbb{F}_p(\zeta _{l^i})_{i\ge 1}$ , so $L(t)\subseteq M$ . Since there are infinitely many roots of unity $\zeta _{l^i}$ in ${\tilde{\mathbb{F}}}_p$ and only finitely many of them belong to each finite field, $L$ is an infinite field. In addition, for each $i\ge 1$ the extension $\mathbb{F}_p(\zeta _{l^{i+1}})/\mathbb{F}_p(\zeta _{l^i})$ is cyclic of degree $l$ or trivial. Hence, $\textrm{Gal}(L/\mathbb{F}_p(\zeta _l))\cong \mathbb{Z}_l$ . Therefore, $L$ is contained in the maximal extension $L'$ of $\mathbb{F}_p(\zeta _l)$ of an $l$ ’th power degree. Since $\textrm{Gal}(L'/\mathbb{F}_p(\zeta _l))\cong \mathbb{Z}_l$ , the restriction map $\textrm{Gal}(L'/\mathbb{F}_p(\zeta _l))\to \textrm{Gal}(L/\mathbb{F}_p(\zeta _l))$ is surjective, and $\mathbb{Z}_l$ is generated by one element, that map is an isomorphism [Reference Fried and Jarden3, p. 331, Cor. 16.10.8]. It follows that $L=L'$ . Therefore, $l$ does not divide the order of $\textrm{Gal}(L)$ .

By [Reference Ribes9, p. 208, Cor. 2.3], $\mathrm{cd}_l(\textrm{Gal}(L))=0$ . Hence, by [Reference Ribes9, p. 272, Prop. 5.2], $\mathrm{cd}_l(\textrm{Gal}(L(t)))=1$ . Since $\textrm{Gal}(M)\le \textrm{Gal}(L(t))$ , we have by [Reference Ribes9, p. 204, Prop. 2.1(a)], that $\mathrm{cd}_l(\textrm{Gal}(M))\le 1$ . As in Case A, this inequality implies that $\textrm{Gal}(M)\not \cong \mathbb{Z}_l\times \mathbb{Z}_l$ , as claimed.

Here is the promised result of Geyer.

Proof. We start the proof with the special case where the torsion group $A_{\mathrm{tor}}$ of $A$ is nontrivial. In this case, there exists a non-unit $\tau \in A$ of finite order. By Lemma 2.2, $\mathrm{char}(K)=0$ and $A\cong \mathbb{Z}/2\mathbb{Z}$ . In particular, $A$ is procyclic.

We may therefore assume that $A$ is a nontrivial torsion-free abelian profinite group. By Proposition 2.1, $A\cong \prod _l\mathbb{Z}_l^{r_l}$ , where $l$ ranges over all prime numbers and for each $l$ , $r_l$ is a cardinal number, so we may assume that $A\cong \mathbb{Z}_l^r$ for a prime number $l$ and a positive cardinal number $r$ and prove that $A\cong \mathbb{Z}_l$ .

Otherwise, $A$ contains a closed subgroup which is isomorphic to $\mathbb{Z}_l\times \mathbb{Z}_l$ . Thus, we may assume that $A\cong \mathbb{Z}_l\times \mathbb{Z}_l$ and prove that this assumption leads to a contradiction.

To this end, we denote the fixed field of $A$ in $K_{\mathrm{sep}}$ by $M$ and identify $\textrm{Gal}(M)$ with $A$ . By Lemma 3.1, $l\ne \mathrm{char}(K)$ .

Claim: $M$ contains a root of unity $\zeta _l$ of order $l$ . Indeed, if $l=2$ , then $\zeta _l=-1\in M$ . Otherwise, $l\gt 2$ and if $\zeta _l\notin M$ , then $[M(\zeta _l)\;:\;M]$ is a divisor of $l-1$ which is greater than $1$ . On the other hand, $[M(\zeta _l)\;:\;M]$ divides the (profinite) order of $A$ which is $l^\infty$ , a contradiction.

