[Paper Review] On Abhyankar's lemma about ramification indices

Jean-Luc Chabert, Emmanuel Halberstadt|arXiv (Cornell University)|May 22, 2018

Algebraic Geometry and Number Theory3 references4 citations

TL;DR

This paper provides a concise, group-theoretic proof that the ramification index of the compositum of two finite extensions of local fields equals the least common multiple (lcm) of the individual ramification indices when at least one extension is tamely ramified. The key contribution is a generalization of Abhyankar's lemma to Dedekind domains with perfect residue fields, establishing that $ e(\mathfrak{q}/\mathfrak{p}) = \mathrm{lcm}\{e(\mathfrak{p}_1/\mathfrak{p}), e(\mathfrak{p}_2/\mathfrak{p})\} $ under tame ramification conditions.

ABSTRACT

We provide a simple proof of the fact that the ramification index of the compositum of two finite extensions of local fields is equal to the least common multiple of the ramification indices when at least one of the extensions is tamely ramified.

Motivation & Objective

• To generalize Abhyankar's lemma to number fields and function fields by establishing a precise formula for the ramification index of a compositum extension.
• To resolve the gap in the literature by providing a direct proof for the case where at least one extension is tamely ramified, which had been previously stated only for function fields.
• To clarify the conditions under which the ramification index of the compositum equals the lcm of individual indices, particularly in the context of Dedekind domains with perfect residue fields.
• To demonstrate that the standard inequality $ \mathrm{lcm}\{e(\mathfrak{p}_1/\mathfrak{p}), e(\mathfrak{p}_2/\mathfrak{p})\} \mid e(\mathfrak{q}/\mathfrak{p}) $ can be upgraded to equality under tame ramification.

Proposed method

• The proof uses the normal closure $ L' $ of the compositum $ L = K_1K_2 $ over $ K $, and analyzes the inertia group $ G_0 = \mathcal{I}_{\mathfrak{q}'}(L'/K) $ at a prime $ \mathfrak{q}' $ above $ \mathfrak{p} $.
• It decomposes the inertia group $ G_0 $ into a semidirect product of a cyclic group of order prime to the residue characteristic $ p $, and a $ p $-group $ G_1 $, leveraging properties of Galois groups and ramification.
• The method applies the formula $ e(\mathfrak{q}/\mathfrak{p}) = |G_0| / |G_0 \cap \Gamma| $, where $ \Gamma = \mathrm{Gal}(L'/L) $, and similarly for $ e(\mathfrak{p}_i/\mathfrak{p}) $, expressing indices in terms of group orders.
• It uses the identity $ \mathrm{lcm}(a,b) \times \gcd(a,b) = ab $ to relate the lcm of ramification indices to the gcd of the corresponding group orders.
• It distinguishes cases based on the characteristic $ p $: when $ p = 0 $, $ G_0 $ is cyclic, so the intersection of subgroups has order equal to the gcd of their orders.
• For $ p > 0 $, it exploits the fact that $ G_1 $ is the unique $ p $-Sylow subgroup and uses the projection $ \pi: G_0 \to G_0/G_1 $ to reduce the problem to the cyclic quotient, proving $ |G_0 \cap \Gamma| = \gcd\{|G_0 \cap \Gamma_1|, |G_0 \cap \Gamma_2|\} $.

Experimental results

Research questions

• RQ1 Under what conditions does the ramification index of the compositum of two local field extensions equal the least common multiple of the individual ramification indices?
• RQ2 Can Abhyankar's lemma be generalized to number fields and Dedekind domains with perfect residue fields, not just function fields?
• RQ3 What happens to the ramification index when both extensions are wildly ramified, and why may the lcm formula fail?
• RQ4 How does the structure of the inertia group in the Galois closure of the compositum determine the ramification index of the compositum?
• RQ5 Is the equality $ e(\mathfrak{q}/\mathfrak{p}) = \mathrm{lcm}\{e(\mathfrak{p}_1/\mathfrak{p}), e(\mathfrak{p}_2/\mathfrak{p})\} $ valid when one extension is tamely ramified?

Key findings

• The ramification index of the compositum $ L = K_1K_2 $ over $ K $ satisfies $ e(\mathfrak{q}/\mathfrak{p}) = \mathrm{lcm}\{e(\mathfrak{p}_1/\mathfrak{p}), e(\mathfrak{p}_2/\mathfrak{p})\} $ when at least one of $ K_1/K $ or $ K_2/K $ is tamely ramified.
• The proof relies on the Galois-theoretic structure of the inertia group in the normal closure, showing that the order of the intersection of Galois groups equals the gcd of their orders.
• When the residue characteristic $ p = 0 $, the inertia group is cyclic, and the result follows directly from the property that the order of the intersection of subgroups in a cyclic group is the gcd of their orders.
• For $ p > 0 $, the result holds because the $ p $-Sylow subgroup $ G_1 $ is normal and contained in one of the Galois groups (e.g., $ \Gamma_1 $) if $ p $ does not divide $ e(\mathfrak{p}_1/\mathfrak{p}) $, allowing reduction to the cyclic quotient.
• The formula fails in general when both extensions are wildly ramified; the paper provides a counterexample with $ K = \mathbb{Q} $, $ K_1 = \mathbb{Q}(\sqrt[3]{3}) $, $ K_2 = \mathbb{Q}(j\sqrt[3]{3}) $, where $ e(\mathfrak{q}/\mathfrak{p}) = 6 $, $ \mathrm{lcm}(3,3) = 3 $, and $ 6 \nmid 9 $, showing that the product bound can also fail.
• The corollary states that if $ e(\mathfrak{p}_1/\mathfrak{p}) $ and $ e(\mathfrak{p}_2/\mathfrak{p}) $ are coprime, then $ e(\mathfrak{q}/\mathfrak{p}) = e(\mathfrak{p}_1/\mathfrak{p}) \times e(\mathfrak{p}_2/\mathfrak{p}) $, which follows from $ \mathrm{lcm}(a,b) = ab $ when $ \gcd(a,b) = 1 $.