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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.

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5 Answers

up vote 25 down vote accepted

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:

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.ajm/1144070587

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://math.uga.edu/~pete/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!

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Thanks very much for the answer! And you are right that I meant $k$-forms, sorry for not making this clear in my question. Ill go away and have a read of some of the references you suggested. –  Daniel Loughran Jul 29 '10 at 13:45
    
@DL: You're quite welcome. Also, in your defense, many people say "rational variety" when they mean "geometrically rational variety". For instance, Colliot-Thelene has a famous, long running seminar on "rational varieties". –  Pete L. Clark Jul 29 '10 at 16:38
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On the positive side, there are many cases where it is known that existence of a degree 1 zero cycle implies existence of a rational point. For a projective homogeneous variety over the function field of a surface (over an algebraically closed field) whose Picard group is a trivial Galois module (e.g., if the Picard group is rank 1), this was proved by de Jong, Xuhua He and myself following a suggestion of Ph. Gille. For homogeneous varieties over more general fields of cohomological dimension 2, the question whether degree 1 zero cycles imply rational points is closely related to a conjecture of Serre, cf. Theorem 3.8 of the following article of Borovoi, Colliot-Thélène and Skorobogatov. "The elementary obstruction and homogeneous spaces", Duke Math. J. 141 (2008) 321-364. http://www.math.u-psud.fr/~colliot/BoCTSk21jan08.pdf

Technically the connection above is between existence of rational points and vanishing of the "elementary obstruction". Existence of a degree 1 zero cycle implies vanishing of the "elementary obstruction". Merkurjev-Suslin proved that for every (perfect) field of cohomological dimension > 2, there exist principal homogeneous varieties for groups of type SL_n which have vanishing elementary obstruction but which have no rational point.

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Ahh thanks I was already aware of the elementry obstruction, however only that it was to do with the existence of universal torsors, it's nice to know that it is related to this as well! –  Daniel Loughran Jul 30 '10 at 11:40
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In addition to Jason's answer, I mention the following result, which I found out to be not known to experts (except Jason).

Theorem. Let $X$ be a homogeneous space of a connected linear algebraic group $G$ over a field $k$, with connected geometric stabilizers. Assume that $X$ has a zero cycle of degree 1. If $k$ is a either a $p$-adic field or a number field, then $X$ has a $k$-point.

I give a proof based on Jason's observation (actually the case of a $p$-adic field is contained in his answer) and use the paper by Borovoi, Colliot-Thélène and Skorobogatov [BCS] that Jason cites.

Proof. If $X$ has a zero cycle of degree 1, then the elementary obstruction for $X$ is 0. If $k$ is a $p$-adic field, then by [BCS], Thm. 3.3, $X$ has a $k$-point. If $k$ is a number field, then for any real place $v$ of $k$, $X$ has a zero cycle of degree 1 over $k_v$, hence $X$ has a $k_v$-point (because $k_v$ is isomorphic to $\mathbf{R}$), and by [BCS], Thm. 3.10, $X$ has a $k$-point.

Another proof of this theorem was recently obtained by Cyril Demarche and Liang Yongqi.

Note that both assumptions of the theorem, namely that geometric stabilizers are connected and that the base field $k$ is either a $p$-adic field or a number field, are important.

Mathieu Florence in the paper Zéro-cycles de degré un sur les espaces homogènes, Int. Math. Res. Not. 2004, no. 54, 2897–2914, http://alg-geo.epfl.ch/~florence/esphomog.pdf, constructed homogeneous spaces $X$ over $p$-adic and number fields with non-connected (finite) geometric stabilizers, such that $X$ has a zero cycle of degree 1, but neither $X$ nor any smooth compactification of $X$ has rational points.

Parimala in the paper Homogeneous varieties — zero cycles of degree one versus rational points, Asian J. Math. 9 (2005), 251–256, see the link in Artie's answer, constructed a projective homogeneous space $X$ (hence with connected geometric stabilizers) over the Laurent series field over a $p$-adic field, such that again $X$ has a zero cycle of degree 1, but no rational points.

Note that Jodi Black http://arxiv.org/abs/1010.1582 recently proved that if a principal homogeneous space $X$ of a connected linear group $G$ over a field $k$ of virtual cohomological dimension $\le 2$ has a zero cycle of degree 1, and $G$ satisfies the Hasse principle, then $X$ has a $k$-point.

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Even for projective homogeneous varieties the converse is not true in general, as shown in http://www.mathcs.emory.edu/~parimala/homogeneous.pdf.

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On the positive side (with no restriction on the field), I don't think anyone has mentioned quadrics. For a smooth quadric, the existence of a 0-cycle of degree one is equivalent to the existence of a point with odd degree. This implies the existence of a rational point, by a theorem of Springer.

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