By Andrew Baker
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Extra info for An Introduction to p-adic Numbers and p-adic Analysis [Lecture notes]
21. The subset Op ⊆ Cp is a subring of Cp . The proof uses the ultrametric inequality and is essentially the same as that for Zp ⊆ Qp . We end with yet another version of Hensel’s Lemma, this time adapted to use in Cp . 22 (Hensel’s Lemma: Cp version). Let f (X) ∈ Op [X]. Suppose that α ∈ Op and d 0 is a natural number satisfying the two conditions 1 1 |f (α)|p , f ′ (α) p = d . p2d+1 p Setting α1 = α − f (α)f ′ (α)−1 , we have 1 |f (α1 )|p p2d+3 . The proof is left as an exercise. This result generalises our earlier versions of Hensel’s Lemma.
Find a generator for the cyclic group of units (Z/n)× in each of the following rings: (i) Z/23, 1-5. (a) For a prime p, n (ii) Z/27, (iii) Z/10. 1 and x≡0, consider p sn = 1 + x + x2 + · · · + xn−1 ∈ Z. What element of Z/pn does sn represent? (b) Let p be an odd prime. Let n 0, x≡0 and a be an integer such that 2a≡n 1. Show that p rn = 1 + p ∑ 1 k n−1 ( ) 2k (a2 x)k k satisﬁes the equation (rn )2 (1 − x)≡n 1. p For p = 2, show that this equation holds if x≡0. 8 55 Problem Set 2 2-1. Use Hensel’s Lemma to solve each of the following equations: X 2 + 6 ≡ 0; (i) 625 X + X + 8 ≡ 0.
11, the p-adic radii of convergence of the p-adic power series expp (X) = ∞ ∑ 1 n X , n! logp (X) = ∞ ∑ (−1)n−1 n=1 n=0 n Xn are p−1/(p−1) and 1 respectively. 33. There are p-adic continuous functions expp : D 0; p1/(p−1) −→ Qp and ( ) logp : D (1; 1/p) −→ Qp , where for x ∈ D 0; p1/(p−1) and y ∈ D (1; 1/p), expp (x) = logp (y) = ∞ ∑ xn n=0 ∞ ∑ n! (−1)n−1 n=1 ∞ ∑ =− ( Furthermore, for x1 , x2 ∈ D 0; p ) −1/(p−1) , n=1 (y − 1)n n (1 − y)n . n and y1 , y2 ∈ D (1; 1), expp (x1 + x2 ) = expp (x1 ) expp (x2 ), logp (y1 y2 ) = logp (y1 ) + logp (y2 ).