Let $k$ be a field (I'm mainly interested in the case where $k$ is a number field, however results for other fields would be interesting), and $X$ a smooth projective variety over $k$. 


By a zero cycle on $X$ over $k$ I mean a formal sum of finitely many (geometric) points on $X$, which is fixed under the action of the absolute Galois group of $k$. We can define the degree of a zero cycle to be the sum of the multiplicities of the points. 


Now, if $X$ contains a $k$-rational point then it is clear that $X$ contains a zero cycle of degree one over $k$. 


What is known in general about the converse? That is, which classes of varieties are known to satisfy the property that the existence of a zero cycle of degree one over $k$ implies the existence of a $k$-rational point? For example what about rational varieties and abelian varieties?


As motivation I shall briefly mention that the case of curves is easy. Since here zero cycles are the same as divisors we can use Riemann-Roch to show that the converse result holds if the genus of the curve is zero or one, and there are plently of counter-examples for curves of higher genus. However in higher dimensions this kind of cohomological argument seems to fail as we don't (to my knowledge) have such tools available to us.



Existence of zero cycles of degree one vs existence of rational points
There has been a lot of work on this problem, although nothing like a general answer is known. By way of abbreviation, the **index** of a nonsingular projective variety is the least positive degree of a $k$-rational zero cycle, so you are asking about the relationship between index one and having a $k$-rational point.


First, you ask whether rational varieties and abelian varieties with index one must have a rational point. Here you probably mean $k$-forms of such things: i.e., *geometrically* rational varieties and *torsors* under abelian varieties. (Both rational varieties and abelian varieties have rational points, the latter by definition, the former e.g. by the theorem of Lang-Nishimura which says that having rational points is a birational invariant of a nonsingular projective variety.) I can answer this:


1. A torsor under an abelian variety has index one iff it has a rational point. This follows from the cohomological interpretation of torsors as elements of $H^1(k,A)$.
2. A geometrically rational surface of index one need not have a rational point: this is a theorem of Colliot-Thelene and Coray. (A reference appears in the link below.)


On to the general question. A very nice recent paper which proves a big result of this type and gives useful bibliographic information about other results is Parimala's 2005 paper on homogeneous varieties:


[Link](https://projecteuclid.org/journals/asian-journal-of-mathematics/volume-9/issue-2/Homogeneous-varieties---zero-cycles-of-degree-one-versus/ajm/1144070587.full)


Finally, there are some fields $k$ for which every geometrically irreducible projective variety has index one -- most notably finite fields. In this case any variety without a rational point over such a field gives a counterexample to "index one implies rational point". For instance, for any finite field $\mathbb{F}\_q$ and all sufficiently large $g$, one can easily write down a hyperelliptic cuve over $\mathbb{F}\_q$ of genus $g$ without rational points. [N.B.: What I had written before was too strong: if instead you fix $g$ and let $q$ be sufficiently large, then by the Weil bounds you must have a rational point.] There are also K3 surfaces over finite fields without rational points, and so forth.


Some further discussion of fields over which every (geometrically irreducible) variety has index one occurs in the appendix of a recent paper of mine:


[http://alpha.math.uga.edu/~pete/trans.pdf](http://alpha.math.uga.edu/%7Epete/trans.pdf)


There are many more results than the ones I've mentioned so far. If you have further questions, please don't hesitate to ask!


