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By Filaseta M.

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The most function of those lectures is first to in brief survey the basic con­ nection among the illustration conception of the symmetric crew Sn and the speculation of symmetric features and moment to teach how combinatorial equipment that come up obviously within the concept of symmetric services result in effective algorithms to precise numerous prod­ ucts of representations of Sn by way of sums of irreducible representations.

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Then q prime in R and q|(a + √ 5)(a − √ Now, assume a √ 5) implies q|(a + 5) or q|(a − 5). This gives a contradiction as neither (a + 5)/q nor √ (a − 5)/q are in R (as q > 2). Thus, there are no solutions to the congruence x2 ≡ 5 (mod q). Lemma 2. Let q be a rational prime with q = 2 and q ≡ ±2 (mod 5). Then 5(q−1)/2 ≡ −1 (mod q). Proof. From Lemma 1, there are no solutions to the congruence x2 ≡ 5 (mod q). Hence, the numbers {1, 2, . . , q −1} can be paired so that each pair (x, y) satisfies xy ≡ 5 (mod q).

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). 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.

N−1 ). The determinant defining ∆(1, α, . . , αn−1 ) is called a Van der Monde determinant, and it will follow by our first lemma below that it is non-zero. This then will imply that ∆(ω (1) , . . , ω (n) ) is non-zero for some algebraic numbers ω (1) , . . , ω (n) in Q(α). Given two bases (not necessarily integral), say {ω (1) , . . , ω (n) } and {ω (1) , . . , ω (n) }, the values of ∆(ω (1) , . . , ω (n) ) and ∆(ω (1) , . . , ω (n) ) differ by the square of a rational number. If the numbers ω (1) , .

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