For a projective variety $X$ over an algebraically closed field of characteristic $0$, why the condition $\mathrm{CH}_0(X)\simeq \mathbb Z$ does not imply that $X$ is rationally (chain) connected ?
3 Answers
I hope that my answer will make it clearer. If $\mathrm{CH}_0X=\mathbb{Z}$, not all points are necesserily connected by rational curves. For instance, fake projective planes satisfy conditions of Bloch's conjecture, but they don't contain rational curve at all (they are quotients of the complex ball). However, points should be "connected by rational curves" in some symmetric power of $X$. I mean that for every two points $P,Q$ of $X$ there are an effective cycle $T$ of degree $n$ and a map $f:\mathbb{P}^1 \to X^{(n+1)}$, such that $f(0)=T+P$ and $f(\infty)=T+Q$. (See Mumford "Rational equivalence of 0cycles on surfaces" and Fulton "Intersection theory").
Let me give a simple example of a nonrationally connected surface with trivial Chow group. I am going to prove Bloch's conjecture for Enriques surfaces (it will be little different from the original proof by Bloch, Kas, Lieberman in "Zero cycles on surfaces with $p_g = 0$"). Let $X$ be an Enriques surface. It is wellknown that $X$ admits an elliptic fibration of index 2 and its Jacobian fibration $J$ is a rational surface [See e.g. Beaville's book on algebraic surfaces]. There is a map $\pi:X^{(2)}_{\mathbb{P}^1} \to J$ from the second relative symmetric power to the relative Jacobian with fibers $\mathbb{P}^1$. Thus, by GraberHarrisStarr theorem $X^{(2)}_{\mathbb{P}^1}$ is rationaly connected. We can map it to $X^{(2)}$ ( nonrelative symmetric power) and we can embed $X$ to $X^{(2)}$ as the diagonal. The variety $X^{(2)}_{\mathbb{P}^1}$ is a rationaly connected divisor in $X^{(2)}$ which contains the diagonal. So, points of $X$ are connected by rational curves in the second symmetric power. Precisely, for every $P,Q \in X$ we have $2P \sim 2Q$. By Roitman's theorem (proof in Bloch's lectures on cycles) $\mathrm{CH}_0X$ is torsion free. Hence, $P \sim Q$ and $\mathrm{CH}_0X= \mathbb{Z}$.
Take a surface $S$ of general type with $p_g(S)=q(S)=0$.
Then $\mathrm{Alb}(S)=0$, hence Bloch's conjecture predicts that $\mathrm{CH}_0(S)= \mathbb{Z}$. But of course $S$ is not rationally connected, because a rationally connected surface is necessarily rational.
Note that Bloch's conjecture has been proven for several classes of surfaces of general type with $p_g=q=0$, for instance it is true for Godeaux, Catanese and Barlow surfaces.
You can look for instance at the paper by Claire Voisin
Bloch’s conjecture for Catanese and Barlow surfaces, J. Differential Geom. 97 (2014), no. 1, 149175
and at the references given therein.

1$\begingroup$ Thank you for your answer (and Paranjape also). But what can be an "intuitive" interpretation of the condition $\mathrm{CH}_0(X)\simeq \mathbb Z$? $\endgroup$– pi_1Sep 13, 2016 at 14:05

1$\begingroup$ I'm afraid that there is no a really "intuitive" interpretation. As you surely know, for a surface as soon as $H^{2,0}(S) \neq 0$ one has that $\textrm{CH}_0(S)$ is infinitedimensional, but it is not intuitive (at least, not for me) that the existence of a nonzero global holomorphic $2$form forces the existence of such a huge Chow group. $\endgroup$ Sep 13, 2016 at 14:19
Bloch's conjecture (that $p_g=0$ implies $\mathrm{CH}_0(X)=\mathbb{Z}$) is known to hold for a number of surfaces of general type. These are not rationally connected. For example, see Voisin's paper on Catenese surfaces.

1$\begingroup$ I think that this paper is quite hard, especially the second part. Probably, it is not the best way to start from scratch. $\endgroup$ Sep 13, 2016 at 20:31