By Filaseta M.

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**Sample text**

N β γ 1 2 1 2 Taking determinants and squaring, the result follows. Theorem 41. Let Q(α) be an algebraic extension of Q of degree n. Let ω (1) , . . , ω (n) be n algebraic integers in Q(α) with |∆(ω (1) , . . , ω (n) )| > 0 as small as possible. Then ω (1) , . . , ω (n) form an integral basis in Q(α). Proof. First, we show that ω (1) , . . , ω (n) form a basis for Q(α). To do this, it suffices to show that det aij = 0 where the numbers aij are the rational numbers uniquely determined by the equations n ω (i) aij αj−1 = for 1 ≤ i ≤ n.

N) is defined by (i) 2 ∆(β (1) , . . , β (n) ) = det(βj ) . Observe that the ordering of the conjugates α1 , . . , αn of α as well as the ordering of β (1) , . . , β (n) does not affect the value of the discriminant. On the other hand, the ordering (1) on the conjugates of the β (i) is important (if βj = h1 (αj ), then we want the jth conjugate of each β (i) to be determined by plugging in αj into hi (x)). Theorem 40. If β (1) , . . , β (n) ∈ Q(α), then ∆(β (1) , . . , β (n) ) ∈ Q. If β (1) , .

Ann β (n) γ (n) where the aij are rational numbers. Then 2 ∆(β (1) , . . , β (n) ) = det aij ∆(γ (1) , . . , γ (n) ). Proof. For i ∈ {1, 2, . . , n}, let hi (x) ∈ Q[x] denote the polynomial of degree ≤ n − 1 such that γ (i) = hi (α). Then the matrix equation implies that β (i) = gi (α) where gi (x) = ai1 h1 (x) + · · · + ain hn (x) ∈ Q[x] for 1 ≤ i ≤ n. It follows that (1) (1) (1) γ (1) γ (1) . . γ (1) β2 . . βn β1 a11 a12 . . a1n n 1 2 β (2) β (2) . . β (2) a21 a22 . .

### Algebraic number theory (Math 784) by Filaseta M.

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