Let X be a smooth variety over a field k. The *index* of X over k is the gcd of the degrees [κ(x) : k] over all closed points x of X. The index is 1 if and only if X has a zero cycle of degree 1. If k is perfect, then the index of X is a birational invariant on smooth varieties over k: The reason is that given a nonempty open U of X and a closed point x in X you can find a curve C ⊂ X with x ∈ C, and it is easy to move zero cycles on curves. (I think the birational invariance also holds over nonperfect fields, but I haven’t checked this.)

Another birational invariant of a d-dimensional variety X over k is the gcd of the degrees of rational maps X —> P^d_k. This is the same as the gcd of closed subvarieties of P^n (any n) birational to X. Let’s temporarily call this the *b-index*. Note that by taking inverse images of k-rational points on P^d_k we see that index | b-index for smooth X (if k is finite you have to look at points over finite extensions). I claim that in fact index = b-index at least over a perfect field. After shrinking X we may assume that X is affine, hence quasi-projective, so X ⊂ P^N_k for some N >> 0 having some (super large and super divisible) degree D. On the other hand, consider the blow up b : X’ —> X of X in x. Then the invertible sheaf b^*O_X(N)(-Exceptional) will be very ample and will embed X’ into a large projective space where it has degree N^dD – [κ(x) : k]. This implies that b-index divides [κ(x) : k] and we win.

Dear Johan,

Is there any possibility for an analogue of Hilbert’s 10th problem for degree 1 zero cycles? In other words, could there be a finite algorithm for computing whether or not a (smooth, projective) scheme over the rationals has index 1? Could there be such a thing “modulo” the computation of finitely many “natural” cohomological obstructions (which perhaps are not effectively computable, but nonetheless satisfy some useful properties)? More generally, what if anything gets easier when trying to determine whether a variety has a degree 1 zero cycle as opposed to a rational point?

Best regards,

Jason