On a conjecture for Rubin-Stark elements in a special case (Q267445)

Let \(K/k\) be an abelian extension of global fields with Galois group \(G\). Let \(S\), \(T\) and \(V\) be three finite sets of places of \(k\) satisfying certain conditions. Here we only mention that \(S\) has to contain all archimedean places (in the number field case) and all places that ramify in \(K/k\). Moreover, \(V\) is a proper subset of \(S\) of cardinality \(r\) such that all \(v \in V\) split completely in \(K\).NEWLINENEWLINEA conjecture of \textit{K. Rubin} [Ann. Inst. Fourier 46, No. 1, 33--62 (1996; Zbl 0834.11044)] predicts the existence of a so-called Rubin-Stark element \(\epsilon_{K,S,T,V}\) which maps under a certain regulator map to an element constructed from \(r\)-th derivatives of \((S,T)\)-modified Artin \(L\)-functions at zero, and which lies in a certain lattice in \(\mathbb Q \otimes_{\mathbb Z} \bigwedge_G^r \mathcal{O}^{\times}_{K,S,T}\). Here, \(\mathcal{O}_{K,S,T}^{\times}\) is a \(T\)-modified subgroup of the \(S\)-units of \(K\) which is of finite index and torsionfree (by a condition on \(T\)).NEWLINENEWLINENow let \(L\) be an intermediate field and let \(V'\) be a proper subset of \(S\) such that \(V \subset V'\) and each \(v' \in V'\) splits completely in \(L\). Assume that the Rubin-Stark conjecture holds. Then the author [Compos. Math. 150, No. 11, 1809--1835 (2014; Zbl 1311.11108)] and, independently, \textit{B. Mazur} and \textit{K. Rubin} [J. Théor. Nombres Bordx. 28, No. 1, 185--211 (2016; Zbl 1414.11155)] have formulated a conjecture that predicts an equality NEWLINE\[NEWLINE \mathcal{N}_{K/L}(\epsilon_{K,S,T,V}) = \pm R_{V',V}(\epsilon_{L,S,T,V'}), NEWLINE\]NEWLINE where \(\mathcal{N}_{K/L}\) is a certain norm map and \(R_{V',V}\) is a regulator map that incorporates the Artin reciprocity maps at the places \(v' \in V' \setminus V\).NEWLINENEWLINEIn this paper the author proves this conjecture in the following special case (still assuming the Rubin-Stark conjecture for a certain class of intermediate field extensions): \(V\) contains all archimedean places; all \(v \in S\) split completely in \(L\); the Galois group \(\mathrm{Gal}(K/L)\) is the direct product of the inertia subgroups at \(v \in S \setminus V\).NEWLINENEWLINEThe proof is by induction on the cardinality of \(S \setminus V\) and follows a strategy of \textit{H. Darmon} [Can. J. Math. 47, No. 2, 302--317 (1995; Zbl 0844.11071)].