Get Algebraic Number Theory [Course Notes] PDF

By Matthew Baker

Path Notes (Fall 2006) Math 8803, Georgia Tech, model 24 Nov 2012
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33, we must have α = ±ζpk for some integer k. We need to show that the plus sign holds. Suppose for the sake of contradiction that α = u/u = −ζpk . Then (raising both sides to the pth power) we have up = −up . 36 that there exists an integer a ∈ Z such that up ≡ a (mod p). Therefore up ≡ a (mod p) also, and we have a ≡ −a (mod p). Since p is odd, this implies that a ≡ 0 (mod p), so that p | up . This contradicts the fact that u is a unit. 38. If p is an odd prime, then every unit in Z[ζp ] can be written as rζpj for some integer 0 ≤ j ≤ p − 1, where r is a real number belonging to the field Q(ζp ).

23. Prove that two quadratic fields K, K are isomorphic if and only if their discriminants are the same. 24. In contrast to the preceding exercise, there exist non-isomorphic cubic fields with the same discriminant. On the one hand, let θ be a root of g(x) = x3 + 10x + 1, K = Q(θ). We know the discriminant of x3 + ax + b is −4a3 − 27b2 . So ∆(θ) = −4027, which is prime (and hence square-free). Thus OK = Z[θ]. It is not hard to verify that g(x) (mod 17) is irreducible so that (17) is a prime ideal in OK .

Therefore pi is a prime ideal with norm pfi . 1) ei fi = deg g(x) = n. i=1 3. THE SPLITTING OF PRIMES 41 We now prove that the pi ’s are distinct. Given i = j, we know that gi (x) and gj (x) are relatively prime in Fp [x]. So there exist a(x), b(x) ∈ Z[x] such that 1 = a(x) · gi (x) + b(x) · gj (x) in Fp [x]. But then 1 ∈ (p, gi (x), gj (x)) in Z[x], from which we deduce that 1 ∈ (p, gi (θ), gj (θ)) ⊆ OK . As (p, gi (θ), gj (θ)) ⊆ (p, gi (θ)) + (p, gj (θ)), it follows that 1 ∈ pi + pj . So pi = pj as desired.

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Algebraic Number Theory [Course Notes] by Matthew Baker


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