By Filaseta M.

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N) (n) (n) (n) (n) (n) an1 an2 . . ann β . . βn γ . . γn 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.

We state the following without proof. √ Theorem 57. Let N be squarefree, and let R be the ring of algebraic integers in Q( N ). For N < 0, R is Euclidean if and only if N = −1, −2, −3, −7, or −11. For N > 0, R is Euclidean with the Euclidean function φ(β) = |N (β)| if and only if N = 2, 3, 5, 6, 7, 11, 13, 17, 19, 21, 29, 33, 37, 41, 55, or 73. Open Problem 1. Is there a positive integer N = 69 such that the ring R is Euclidean and N is not in the list above? Open Problem 2. Do there exist infinitely many positive integers N for which the ring R is a UFD?

N) ) = det aij ∆(1, α, . . , αn−1 ). Since |∆(ω (1) , . . , ω (n) )| > 0, we deduce that det aij = 0. Thus, {ω (1) , . . , ω (n) } is a basis for Q(α). Now, let β be an algebraic integer in Q(α). Let b1 , . . , bn be rational numbers such that β = b1 ω (1) + b2 ω (2) + · · · + bn ω (n) . We want to show that each bi is in Z. Assume otherwise so that for some k ∈ {1, 2, . . , n}, we have bk = u + θ where u ∈ Z and 0 < θ < 1. Define ω (k) = b1 ω (1) + · · · + bk−1 ω (k−1) + θω (k) + bk+1 ω (k+1) + · · · + bn ω (n) 36 and ω (j) = ω (j) for j = k with 1 ≤ j ≤ n.