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Relative Galois module structure of integers of local abelian fields. (English) Zbl 0910.11050
This nice paper shows that for any abelian extension \(N/K\) with \(N\) abelian over \({\mathbb Q}_p\), the ring of integers \({\mathcal O}_N\) is free (of rank one) over the associated order \({\mathcal A}_{N/K} \subset K[\text{Gal} (N/K)]\). Some context: If \(N/K\) is at most tamely ramified, the result is clear since \({\mathcal O}_N\) is projective, hence free over \({\mathcal O}_K[\text{Gal} (N/K)]\), and consequently \({\mathcal A}_{N/K} = K[\text{Gal} (N/K)]\); and if \(N\) is not required to be abelian over \({\mathbb Q}_p\), the result ceases to be true.
The main line of the argument goes as follows: Since associated orders behave as well as possible under base change with unramified extensions (see Proposition 1 (a)), the problem may be reduced to the case where both \(N\) and \(K\) sit inside an extension \({\mathbb Q}_p(\zeta_{p^n})\). In this case, an explicit generator for \({\mathcal O}_N\) over \({\mathcal A}_{N/K}\) is given, somewhat in the spirit of Leopoldt’s theorem. Actually in this setting, the associated order is the maximal order if \(p\) is different from \(2\) (and also in some cases with \(p=2\)).
One peripheral remark. Even if the proof of Proposition 1 (b) (“freeness descends”) is quite short, it does use the Krull-Schmidt-Azumaya theorem. Alternatively, one might just say that faithfully flat base change reflects finitely generated projective modules (true in full generality and easy), and all projective modules are free in our local situation. To the reviewer this would appear to be a more direct argument.

11S23 Integral representations
11S15 Ramification and extension theory
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