Hw8-Solution1 - Math 567 Number Theory I Homework 8 Name Zhaoning Yang November 3 2013 Problem 1(Textbook P.234#9.13 Let K be an algebraic number eld

# Hw8-Solution1 - Math 567 Number Theory I Homework 8 Name...

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Math 567 Number Theory I Homework 8 Name: Zhaoning Yang November 3, 2013 Problem 1 (Textbook P.234 #9.13) Let K be an algebraic number field and O K be its ring of integers. Let m Z \{ 0 } . Prove that there exists only finitely many integral ideals of O K to which m belongs. Solution: Proof. Let K be an algebraic number field. Let I be an integral ideal of O K . SInce O K is a Dedekind domain, m I ⇐⇒ h m i ⊆ I ⇐⇒ I |h m i . So it suffice to prove there are only finitely many ideals which divides h m i . Because m Z \{ 0 } , m ∈ O K \{ 0 } . It follows that h m i is a nonzero integral ideal in O K . Then it has a unique factorization as products of prime ideals h m i = n Y i =1 p a i i where a i Z + , p i are distinct prime ideals of O K for all 1 i n . So if I is a ideal that divides h m i , from above factorization I = Q n i =1 p b i i where 0 b i a i , b i Z for all 1 i n with b i s not all zero. Since for each b i there are only finitely many choices, we can say I = Q n i =1 p b i i also has finitely many choices. Therefore there are only finitely many integral ideals that divides h m i . Problem 2 (Textbook P.234 #9.17) Let K be an algebraic number field. Let O K be its ring of integers and p Spec O K . Let a ∈ O K be such that p does not divide h a i . Prove that a N ( p ) - 1 1(mod p ). Solution: Proof. Since p is a prime ideal in O K . We know O K / p is an integral domain. Moreover, since O K is a Dedekind domain, p is also a maximal ideal of O K , hence O K / p is a field. From now on, we use the notation b denote the coset representative b + p in O K / p . Let G = ( O K / p ) * = O K / p - { 0 } be the multiplicative group of O K / p . Since p does not divide h a i , we know a / p for a p ⇐⇒ h a i ⊆ p ⇐⇒ p |h a i . Therefore a = a + p ∈ O K / p - { 0 } = ( O K / p ) * = G . From Theorem 9.1.3 since p is a nonzero ideal in O K , we know N ( p ) = [ O K : p ] Z + . It follows that | G | = |O K / p | - 1 = [ O K : p ] - 1 = N ( p ) - 1 Since a G , it follows from basic group theory that a | G | = id G . (i.e. in a finite group, the order of any elements must divide the order of the group since by Lagrange’s Theorem | G | = [ G : H ] · | H | for any H subgroup of G ) Because id G = 1 + p ∈ O K / p , and a | G | = ( a + p ) | G | = a | G | + p = a N ( p ) - 1 + p , we can say a | G | = id G ⇐⇒ a N ( p ) - 1 + p = 1 + p ⇐⇒ a N ( p ) - 1 - 1 p ⇐⇒ a N ( p ) - 1 1( mod p ) So we have shown a N ( p ) - 1 1(mod p ) as expected. 1
Problem 3 (Textbook P.234 #9.19) Determine all complex quadratic fields K for which O K possesses elements of norm 38 and trace 11. Solution: We want to find m Z square free such that when K = Q ( m ), O K has elements of norm 38 and trace 11. First, note this m must congruent 1 modulo 4, for if m 6 = 1(mod 4), then O K = Z + Z m .

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• Fall '14
• YANG
• Algebra, Number Theory, Integers, Prime number, Algebraic number theory, Integral domain, Principal ideal domain, prime ideal