2020
DOI: 10.1142/s1793042121500251
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Fields generated by finite rank subgroups of ℚ¯∗

Abstract: Let [Formula: see text] be a finite rank subgroup of [Formula: see text]. We prove that the multiplicative group of the field generated by all elements in the divisible hull of [Formula: see text] is free abelian modulo this divisible hull. This proves that a necessary condition for Rémond’s generalized Lehmer conjecture is satisfied.

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Cited by 4 publications
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“…Clairement, 0trueprefixinf{hfalse(γαϕfalse(1false)false),γbsat}goodbreak=trueprefixinf{hfalse(γαϕfalse(nfalse)false),γbsat}h(αϕ(n)).$$\begin{equation*} 0 \leqslant \inf \lbrace h(\gamma \alpha _{\phi (1)}), \gamma \in \langle b \rangle _\mathrm{sat}\rbrace = \inf \lbrace h(\gamma \alpha _{\phi (n)}), \gamma \in \langle b \rangle _\mathrm{sat}\rbrace \leqslant h(\alpha _{\phi (n)}). \end{equation*}$$Par le théorème des gendarmes, on en conclut que trueprefixinffalse{h(γαϕ(1)),γfalse⟨bfalse⟩satfalse}=0,$$\begin{equation*} \inf \lbrace h(\gamma \alpha _{\phi (1)}), \gamma \in \langle b \rangle _\mathrm{sat}\rbrace =0, \end{equation*}$$ce qui ne peut se produire que si αϕfalse(1false)false⟨bfalse⟩sat$\alpha _{\phi (1)}\in \langle b \rangle _\mathrm{sat}$ d'après [21, Lemma 2.4]. On a donc βi=αϕfalse(1false)ζ(ϕfalse(1false))byϕ(1)false⟨bfalse⟩sat$\beta _i= \alpha _{\phi (1)} \;\zeta (\phi (1)) \; b^{y_{\phi (1)}} \in \langle b\rangle _\mathrm{sat}$.…”
Section: Preuve Du Théorème 18unclassified
“…Clairement, 0trueprefixinf{hfalse(γαϕfalse(1false)false),γbsat}goodbreak=trueprefixinf{hfalse(γαϕfalse(nfalse)false),γbsat}h(αϕ(n)).$$\begin{equation*} 0 \leqslant \inf \lbrace h(\gamma \alpha _{\phi (1)}), \gamma \in \langle b \rangle _\mathrm{sat}\rbrace = \inf \lbrace h(\gamma \alpha _{\phi (n)}), \gamma \in \langle b \rangle _\mathrm{sat}\rbrace \leqslant h(\alpha _{\phi (n)}). \end{equation*}$$Par le théorème des gendarmes, on en conclut que trueprefixinffalse{h(γαϕ(1)),γfalse⟨bfalse⟩satfalse}=0,$$\begin{equation*} \inf \lbrace h(\gamma \alpha _{\phi (1)}), \gamma \in \langle b \rangle _\mathrm{sat}\rbrace =0, \end{equation*}$$ce qui ne peut se produire que si αϕfalse(1false)false⟨bfalse⟩sat$\alpha _{\phi (1)}\in \langle b \rangle _\mathrm{sat}$ d'après [21, Lemma 2.4]. On a donc βi=αϕfalse(1false)ζ(ϕfalse(1false))byϕ(1)false⟨bfalse⟩sat$\beta _i= \alpha _{\phi (1)} \;\zeta (\phi (1)) \; b^{y_{\phi (1)}} \in \langle b\rangle _\mathrm{sat}$.…”
Section: Preuve Du Théorème 18unclassified