Since $\textrm{Gal}(M)\cong \mathbb{Z}_l\times \mathbb{Z}_l$ , Lemma 3.2 implies that not all roots of unity of order $l^i$ with $i\ge 1$ belong to $M$ . Let $n$ be the smallest positive integer such that $M$ contains a root of unity of order $l^{n-1}$ but does not contain a root of unity of order $l^n$ . Choose a root of unity $\zeta _{l^n}$ and set $M_1=M(\zeta _{l^n})$ . Then, $\zeta _{l^n}^l\in M$ but $\zeta _{l^n}\notin M$ . Hence, $[M_1\;:\;M]|l$ and $[M_1\;:\;M]\ne 1$ (by the Claim and [Reference Lang7, p. 289, Thm. 6.2(ii)]), so $[M_1\;:\;M]=l$ .

Let $U$ be the open subgroup of $\mathbb{Z}_l$ of index $l$ . Then, the index of each of the subgroups $\mathbb{Z}_l\times U$ and $U\times \mathbb{Z}_l$ of $\textrm{Gal}(M)$ is $l$ . We choose one of them which is different from $\textrm{Gal}(M_1)$ and denote its fixed field in $K_{\mathrm{sep}}$ by $M_2$ . Then, $M_2$ is a cyclic extension of $M$ of degree $l$ and $M_1\ne M_2$ .

Since $\zeta _l\in M$ , [Reference Lang7, p. 289, Thm. 6.2(i)] implies the existence of $a,x\in K_{\mathrm{sep}}$ with $M_2=M(x)$ and $a\;:\!=\;x^l\in M$ . Choose $b\in K_{\mathrm{sep}}$ with $b^{l^{n-1}}=x$ , so $b^{l^n}=a$ . In particular, $M_2=M(b^{l^{n-1}})\subseteq M(b)$ and $[M(b)\;:\;M_2]\le l^{n-1}$ . It follows from the preceding paragraph that

(3.1) \begin{equation} [M(b)\;:\;M]\le l^n. \end{equation}

Next choose $\sigma \in A$ such that $\sigma |_{M_1}={\mathrm{id}}$ and $\sigma |_{M_2}\ne{\mathrm{id}}$ . In particular, $\sigma x\ne x$ , so $\zeta \;:\!=\;(\sigma b)b^{-1}$ satisfies

\begin{align*} \zeta ^{l^n}=\sigma b^{l^n}\cdot b^{-l^n} =\sigma a\cdot a^{-1}=aa^{-1}=1 \hbox{ and } \zeta ^{l^{n-1}} =\sigma b^{l^{n-1}}\cdot b^{-l^{n-1}} =\sigma x\cdot x^{-1}\ne 1, \end{align*}

thus $\zeta$ is a primitive root of $1$ of order $l^n$ .

The definition of $M_1$ implies that $M_1=M(\zeta )$ . But $M(b)$ is a Galois extension of $M$ (because $\textrm{Gal}(M)$ is abelian). Hence, $\zeta =(\sigma b)b^{-1}\in M(b)$ , so $M_1\subseteq M(b)$ . Since $[M_1\;:\;M]=l$ , we have by (3.1) that $[M(b)\;:\;M_1]\le l^{n-1}$ . Since $\sigma$ is the identity on $M_1$ , the latter inequality implies that ${\mathrm{ord}}(\sigma |_{M(b)})\le l^{n-1}$ .

On the other hand, the relation $\sigma b=b\zeta$ implies by induction on $i$ that $\sigma ^ib=b\zeta ^i\ne b$ for each $1\le i\le l^{n-1}$ . Hence, ${\mathrm{ord}}(\sigma |_{M(b)})\gt l^{n-1}$ . This contradicts the conclusion of the preceding paragraph, as required.

4. Generalization of Geyer’s theorem

The central part of the proof of Geyer’s theorem says that for each prime number $l$ , the largest positive integer $n$ for which $\mathbb{Z}_l^n$ is a closed subgroup of $\textrm{Gal}(\mathbb{Q})$ or of $\textrm{Gal}(\mathbb{F}_p(t))$ is $1$ . The next lemma will allow us to generalize that statement to each finitely generated extension of a global field.

