# Criterion for R-equivalence of two points on cubic surfaces over $\mathbb{Q}$

The definition of R-equivalence is given in the paper as Definition 4.1. Coarsely speaking, given a field $K$ and a cubic surface over $K$, two points $x,y$ are R-equivalent over $K$ if they can be covered by a chain of rational curves over $K$ in the cubic surface.

Question: Is there a criterion for the R-equivalence of two points on a given surface (for example, $x^3+y^3+z^3=3$ and a pair of points $(1,1,1),(17/12,5/6,-3/4)$) over $\mathbb{Q}$?

There is a general necessary condition, and then there are ad hoc sufficient conditions. The necessary condition is related to the Brauer-Manin obstruction. Via pullback of Brauer classes, there is a pairing, $$X(K)\times \text{Br}(X) \to \text{Br}(K), \ (x,\alpha) \mapsto x^*\alpha.$$ Said differently, there is a set map, $$X(K)\to \text{Hom}_{\mathbb{Z}}(\text{Br}(X),\text{Br}(K)).$$ If $x,y\in X(K)$ are $R$-equivalent, then they are in the same fiber of this set map. Thus, if there exists $\alpha\in \text{Br}(X)$ such that $x^*\alpha$ does not equal $y^*\alpha$, then $x$ and $y$ are not $R$-equivalent.
On the other hand, proving that points are $R$-equivalent is much trickier. Regarding cubic surfaces, one important reference is the following.
For trying to prove direct $R$-equivalence, the simplest guess is that there is a singular plane cubic in the cubic surface that contains $x$ and $y$. If $L$ is the line spanned by $x$ and $y$, then either $L$ is contained in $X$ (in which case you are done), or else projection of $X$ away from $L$ defines an elliptic fibration over a complementary $\mathbb{P}^1$. In this case, there is a degree $12$ divisor in $\mathbb{P}^1$ that is the discriminant of the fibration. If that degree $12$ divisor happens to have a rational point, then you are usually done: the normalization of the corresponding singular plane cubic will be what you want. The one caveat is that the singular set of that curve may contain $x$ or $y$, in which case you may not yet be done: the inverse image of that point in the normalization may be a closed point whose residue field has degree $2$ over $K$.