Quadratic Fields and Transcendental Numbers

Mohammad Zaki, MN State Univ, Mankato We define an algebraic number α as a root of an algebraic equation, a0∗x+a1∗x+· · ·+an−1∗x+an = 0 where ao, a1, · · · , an are rational integers, not all zero. We say α is an algebraic integer if a0 = 0. If an algebraic number α satisfies an algebraic equation of degree n with rational coefficients, and none of lower degree, then we say α is of degree n. If α is an algebraic number, then we define an algebraic field as the aggregate of all numbers R(α) = P (α) Q(α) , where P,Q are polynomials with rational coefficients, and Q(α) 6= 0. We denote this field by K(α). It is easy to verify that the sum and product of any two members of K(θ) belong to K(θ). Also if α, β belong to K(θ) and β 6= 0 then αβ belongs to K(θ). If an algebraic number ε is of degree 1, then ε is a rational number, and it is plain that for any rational ε K(ε) is the aggregate of rational numbers. We denote this field by K(1). It is a part of every algebraic field. If the degree of ε is 2, then ε is said to be quadratic, and K(ε) is called a quadratic field. Since the degree of ε is 2 ε satisfies a quadratic equation, a0 ∗x +a1 ∗x+a2 = 0. so, ε = a+b∗ √ m/c for some integers a, b, c,m where m doesn’t have a squared factor. It is easily verified that K(ε) is the same as K( √ m) for some square-free rational integer m, positive or negative, apart from 1. Here we concentrate on quadratic fields. We will try to find the quadratic fields where the fundamental theorem of arithmetic holds. But before we get there we need to develop the notion of primes, and the Euclidean Property that will be described in a moment. If ε ∈ K( √ m) then ε = P ( √ m)