Algebraic number theory (Math 784) by Filaseta M.

By Filaseta M.

Show description

Read or Download Algebraic number theory (Math 784) PDF

Best algebra books

Spinors, Clifford, and Cayley Algebras (Interdisciplinary Mathematics Series Vol 7)

Hermann R. Spinors, Clifford and Cayley algebras (Math Sci Press, 1974)(ISBN 0915692066)(600dpi)(T)(280s)_MAr_

A Course in Algebra

This quantity is predicated at the lectures given via the authors at Wuhan collage and Hubei college in classes on summary algebra. It offers the basic recommendations and simple homes of teams, jewelry, modules and fields, together with the interaction among them and different mathematical branches and utilized points.

Additional info for Algebraic number theory (Math 784)

Example text

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.

Download PDF sample

Rated 4.90 of 5 – based on 27 votes