By Matthew Baker

Direction Notes (Fall 2006) Math 8803, Georgia Tech, model 24 Nov 2012

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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 ). Proof. Let u be a unit in Z[ζp ]. 37, we know that u/u = ζpk for some integer k. Choose an integer 0 ≤ j ≤ p − 1 such that 2j ≡ k (mod p). Then uζp−j = uζpj . , r ∈ R. Therefore u = rζpj with r ∈ R, as desired. We now turn to the “first case” of Fermat’s Last Theorem for regular primes.

To see this, first note that there is a (non-canonical) group homomorphism φ : OK → pk /pk+1 given by choosing an element γ ∈ pk \pk+1 and sending x ∈ OK to γx. (We know that pk = pk+1 by unique factorization, so such an element γ certainly exists). If x ∈ p, then γx ∈ pk+1 , so φ induces a map (which we continue to call φ) φ : OK /p → pk /pk+1 . We will show that this map is an isomorphism. Claim 1: (γ) + pk+1 = pk . Given Claim 1, it follows immediately that φ is surjective. Claim 2: (γ) ∩ pk+1 = γp.

N } = {αω1 , . . , αωn } clearly forms a Z-basis for a. Write α1 ω1 ... = A ... αn ωn with A an invertible n × n matrix with integer coefficients. 11, N (a) = | det(A)|. We also have ∆(α1 , . . , αn ) = det(A)2 ∆(ω1 , . . , ωn ) by linear algebra. 3. THE IDEAL CLASS GROUP 25 On the other hand, it follows from the definition of the discriminant that ∆(αω1 , . . , αωn ) is the square of the determinant of the matrix σ1 (αω1 ) · · · σ1 (αωn ) .. ... , . σn (αω1 ) · · · σn (αωn ) This determinant is σ1 (α) · · · σn (α) times the determinant of the matrix (σi (ωj )).