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Algebraic number theory (Math 784) by Filaseta M.

By Filaseta M.

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Qsfs can be expressed as a sum of two squares. It suffices, therefore, to prove that for any integers x1 , y1 , x2 , and y2 , there exist integers x3 and y3 satisfying (x21 + y12 )(x22 + y22 ) = x23 + y32 . This easily follows by setting x3 + iy3 = (x1 + iy1 )(x2 + iy2 ) and taking norms. Homework: (1) Suppose n is a positive integer expressed in the form given in Theorem 60 with each fj even. Let r(n) denote the number of pairs (x, y) with x and y in Z and n = x2 + y 2 . Find a formula for r(n) that depends only on t and r.

Det  .. . (n) (n) β1 β2 ... (i) (j)   (1) (1) β1 βn  .  .  det  .. (n) (1) βn βn (2) β1 .. (2) βn ... .  (n) β1 ..  .  (n) βn (i) (j) = det β1 β1 + β2 β2 + · · · + βn(i) βn(j) = det T r(β (i) β (j) ) , where the last equation follows from an application of the last lemma. • Integral bases. The numbers 1, α, α2 , · · · , αn−1 form a basis for Q(α) over Q. It follows that every bases for Q(α) over Q consists of n elements. Let R be the ring of algebraic integers in Q(α).

C) What is the field polynomial for 2 in Q( 2 + 3)? Simplify your answer. √ √ (d) Calculate NQ(√2+√3) ( 2) and T rQ(√2+√3) ( 2). (2) Prove Theorem 39. Discriminants and Integral Bases: • Definition. Let α be an algebraic number with conjugates α1 , . . , αn . Let β (1) , . . , β (n) ∈ Q(α). For each i ∈ {1, . . , n}, let hi (x) ∈ Q[x] be such that β (i) = hi (α) and (i) hi (x) ≡ 0 or deg hi ≤ n − 1. For each i and j in {1, . . , n}, let βj = hi (αj ). The discriminant of β (1) , . .

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