In his work on rational equivalence [5] Severi often raised this question: if the points of a nonsingular algebraic variety V are all rationally equivalent to each other, is F a unirational variety? A variety V is said to be unirational (over some field k) if it is the image of a projective space £ under a generically surjective rational map s: £-*V which is defined over k, of finite degree, and separable. If F is unirational, it is easily seen that all its points are rationally equivalent;

more »

... ri's question asks whether the converse is true.2 Now if it is when F is a surface, an easy but interesting consequence would be the following theorem for which we will offer a direct proof. We work always over an algebraically closed field k. Theorem. Let V be a surface over k and let V{n) denote its n-fold symmetric product. If V(n) is unirational for some n, then V is a rational surface. Before proving this, we comment on a few aspects. The theorem has birational character, so we may assume Fis a nonsingular projective surface. If V(n) is unirational, any two points are rationally equivalent and therefore any two positive 0-cycles of degree « on F are rationally equivalent. It follows easily that any two points of F are rationally equivalent and then an affirmative answer to Severi's question would imply that F is unirational. This reduces us to the case « = 1, in which case the result is a well-known consequence of the Castelnuovo-Zariski criterion for rationality (see below). In another direction, it is classical that if C is a curve, the map C(n)-*J of the symmetric product onto the Jacobian has rational varieties for its fibers. The theorem shows this cannot be true for the corresponding map V(n)->A onto the Albanese variety (as has occasionally been conjectured), since a surface having trivial Albanese variety need not be rational.