Proof. Suppose that $A\;:\!=\;\mathbb{Z}_l^{n'}$ is a closed subgroup of $\textrm{Gal}(K(t))$ for some positive integer $n'$ . Let $\varphi\;:\;\textrm{Gal}(K(t))\to \textrm{Gal}(K)$ be the restriction map. Then, ${\mathrm{Ker}}(\varphi )=\textrm{Gal}(K_{\mathrm{sep}}(t))$ . Setting ${\bar{A}}=\varphi (A)$ and $A_0={\mathrm{Ker}}(\varphi )\cap A$ , we get the following commutative diagram of profinite groups:

where $\bf 0$ stands for the trivial group of an additive abelian group. Since $\mathbb{Z}_l$ is a principal ideal domain and $A$ is a free $\mathbb{Z}_l$ -module of rank $n'$ , $A_0$ is a free $\mathbb{Z}_l$ -module, by [Reference Lang7, p. 146, Thm. 7.1]. Also, by [Reference Lang7, p. 148, Lemma 7.4], $\bar{A}$ is a free $\mathbb{Z}_l$ -module and $n'={\mathrm{rank}}(A_0)+{\mathrm{rank}}({\bar{A}})$ .

By [Reference Ribes9, p. 272, Prop. 5.2], $\textrm{Gal}(K_{\mathrm{sep}}(t))$ is a projective group, so also $A_0$ is a projective group. In other words, ${\mathrm{rank}}(A_0)\le 1$ . Also, by Corollary 2.3 and the assumption of the lemma, ${\bar{A}}=\mathbb{Z}_l^m$ with $m\le n$ or $l=2$ and ${\bar{A}}\cong \mathbb{Z}/2\mathbb{Z}$ . In each case, ${\mathrm{rank}}({\bar{A}})\le n$ , hence ${\mathrm{rank}}(A)={\mathrm{rank}}({\bar{A}})+{\mathrm{rank}}(A_0)\le n+1$ , as claimed.

Proof. By Corollary 2.3, $A\cong \mathbb{Z}/2\mathbb{Z}$ or $A\cong \prod _l\mathbb{Z}_l^{r_l}$ , with cardinal numbers $r_l$ . Assume the latter case. If $K$ is a global field, then $r=0$ . Hence, by Theorem 3.3, $r_l\le 0+1$ for each $l$ .

Otherwise, $r\ge 1$ and $K$ is a finitely generated extension of transcendence degree $1$ of a finitely generated extension $K'_0$ of transcendence degree $r-1$ of $K_0$ . By induction, for each prime number $l$ , $r$ is the largest positive integer such that $\mathbb{Z}_l^r$ is a closed subgroup of $\textrm{Gal}(K'_0)$ . Hence, by Lemma 4.2, $r+1$ is the largest positive number for which $\mathbb{Z}_l^{r+1}$ is a closed subgroup of $\textrm{Gal}(K)$ . In particular, $r_l\le r+1$ , as claimed.

5. Realizing ${\hat{\mathbb{Z}}}^{r+1}$ as a closed subgroup of $\textrm{Gal}(K)$

Let $K$ be a finitely generated extension of $\mathbb{Q}$ of transcendence degree $r$ . We complete Proposition 4.3 in this section by proving that ${\hat{\mathbb{Z}}}^{r+1}$ is a closed subgroup of $\textrm{Gal}(K)$ . An analogous result holds for a finitely generated extension $K$ of transcendence degree $r$ of $\mathbb{F}_p(t)$ , in which case $\prod _{l\ne p}\mathbb{Z}_l^{r+1}$ replaces ${\hat{\mathbb{Z}}}^{r+1}$ .

