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By Filaseta M.

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R ) defined as the smallest field containing Q and some algebraic numbers α1 , α2 , . . , αr . Observe that Q(α1 , α2 , . . , αr ) = Q(α1 , α2 , . . , αr−1 )(αr ), the smallest field which contains Q(α1 , α2 , . . , αr−1 ) and αr (this equality should be justified). We show that in fact such a field is an algebraic number field. Theorem 36. Let α1 , α2 , . . , αr 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).

P−1}, T r(ζ j ) = −1 (since ζ j has minimal polynomial Φp (x)). Hence, T r(βζ −k − βζ) = T r(a0 ζ −k + · · · + ak + · · · + ap−2 ζ p−k−2 − a0 ζ − · · · − ap−2 ζ p−1 ) = (p − 1)ak − a0 − · · · − ak−1 − ak+1 − · · · − ap−2 + a0 + · · · + ap−2 = pak . Hence, pak ∈ Z for all k ∈ {0, 1, . . , p − 2}. Let λ = 1 − ζ. Then p−2 (∗) p−2 pβ = k (pak )ζ = k=0 p−2 (pak )(1 − λ) = k cj λj , j=0 k=0 where for each j ∈ {0, 1, . . , p − 2} we have p−2 (−1)j cj = k=j k pak ∈ Z. j Also, since λ = 1 − ζ, an analogous argument gives that for each j ∈ {0, 1, .

N ) = 0. On the other hand, by the lemma to Theorem 41, we deduce ∆(β1 , β2 , . . , βn ) = a211 a222 · · · a2nn ∆(ω1 , ω2 , . . , ωn ) = 0. This contradiction implies the uniqueness condition in the theorem holds. Comment: Given the condition β1 , β2 , . .

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