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Q(α, β) is a field since it is the intersection of all the subfields of C containing Q, α, and β. The intersection of a finite number of subfields in a fixed field is again a field. 2 (Theorem of the Primitive Element) If α and β are algebraic numbers, then ∃ θ, an algebraic number, such that Q(α, β) = Q(θ). Proof. Let f be the minimal polynomial of α and let g be the minimal polynomial of β. We want to show that we can find λ ∈ Q such that θ = α + λβ and Q(α, β) = Q(θ). We will denote Q(θ) by L. Clearly L = Q(θ) ⊆ Q(α, β).

N is an integral basis for K. 5 Show that the discriminant is well-defined. In other words, show that given ω1 , ω2 , . . , ωn and θ1 , θ2 , . . , θn , two integral bases for K, we get the same discriminant for K. We can generalize the notion of a discriminant for arbitrary elements of K. Let K/Q be an algebraic number field, a finite extension of Q of degree n. Let σ1 , σ2 , . . , σn be the embeddings of K. For a1 , a2 , . . , an ∈ K we 2 can define dK/Q (a1 , . . , an ) = det(σi (aj )) . 6 Show that 2 dK/Q (1, a, .

Considering this equation mod λ3 , and recalling that n ≥ 2, we get that ±1 ± e4 ≡ 0 (mod λ3 ) so e4 = ∓1, and we rewrite our equation as C33 + (∓C2 )3 + e5 λ3(n−1) C13 = 0. 4. 1) with n = m, then there exists a solution with n = m − 1. 1) is possible in Z[ρ]. 1 Solve the equation y 2 + 4 = x3 for integers x, y. Solution. We first consider the case where y is even. It follows that x must also be even, which implies that x3 ≡ 0 (mod 8). Now, y is congruent to 0 or 2 (mod 4). If y ≡ 0 (mod 4), then y 2 + 4 ≡ 4 (mod 8), so we can rule out this case.

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