Remark 5.1 (Valued fields). We denote the residue field of a valued field $(F,v)$ by ${\bar{F}}_v$ and its value group by $v(F^\times )$ . In addition, we extend $v$ to a valuation of $F_{\mathrm{sep}}$ that we also denote by $v$ , consider its valuation ring $O_{v,\mathrm{sep}}$ , and let $D_{v,\mathrm{sep}}=\{\sigma \in \textrm{Gal}(F)\mathop{|\;}\sigma O_{v,\mathrm{sep}}=O_{v,\mathrm{sep}}\}$ be the corresponding decomposition group . Then, we let $F_v$ be the fixed field of $D_{v,\mathrm{sep}}$ in $F_{\mathrm{sep}}$ . Abusing our notation, we also let $v$ be the restriction of $v$ to $F_v$ . Then, $(F_v,v)$ is the Henselization of $(F,v)$ .

One knows that $(F_v,v)$ has the same residue field and value group as those of $(F,v)$ [ Reference Efrat2, p. 138, Prop. 15.3.7]. Moreover, the valued fields $(F_{\mathrm{sep}},v)$ and $(F_v,v)$ depend on the extension of $v$ to $F_{\mathrm{sep}}$ up to isomorphism [ Reference Efrat2, p. 138, Cor. 15.3.6].

If $v$ is a rank- $1$ valuation, then so is its extension to $F_v$ . In this case, the completion $({\hat F}_v,v)$ of $(F,v)$ is also discrete with the same value group and residue field as those of $(F,v)$ . Moreover, $({\hat F}_v,v)$ is also the completion of $(F_v,v)$ . By Hensel’s lemma, $({\hat F}_v,v)$ is also Henselian [ Reference Efrat2, p. 167, Cor. 18.3.2]. We embed $F_{\mathrm{sep}}$ into ${\hat F}_{v,\mathrm{sep}}$ and observe that $F_{\mathrm{sep}}\cap{\hat F}_v=F_v$ (since $(F_{\mathrm{sep}}\cap{\hat F}_v,v)$ is an immediate separable algebraic extension of $(F_v,v)$ ) and $F_{\mathrm{sep}}{\hat F}_v={\hat F}_{v,\mathrm{sep}}$ (by the Krasner-Ostrowski lemma [ Reference Efrat2, p. 172, Cor. 18.5.3]). Thus, restriction gives an isomorphism $\textrm{Gal}({\hat F}_v)\cong \textrm{Gal}(F_v)$ of the corresponding absolute Galois groups.

We denote the maximal unramified extension of $F_v$ (resp. ${\hat F}_v$ ) by $F_{v,\mathrm{ur}}$ (resp. ${\hat F}_{v,\mathrm{ur}}$ ) and the maximal tamely ramified extension by $F_{v,{\mathrm{tr}}}$ (resp. ${\hat F}_{v,{\mathrm{tr}}}$ ). These fields are Galois extensions of $F_v$ (resp. ${\hat F}_v$ ). As in [ Reference Efrat2, p. 133, p. 141, and p. 145], we set $Z(v)=\textrm{Gal}(F_v)$ for the decomposition group , $T(v)=\textrm{Gal}(F_{v,\mathrm{ur}})$ for the inertia group , and $V(v)=\textrm{Gal}(F_{v,{\mathrm{tr}}})$ for the ramification group of $(F,v)$ . The letters $Z$ , $T$ , and $V$ are borrowed from the German translations Zerlegsungruppe, Trägheitsgruppe, and Verzweigungsgruppe of the English expressions decomposition group, inertia group, and ramification group,

(5.1)

Each of the fields $F_{v,\mathrm{ur}}$ , $F_{v,{\mathrm{tr}}}$ , and $F_{\mathrm{sep}}$ is a Galois extension of $F_v$ . By [ Reference Efrat2, p. 199, Thm. 22.1.1] and [ Reference Kuhlmann, Pank and Roquette6, Thm. 2.2] (resp. [ Reference Efrat2, p. 203, Thm. 22.2.1]) both restriction maps

\begin{align*} \textrm{Gal}(F_{v,{\mathrm{tr}}}/F_v)\to \textrm{Gal}(F_{v,\mathrm{ur}}/F_v) \quad \hbox{and} \quad \textrm{Gal}(F_v)\to \textrm{Gal}(F_{v,{\mathrm{tr}}}/F_v) \end{align*}

