On Moduli in Conformal Mapping.

Geometrically, the theorem states that if a variety U/k becomes birationally equivalent to an Abelian variety over the algebraic closure of k, then it is birationally equivalent to an Abelian variety over k. (If k is a field with a discrete valuation, and ir is a prime element, then the curve defined by the equation X" + 7rY" + 7r2Zn = 0, with n = 3, shows that the conclusion of the theorem does not remain valid. In fact, if the curve has a rational point in a separable extension k1 of k, then n divides [k1: k ] because two terms of the equation must have the same absolute value in k1.) Proof of Theorem 3:4 Let V be the model of K given by Theorem 2. According to the lemma, it suffices to prove that V has a rational point. Let x be a generic point of V/k, and let x5 be obtained from x by raising all co-ordinates of x to the qth power. Then x4 is also a generic point of V/k, and (x X x2) has a locus over k which is the graph of a rational map so: V -V (onto). But over k, V is an Abelian variety, and hence so = (oo + c, where (po is an endomorphism of V and c is a constant. Hence the map x xq c is an endomorphism of V. The map x x c is an endomorphism of V, which is easily seen to have finite kernel. It is therefore onto V, and hence Sqc = -c for some t e V. Hence we have (q = i, and t is therefore a rational point. COROLLARY. If a variety Ulk becomes birationally equivalent to an Abelian variety over any extension of k (algebraic or not), then U is birationally equivalent to an Abelian variety over k. Proof: Standard specialization techniques.