Power integral bases in a parametric family of totally real cyclic quintics

We consider the totally real cyclic quintic fields Kn = Q(θn), generated by a root θn of the polynomial fn(x) = x 5 + nx − (2n + 6n + 10n + 10)x + (n + 5n + 11n + 15n + 5)x + (n + 4n + 10n + 10)x + 1. Assuming that m = n4 + 5n3 + 15n2 + 25n + 25 is square free, we compute explicitly an integral basis and a set of fundamental units of Kn and prove that Kn has a power integral basis only for n = −1,−2. For n = −1,−2 (both values presenting the same field) all generators of power integral bases are computed. Introduction Let n ∈ Z and denote by θn a root of the polynomial fn(x) = x 5 + nx − (2n + 6n + 10n+ 10)x + (n + 5n + 11n + 15n+ 5)x + (n + 4n + 10n+ 10)x+ 1. (1) These polynomials were discussed by Emma Lehmer (cf. [8]). The corresponding parametric family of cyclic quintic fields Kn = Q(θn), θn a root of fn, was also investigated by Schoof and Washington [10] and Darmon [2] for prime conductors m = n + 5n + 15n + 25n+ 25. Assuming only that m = n+5n+15n+25n+25 is square free, we describe an integral basis and a set of fundamental units of the field Kn = Q(θn). We construct explicitly the index form corresponding to that integral basis. The coefficients in the integral basis of those ξ ∈ Kn generating a power integral basis {1, ξ, ξ, ξ, ξ4} of Kn can be obtained as solutions of the index form equation. The index form equation reduces to a unit equation in two variables over Kn. By using congruence considerations modulo m we show, that this unit equation is only solvable for n = −1,−2. For n = −1,−2 the fields Kn coincide. This field indeed admits power integral bases, generated e.g. by the roots of the polynomial fn. In fact in this case there Received by the editor August 13, 1996. 1991 Mathematics Subject Classification. Primary 11Y50; Secondary 11Y40, 11D57.