split. In particular, each closed subgroup of $\textrm{Gal}(F_{v,\mathrm{ur}}/F_v)$ ; hence, each closed subgroup of $\textrm{Gal}({\bar{F}}_v)$ is isomorphic to a closed subgroup of $\textrm{Gal}(F_{v,{\mathrm{tr}}}/F_v)$ . Also, each closed subgroup of $\textrm{Gal}(F_{v,{\mathrm{tr}}}/F_v)$ is isomorphic to a closed subgroup of $\textrm{Gal}(F_v)$ .

Note that $E$ in Theorem 22.1.1 of [Reference Efrat2] is $F_{\mathrm{sep}}$ , in our notation, so it satisfies the condition $E=E^l$ for all prime numbers $l\ne \mathrm{char}({\bar{F}}_v)$ needed in that theorem.

Remark 5.3. Given a field $K$ , the field of formal power series $K((t))$ in the variable $t$ with coefficients in $K$ , also called the field of Laurent series over $K$ , is the field of all formal power series $\sum _{i=m}^\infty a_it^i$ with $m\in \mathbb{Z}$ and $a_i\in K$ for all $i\ge m$ . If $l\lt m$ , then $\sum _{i=m}^\infty a_it^i$ is identified with $\sum _{i=l}^\infty a_it^i$ with $a_i=0$ for each $l\le i\lt m$ . Summation and multiplication in $K((t))$ are defined by the following rules:

\begin{align*} \begin{aligned} \sum _{i=m}^\infty a_it^i +\sum _{i=m'}^\infty a'_it^i &=\sum _{i=\min (m,m')}^\infty (a_i+a'_i)t^i, \cr \big (\sum _{i=m}^\infty a_it^i\big ) \big (\sum _{j=m'}^\infty a'_jt^j\big ) &=\sum _{k=m+m'}^\infty \big (\sum _{i+j=k}a_ia'_j)t^k. \end{aligned} \end{align*}

Let $v$ be the unique discrete valuation of $K(t)$ with $v(a)=0$ for each $a\in K$ and $v(t)=1$ . Then, $(K((t)),v)$ is the completion of $(K(t),v)$ , where $v(\sum _{i=m}^\infty a_it^i)=m$ whenever $a_m\ne 0$ . By [ Reference Efrat2, p. 167, Cor. 18.3.2], $K((t))$ is Henselian with respect to $v$ .

By [ Reference Cassels and Fröhlich1, p. 28, Cor. 2] (or [ Reference Efrat2, p. 141, Thm. 16.1.1]),

\begin{align*} \textrm{Gal}(K((t))_{\mathrm{ur}}/K((t)))\cong \textrm{Gal}(K). \end{align*}

Replacing $K$ by $K_{\mathrm{sep}}$ , we have that $K_{\mathrm{sep}}((t))_{\mathrm{ur}}=K_{\mathrm{sep}}((t))$ . Since the roots of unity of order $n$ with $\mathrm{char}(K)\nmid n$ are in $K_{\mathrm{sep}}$ , we have that $K_{\mathrm{sep}}((t))$ has a cyclic extension of degree $n$ in $K_{\mathrm{sep}}((t))_{\mathrm{tr}}$ . Indeed, that extension is $K_{\mathrm{sep}}((t^{1/n}))$ .

Going to the limit of these extensions, we obtain with $p\;:\!=\;\mathrm{char}(K)$ that $K_{\mathrm{sep}}((t))_{\mathrm{tr}} =\bigcup _{p\nmid n}K_{\mathrm{sep}}((t^{1/n}))$ and $\textrm{Gal}(K_{\mathrm{sep}}((t))_{\mathrm{tr}}/K_{\mathrm{sep}}((t))) \cong \prod _{l\ne p}\mathbb{Z}_l$ .

