Filaseta M.'s Algebraic number theory (Math 784) PDF

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This contradicts the minimality of |∆(ω (1) , . . , ω (n) )|, completing the proof. Homework: (1) Let ω (1) , . . , ω (n) be an integral basis in Q(α). Prove that |∆(ω (1) , . . , ω (n) )| is > 0 and as small as possible. (2) Compute ∆(1, α) where α is a root of ax2 + bx + c = 0 where a, b, and c are in Z and α is irrational. Comments and Definitions: By the first problem above, it follows that the discriminants of any two integral bases for a given number field Q(α) have the same absolute value.

R be any algebraic numbers. Then there exists an algebraic number γ such that Q(γ) = Q(α1 , α2 , . . , αr ). Proof. It suffices to show that if α and β are algebraic, then there exists an algebraic number γ for which Q(γ) = Q(α, β). Let α1 = α and α2 , . . , αn be the distinct roots of the minimal polynomial f (x) for α; and let β1 = β and β2 , . . , βm be the distinct roots of the minimal polynomial g(x) for β. Note that for i ∈ {1, 2, . . , n} and j ∈ {2, 3, . . , m}, there exists a unique x = x(i, j) such that αi + xβj = α + xβ.

G(αn ) are the field conjuSet g(x) = gates of β, and h(α1 ), . . , h(αn ) are the field conjugates of γ. Let w(x) = g(x)h(x) ∈ Q[x] so that βγ = w(α). Then the last lemma implies N (βγ) = w(α1 ) · · · w(αn ) = g(α1 )h(α1 ) · · · g(αn )h(αn ) = N (β)N (γ), completing the proof. 34 Theorem 39. Let β ∈ Q(α). Then N (β) ∈ Q and T r(β) ∈ Q. If β is an algebraic integer, then N (β) ∈ Z and T r(β) ∈ Z. Homework: √ √ √ √ (1) (a) Prove√that Q(√ 2, √3) = Q( 2 + 3). √ √ √ (b) Since 2 ∈ Q( 2 + 3), there is an h(x) ∈ Q[x] such that 2 = h( 2 + 3).

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Algebraic number theory (Math 784) by Filaseta M.


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