1 results
Ramification des séries formelles
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- Journal:
- Canadian Mathematical Bulletin / Volume 47 / Issue 2 / 01 June 2004
- Published online by Cambridge University Press:
- 20 November 2018, pp. 237-245
- Print publication:
- 01 June 2004
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- Article
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Let
$p$ be a prime number. Let 
$k$ be a finite field of characteristic $p$. The subset 
$X\,+\,{{X}^{2}}k[[X]]$ of the ring 
$k\left[\!\left[ X \right]\!\right]$ is a group under the substitution law 
$\circ $ sometimes called the Nottingham group of $k$, it is denoted by 
${{\mathcal{R}}_{k}}$. The ramification of one series 
$\gamma \,\in \,{{\mathcal{R}}_{k}}$ is caracterized by its lower ramification numbers: 
${{i}_{m}}(\gamma )\,=\,\text{or}{{\text{d}}_{X}}({{\gamma }^{{{p}^{m}}}}\,(X)/X-1)\,$, as well as its upper ramification numbers:
$${{u}_{m}}(\gamma )\ =\ {{i}_{0}}(\gamma )+\frac{{{i}_{1}}(\gamma )-{{i}_{0}}(\gamma )}{p}\,+\,.\,.\,.\,+\,\frac{{{i}_{m}}(\gamma )-{{i}_{m-1}}(\gamma )}{{{p}^{m}}},\,\,\,\,\,\,(m\,\in \,\mathbb{N}).$$By Sen's theorem, the
${{u}_{m}}(\gamma )$ are integers. In this paper, we determine the sequences of integers (
${{u}_{m}}$) for which there exists $\gamma \,\in \,{{\mathcal{R}}_{k}}$ such that 
${{u}_{m}}(\gamma )\,=\,{{u}_{m}}$ for all integer 
$m\,\ge \,0$.