Moreover, if $\mathrm{char}(K)=0$ , then the ramification group $\textrm{Gal}({\tilde{K}}((t))_{\mathrm{tr}})$ of ${\tilde{K}}((t))$ is trivial [ Reference Efrat2, p. 145, Thm. 16.2.3], so ${\tilde{K}}((t))_{\mathrm{tr}}=\widetilde{K((t))}$ . Thus, by the preceding paragraph, in this case, $\textrm{Gal}({\tilde{K}}((t)))\cong{\hat{\mathbb{Z}}}$ .

Proof. By assumption, the field $K_0$ has a separable algebraic extension $K$ with $\textrm{Gal}(K)\cong \prod _{l\ne p}\mathbb{Z}_l^r$ . Let $v$ be the discrete $K$ -valuation of $K(t)$ with $v(t)=1$ and choose a Henselization $M\;:\!=\;K(t)_v$ of $K(t)$ with respect to $v$ . Then,

(5.2) \begin{equation} {\bar{M}}\;:\!=\;\overline{K(t)}_v=K \end{equation}

is the residue field of both $K(t)$ and $M$ with respect to $v$ .

Claim: $M$ is linearly disjoint from $\tilde{K}$ over $K$ . Indeed, let ${\tilde{K}}_1,\ldots,{\tilde{K}}_n$ be linearly independent elements of $\tilde{K}$ over $K$ . Assume toward contradiction that there exist $m_1,\ldots,m_n\in M$ not all zero with $\sum _{i=1}^nm_i{\tilde{K}}_i=0$ . Dividing $m_1,\ldots,m_n$ by the element with the least $v$ -value, we may assume that the $v$ -residues ${\bar m}_1,\ldots,{\bar m}_n$ are elements of $K$ and one of them is non-zero. Thus, $\Sigma _{i=1}^n{\bar m}_i{\tilde{K}}_i=0$ , contradicting the assumption on ${\tilde{K}}_1,\ldots,{\tilde{K}}_n$ . This proves our claim.

By [Reference Efrat2, p. 200, Cor. 22.1.2],

(5.3) \begin{equation} Z(v)/V(v)\cong \chi (v)\rtimes \textrm{Gal}({\bar{M}}) \mathop =^{(5.2)} \chi (v)\rtimes \textrm{Gal}(K), \end{equation}

where $Z(v)=\textrm{Gal}(M)$ and $V(v)$ are respectively the corresponding decomposition and the ramification groups of $M$ and

(5.4) \begin{equation} \chi (v) ={\mathrm{Hom}}(v(M_{\mathrm{sep}}^\times )/v(M^\times ),\mu (K_{0,\mathrm{sep}})). \end{equation}

See [Reference Efrat2, last line of page 144] with $\bar \mu$ in that line being $\mu (K_{0,\mathrm{sep}})$ , as introduced in the first paragraph of [Reference Efrat2, p. 143, Sec. 16.2].

The action of $\textrm{Gal}(K)$ on $\chi (v)$ is given for each $\tau \in \textrm{Gal}(K)$ , each homomorphism $h\;:\; v(M_{\mathrm{sep}}^\times )/v(M^\times ) \to \mu (K_{0,\mathrm{sep}})$ , and every $\gamma \in v(M_{\mathrm{sep}}^\times )$ , by

\begin{align*} \tau (h)(\gamma +v(M^\times )) =\tau (h(\gamma +v(M^\times ))) =h(\gamma +v(M^\times )), \end{align*}

where the latter equality holds because $\mu (K_{0,\mathrm{sep}})\subseteq K_0\subseteq K$ . In other words, that action is trivial. It follows that

(5.5) \begin{equation} \textrm{Gal}(M_{\mathrm{tr}}/M) \mathop{\cong }^{(5.1)}Z(v)/V(v)\mathop{\cong }^{(5.3)}\chi (v)\times \textrm{Gal}(K). \end{equation}

By [Reference Efrat2, p. 147, Cor. 16.2.7], there is a short exact sequence

\begin{align*} \textbf{1}\buildrel \over \longrightarrow V(v)\buildrel \over \longrightarrow T(v)\buildrel \over \longrightarrow \chi (v)\buildrel \over \longrightarrow \textbf{1}. \end{align*}

Hence, $\chi (v)\cong T(v)/V(v)$ .

