{"paper":{"title":"On Galois realizations of the 2-coverable symmetric and alternating groups","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.NT","authors_text":"Daniel Rabayev, Jack Sonn","submitted_at":"2010-08-18T09:45:17Z","abstract_excerpt":"Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q_p for all p, or equivalently, with roots mod n for all n. It is known that f(x) cannot be irreducible but can be a product of two or more irreducible polynomials, and that if f(x) is a product of m>1 irreducible polynomials, then its Galois group must be \"m-coverable\", i.e. a union of conjugates of m proper subgroups, whose total intersection is trivial. We are thus led to a variant of the inverse Galois problem: given an m-coverable finite group G, find a Galois realization of G over the rationals Q by a polynom"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1008.3061","kind":"arxiv","version":4},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"}