Algebraic number theory (Math 784) by Filaseta M.

By Filaseta M.

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4) Let β ∈ R. Prove that if N (β) is a rational prime, then β is irreducible in R. Euclidean Domains, PID’s, and UFD’s: • Definitions. Let Q(α) be an algebraic number field and R its ring of algebraic integers. The results in this section will, however, hold in the more general situation of R being a domain (a commutative ring with a non-zero multiplicative identity and having no zero divisors). We say that R is a Euclidean domain if there exists a function φ : R − {0} → Z+ such that (i) if β and γ are in R − {0} and β|γ in R, then φ(β) ≤ φ(γ); and (ii) if β and γ are in R − {0}, then there are q and r in R such that β = γq + r with either r = 0 or φ(r) < φ(γ).

B) Since 2 ∈ Q( 2 + 3), there is an h(x) ∈ Q[x] such that 2 = h( 2 + 3). Find such an h(x). √ √ √ (c) What is the field polynomial for 2 in Q( 2 + 3)? Simplify your answer. √ √ (d) Calculate NQ(√2+√3) ( 2) and T rQ(√2+√3) ( 2). (2) Prove Theorem 39. Discriminants and Integral Bases: • Definition. Let α be an algebraic number with conjugates α1 , . . , αn . Let β (1) , . . , β (n) ∈ Q(α). For each i ∈ {1, . . , n}, let hi (x) ∈ Q[x] be such that β (i) = hi (α) and (i) hi (x) ≡ 0 or deg hi ≤ n − 1.

42 Proof. Let β ∈ R − {0} with β not a unit. By Theorem 49, |N (β)| > 1. If β is irreducible, then we’re through. Otherwise, there exist γ and δ in R with γ and δ nonunits such that β = γδ. Then N (β) = N (γ)N (δ) and |N (γ)| > 1 and |N (δ)| > 1 so that 1 < |N (γ)| < |N (β)| and 1 < |N (δ)| < |N (β)|. By Theorem 39, |N (β)|, |N (γ)|, and |N (δ)| are in Z. If γ or δ is not irreducible, we may repeat the above procedure to obtain numbers in R with smaller norms in absolute value. The procedure may be repeated again and again but must eventually end resulting in a factorization of β into irreducibles.

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