By our choice of $v$ , the completion of $K(t)$ with respect to $v$ (which is also the completion of the Henselian field $M$ ) is the field $K((t))$ of formal power series in $t$ with coefficients in $K$ [Reference Efrat2, p. 83, Example 9.2.2]. The maximal unramified extension of $K((t))$ is $K_{\mathrm{sep}}((t))$ and by Remark 5.3, $\chi (v)\cong T(v)/V(v) \cong \textrm{Gal}(M_{\mathrm{tr}}/M_{\mathrm{ur}})\cong \prod _{l\ne p}\mathbb{Z}_l$ .

By the definition of $K$ , $\textrm{Gal}(K)\cong \prod _{l\ne p}\mathbb{Z}_l^r$ . Hence, by the preceding paragraph,

\begin{align*} \textrm{Gal}(M_{\mathrm{tr}}/M) \mathop{\cong }^{(5.5)} \chi (v)\times \textrm{Gal}(K) \cong \prod _{l\ne p}\mathbb{Z}_l\times \prod _{l\ne p}\mathbb{Z}_l^r =\prod _{l\ne p}\mathbb{Z}_l^{r+1}. \end{align*}

Since by [Reference Kuhlmann, Pank and Roquette6, Thm. 2.2], the epimorphism $\textrm{Gal}(M)\to \textrm{Gal}(M_{\mathrm{tr}}/M)$ splits, $\prod _{l\ne p}\mathbb{Z}_l^{r+1}$ is a closed subgroup of $\textrm{Gal}(M)$ . Since $M$ is a separable algebraic extension of $K_0(t)$ [Reference Efrat2, p. 137, Thm. 15.3.5], $\prod _{l\ne p}\mathbb{Z}_l^{r+1}$ is also a closed subgroup of $\textrm{Gal}(K_0(t))$ , as claimed.

The following result will be needed in Theorem 5.7.

Proof. We set $Z\;:\!=\;\prod _{l\in L}\mathbb{Z}_l$ and consider all the groups appearing in this proof as additive groups. Since $H$ is open in $Z$ , its index $n\;:\!=\;(Z\;:\;H)$ is a positive integer. Since $Z$ is abelian, $H$ is normal in $Z$ , so $nZ\le H$

By [Reference Fried and Jarden3, p. 13, Lemma 1.4.2(e)], $n\mathbb{Z}_l\cong \mathbb{Z}_l$ for each $l\in L$ . Hence, $nZ=\prod _{l\in L}n\mathbb{Z}_l \cong \prod _{l\in L}\mathbb{Z}_l=Z$ .

Let $n=\prod _{l\in L'}l^{i(l)}$ be the decomposition of $n$ into a product of prime powers. If $l$ and $l'$ are distinct prime numbers, then $l'$ is a unit of the ring $\mathbb{Z}_l$ , so $l'\mathbb{Z}_l=\mathbb{Z}_l$ . Hence, $nZ=\prod _{l\in L\cap L'}l^{i(l)}\mathbb{Z}_l \times \prod _{l\in L \newcommand{\hefreshD }{\mathop{\raise 1.5pt\hbox{${\smallsetminus }$}}} \newcommand{\hefreshS }{\mathop{\raise 0.85pt\hbox{$\scriptstyle \smallsetminus $}}} \mathchoice{\hefreshD }{\hefreshD }{\hefreshS }{\hefreshS } L'}\mathbb{Z}_l$ . Therefore, $(Z\;:\;nZ) =\prod _{l\in L\cap L'}(\mathbb{Z}_l\;:\;l^{i(l)}\mathbb{Z}_l) =\prod _{l\in L\cap L'}l^{i(l)} \le n =(Z\;:\;H)$ . Combining this result with the result of the first paragraph of the proof, we have $H=nZ$ . Therefore, by the second paragraph of the proof, $H\cong Z$ , as claimed.

