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TauCeti.NumberTheory.LocalField.GeneratorRank

The lower bound for the generator rank of a local absolute Galois group #

Let K be a finite extension of ℚ_p and N = [K : ℚ_p]. This file proves that the absolute Galois group G_K needs at least N + 2 topological generators:

N + 2 ≤ d(G_K).

When μ_p ⊆ K this is visible on the maximal pro-p quotient, whose generator rank is exactly N + 2. When μ_p ⊄ K the maximal pro-p quotient only needs N + 1 generators, and the bound comes from the open subgroup G_L for L = K(μ_p). With m = [L : K], the field L contains μ_p, so d(G_L) ≥ d(G_L(p)) = [L : ℚ_p] + 2 = mN + 2, while Schreier's bound for an open subgroup of index m gives d(G_L) ≤ 1 + m (d(G_K) - 1). Hence m (d(G_K) - 1) > mN, that is d(G_K) ≥ N + 2. The argument with L = K(μ_p) covers both cases at once, with m = 1 when μ_p ⊆ K.

The bound is stated twice: unconditionally, as a lower bound on the size of every finite set generating a dense subgroup of G_K, and for the natural-number generator rank, which takes a proof of topological finite generation as an argument.

Main results #

References #

The lower bound for the generator rank of a local absolute Galois group. For a finite extension K of ℚ_p, the absolute Galois group G_K needs at least [K : ℚ_p] + 2 topological generators. The finite-generation witness is an argument, so that the natural-valued rank is only read on a topologically finitely generated group.

Every finite set topologically generating a local absolute Galois group has at least [K : ℚ_p] + 2 elements. This is the lower half of the exact generator rank of G_K, stated without assuming topological finite generation: a finite generating set is itself a witness.