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Extra info for An Introduction to p-adic Numbers and p-adic Analysis [Lecture notes]

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Then in general f need not have any roots in Qp . 1. For a prime p, consider the polynomial X 2 − p. If α ∈ Qp were a root then we would have α2 = p and so |α|p 2 = 1/p. But we know that the norm of a p-adic number has to have the form 1/pk with k ∈ Z, so since |α|p = p−1/2 this would give a contradiction. We will not prove the next result, the interested reader should consult [5]. 2. There exists a field Qalg p containing Qp as a subfield and having the following properties: (a) every α ∈ Qalg p is algebraic over Qp ; alg (b) every polynomial f (X) ∈ Qalg p [X] has a root in Qp .

Taking αn = npn we have sm = m ∑ npn n=1 and sn+1 − sn = (n + 1)pn+1 . 29 This has norm (n + 1)pn+1 p = |n + 1|p pn+1 p 1 , pn+1 which clearly tends to 0 as n → ∞ in the real numbers. 1, (sn ) is a Cauchy sequence and therefore it has a limit in Qp . In real analysis, there are series which converge but are not absolutely convergent. For ∑ ∑ example, the series (−1)n /n converges to − ln 2 but 1/n diverges. Our next result shows that this cannot happen in Qp . 4. The series αn in Qp converges if and only if (αn ) is a null sequence.

The expansion in this result is called the Mahler expansion of f and the coefficients αn are the Mahler coefficients of f . We need to understand how to determine these coefficients. Consider the following sequence of functions f [n] : Zp −→ Qp : f [0] (x) = f (x) f [1] (x) = f [0] (x + 1) − f [0] (x) f [2] (x) = f [1] (x + 1) − f [1] (x) .. f [n+1] (x) = f [n] (x + 1) − f [n] (x) .. f [n] is called the n-th difference function of f . 30. The Mahler coefficients are given by αn = f [n] (0) Proof. (Sketch) Consider f (0) = ∑ (n 0).

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