This brings us to the main result of the current section.

Proof. In the case where $r=0$ , $F$ itself is a global field, hence Hilbertian [Reference Fried and Jarden3, p. 242, Thm. 13.4.2]. Since $F'$ is an abelian extension of $F$ , a theorem of Kuyk asserts that $F'$ is also Hilbertian [Reference Fried and Jarden3, p. 333, Thm. 16.11.3]. Since $F$ is countable, so is $F'$ . By [Reference Fried and Jarden3, p. 379, Thm. 18.5.6], for almost all $\sigma \in \textrm{Gal}(F')$ (in the sense of the Haar measure of $\textrm{Gal}(F')$ ) the closed subgroup $\langle \sigma \rangle$ of $\textrm{Gal}(F')$ generated by $\sigma$ is isomorphic to $\hat{\mathbb{Z}}$ . Since $\prod _{l\ne p}\mathbb{Z}_l$ is a closed subgroup of $\prod _l\mathbb{Z}_l$ and $\prod _l\mathbb{Z}_l\cong{\hat{\mathbb{Z}}}$ [Reference Fried and Jarden3, p. 15, Lemma 1.4.5], $\prod _{l\ne p}\mathbb{Z}_l$ is a closed subgroup of $\textrm{Gal}(F')$ .

Alternatively, by a theorem of Whaples, for each $l\ne p$ the field $F'$ has a Galois extension $F'_l$ with $\textrm{Gal}(F'_l/F')\cong \mathbb{Z}_l$ [Reference Fried and Jarden3, p. 314, Cor. 16.6.7]. Then, $F''\;:\!=\;\prod _{l\ne p}F'_l$ is a Galois extension of $F'$ with $ \textrm{Gal}(F''/F')\cong \prod _{l\ne p}\mathbb{Z}_l$ . Since $\prod _{l\ne p}\mathbb{Z}_l$ is projective [Reference Fried and Jarden3, p. 507, Cor. 22.4.6], the restriction map $\textrm{Gal}(F')\to \textrm{Gal}(F''/F')$ splits [Reference Fried and Jarden3, p. 506, Remark 22.4.2]. Hence, again, $\prod _{l\ne p}\mathbb{Z}_l$ is a closed subgroup of $\textrm{Gal}(F')$ .

Next assume by induction that $r\ge 1$ and the theorem holds for $r-1$ . Choose a finitely generated extension $F_{r-1}$ of transcendence degree $r-1$ of $F_0$ in $F$ and let $F'_{r-1}=F_{r-1}(\mu (F_{0,\mathrm{sep}}))$ . Since $F$ is finitely generated over $F_0$ of transcendence degree $r$ , we may choose $t$ in $F$ which is transcendental over $F_{r-1}$ and $[F\;:\;F_{r-1}(t)]\lt \infty$ . Then, $F'=F'_{r-1}F$ is a finite extension of $F'_{r-1}(t)$ . Let $L$ be the maximal separable extension of $F'_{r-1}(t)$ in $F'$ , so $F'/L$ is a purely inseparable extension of $L$ . Then, $L$ is a finite separable extension of $F'_{r-1}(t)$ .

Hence,

(5.6) \begin{equation} \textrm{Gal}(L) \text{ is an open subgroup of } \textrm{Gal}(F'_{r-1}(t)). \end{equation}

By the induction hypothesis, $\prod _{l\ne p}\mathbb{Z}_l^r$ is a closed subgroup of $\textrm{Gal}(F'_{r-1})$ . Therefore, by (5.6), Lemma 5.4, and Lemma 5.6, $\prod _{l\ne p}\mathbb{Z}_l^{r+1}$ is a closed subgroup of $\textrm{Gal}(L)$ . Since $F'/L$ is a purely inseparable extension (in particular $F'=L$ if $\mathrm{char}(F_0)=0$ ), $\prod _{l\ne p}\mathbb{Z}_l^{r+1}$ is a closed subgroup of $\textrm{Gal}(F')$ , hence also of $\textrm{Gal}(F)$ , as claimed.