On integral bases in relative quadratic extensions

Abstract

Abstract. Let F be an algebraic number field and E a quadratic extension with E = F( √ µ). We describe a minimal set of elements for generating the integral elements o E of E as an o F module. A consequence of this theoretical result is an algorithm for constructing such a set. The construction yields a simple procedure for computing an integral basis of E as well. In the last section, we present examples of relative integral bases which were computed with the new algorithm and also give some running times. PreliminariesThe computation of integral bases of algebraic number fields is one of the basic tasks in computational algebraic number theory. Nonetheless, the existing algorithms for this problem tend to be very slow for fields of higher degree. In this paper, we therefore outline a new algorithm for the computation of an integral basis for those fields E which contain a subfield F of index 2.Quadratic extensions have been extensively studied in the past [8,6] and the main result of this paper is a generalization of a result of Sommer [10], who investigated biquadratic extensions.In the sequel we consider number fields F with [F : Q] = n and E subject towith an integral nonsquare element µ of F. It is well known that the ring of integers o E of E is not a free o F module, in general. The following theorem gives a necessary and sufficient criterion for the existence of a relative integral basis [1].

Arithmetic of quasi-cyclotomic fields

“…For (1), see Theorem 119 in [5], by noticing that π 2 O K and v pq O K are coprime. We now show (2). First we remark that if k < 2e, then k must be odd, since if…”

“…Notice that for a general relative quadratic extension of number fields, a relative integral basis may not exist, see [2]. But our quasi-cyclotomic fields always have relative integral bases.…”

Irreducible representations and Artin L-functions of quasi-cyclotomic fields

“…we have that 2 is ramified in K/K if and only if x 2 ≡ √ 2modπ 10 is not solvable in the ring O K of the integers of K by [2], which is equivalent to that (1 + 2 π ζ 8 )ζ 8 modπ 8 is not a square. Since 2 = uπ 4 for some unit u, we have…”