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By Matthew Baker

Direction Notes (Fall 2006) Math 8803, Georgia Tech, model 24 Nov 2012
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EXAMPLES AND COMPUTATIONAL METHODS (8) (9) (10) (11) (b) Show that √ Cl(OK ) is trivial. Let K = Q( −6). Determine which rational primes p split, ramify, and remain inert in K. Your answer should be expressed in terms of congruence conditions on p. ) Let p be a prime, and let a be a squarefree integer √ which is relatively prime to p. Let K = Q(θ), where θ = p a. Show that OK = Z[θ] if and only if ap−1 ≡ 1√ (mod p2 ). Determine the ideal class group of Z[ 3 2]. Determine √ the ideal class groups (not just their orders) of: (a) Z[√−14].

Let K, K be number fields of degrees m and m , respectively. Assume that: (i) K and K are Galois over Q (ii) K ∩ K = Q. (iii) The discriminants d and d of K and K are relatively prime. If α1 , . . , αm (resp. α1 , . . , αm ) is an integral basis for OK (resp. OK ), then αi αj (1 ≤ i ≤ m, 1 ≤ j ≤ m ) is an integral basis for OKK . 43. , the smallest subfield of L containing both K and K ). Then by Galois theory, the hypothesis K ∩ K = Q implies that: (a) KK is Galois and [KK : Q] = mm . In fact, there is a natural isomorphism Gal(KK /Q) ∼ = Gal(K/Q) × Gal(K /Q).

N . Let mj := δβj . Then since δ 2 = d, we have     m1 c1  ...  = d  ...  . mn cn For any j, we see that δ and βj are algebraic integers, and thus mj is an algebraic integer. Furthermore cj and d are both rational numbers, so dcj = mj is an algebraic integer that is also in Q. Thus dcj is in Z for all j. This proves that α is in the Z-module spanned by αd1 , . . , αdn as claimed. 1 provides another proof of the fact that if K/Q is a finite extension of degree n, then OK is a lattice in K.

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