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Share Ostrowski's theorem (sketching proof) #ALNT-LB 1.2.SP Chapter 1 Section 2 #Algebraic Number Theory # Lecture note Benois
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Ostrowski's theorem (sketching proof) #ALNT-LB 1.2.SP Chapter 1 Section 2 #Algebraic Number Theory # Lecture note Benois
Theorem (Ostrowski) Every nontrivial norm in Q is equivalent to ||-||_p for some prime p, or equivalent to |-|.
Proof:
Case (A): Assume that ||n|| leq 1 for all natural numbers n， then we will show that for any x in Q, ||x||=||x||^lambda for some positive real number lambda, hence it is equivalent to some ||-||_p.
Case (B): Assume that there exists some natural number n such that ||n||>1, then we will show that for any natural number n, ||n||=n^s for some positive real s, hence it is equivalent to the absolute value (norm) |-|.
Proposition 2.2 Prod_{p leq +infty} ||x||_p = 1.
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By LucTheorem (Ostrowski) Every nontrivial norm in Q is equivalent to ||-||_p for some prime p, or equivalent to |-|.
Proof:
Case (A): Assume that ||n|| leq 1 for all natural numbers n， then we will show that for any x in Q, ||x||=||x||^lambda for some positive real number lambda, hence it is equivalent to some ||-||_p.
Case (B): Assume that there exists some natural number n such that ||n||>1, then we will show that for any natural number n, ||n||=n^s for some positive real s, hence it is equivalent to the absolute value (norm) |-|.
Proposition 2.2 Prod_{p leq +infty} ||x||_p = 1.