Euclid's Algorithm in Algebraic Function Fields

Suppose that g = 0 and that s x > 1. Then h = 1 and r x = 1; so h x = s x > 1. This settles one of the cases noted in the paper of the title (see. p. 56). Now suppose that g > 0. We show that the methods of §7 can be improved to give: if h = 1, then q ⩽ 4. We have (see, for example [ 2 ; pp. 315–322]) h = L (l) where L ( u ) = L ( q − s ) is a polynomial of degree 2 g whose zeros all have absolute value q ½ . Henceh = ∏ m = 1 g( 1 − q 1 2e i α m)( 1 − q 1 2e − i α m)= ∏ m = 1 g( 1 − 2 q 1 2cos α m + q ) ⩾( 1 − q 1 2)2 g.It follows that if h = 1, then 1 ⩾ q ½ −1 and so q ⩽ 4. The question of whether or not h x = 1 now depends on the value of r x . It seems plausible that if [ K : k ( x )] is small compared with q , then r x < h and so h x > 1. Added 3 February , 1966. On the other hand, Professor J.‐P. Serre has pointed out to me that one can always choose a finite set, S , of places and a corresponding I , which is the integral closure of some k [ x ] in K , for which h x = 1. I intend to discuss this problem in a future paper.