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Normes (detailed proof) #ALNT-LB 1.1.DP Chapter 1 Section 1 #Algebraic Number Theory # Lecture note Benois
The proof of p-adic norm is non-Archimedeam: write x and y in the general norm.
The proof of Proposition 1.1: Step (a) ||x+y|| leq 2 max {||x||, ||y||}; Step (b) ||sum x_i|| leq 4n max {||x_i||}; Step (c) Use Newton-expansion and use (a) as an approximation of triangular inequality.
The proof of Proposition 1.2: Use Newton-expansion and triangular inequality.
The proof of Proposition 1.4: For (2) => (3): there exists ||x_0||_1 >1, then for any x, there exists alpha such that ||x||_1 = ||x_0||_1^alpha . The condition (2) can be used as a bridge between ||-||_1 and ||-||_2 and one can show that ||x||_2 = ||x_0||_1^alpha. Now define lambda to be the real number such that ||x_0||_2=||x_0||_1^alpha and conclude easily ||x||_1^lambda=||x||_2.
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By LucThe proof of p-adic norm is non-Archimedeam: write x and y in the general norm.
The proof of Proposition 1.1: Step (a) ||x+y|| leq 2 max {||x||, ||y||}; Step (b) ||sum x_i|| leq 4n max {||x_i||}; Step (c) Use Newton-expansion and use (a) as an approximation of triangular inequality.
The proof of Proposition 1.2: Use Newton-expansion and triangular inequality.
The proof of Proposition 1.4: For (2) => (3): there exists ||x_0||_1 >1, then for any x, there exists alpha such that ||x||_1 = ||x_0||_1^alpha . The condition (2) can be used as a bridge between ||-||_1 and ||-||_2 and one can show that ||x||_2 = ||x_0||_1^alpha. Now define lambda to be the real number such that ||x_0||_2=||x_0||_1^alpha and conclude easily ||x||_1^lambda=||x||_2.