Why is $2A_0(X)=0$ for a cubic threefold $X$ containing a line, over an arbitrary field $k$

Question

I can't quite follow Proposition $2.1$ of "UNIVERSAL UNRAMIFIED COHOMOLOGY OF CUBIC FOURFOLDS CONTAINING A PLANE". I posted this on Math stackexchange but got no answer.

Let $X$ be a smooth cubic threefold over a field $k$, and let $\sim$ denote rational equivalence of cycles. Let $A_0(X)$ denote the group of $0$-cycles algebraically equivalent to $0$ modulo rational equivalence. It is claimed in the above that $2A_0(X) = 0$. It suffices to show for points $p,q$ on $X$ that $2(p−q)\sim 0$. If $p,q$ lie on a line, we are done. In general, we can reduce to the case where $p$ lies on the line $L$. Let $P$ be the $2$-plane spanned by $L$ and $q$. Then $P=L\cup C$ for some conic $C$ containing a point $q$. Thus $C$ is rational, unless it is the union of two lines $L_1\cup L_2$ over some finite extension of $k$.

In any case $p \sim p′$ for some point $p′$ which lies on $C\cap L$. Then $p′\sim q$ in the case that $C$ is geometrically irreducible. So the only issue is if $C$ is a union of two lines? Then:

1. How does one conclude in this case that $2(p′−q)\sim 0$
2. Generically in the space of $2$-planes through $L$, the situation is that $C$ is geometrically irreducible, so can we not always reduce to this case?

Further, this case is only an issue when the two lines are non-split. Thus the Galois action of the degree two extension $L/k$ which splits the lines acts by swapping the lines. Therefore it is only an issue when $q=L_1\cap L_2$. Let $\pi:Bl_L(X)→\mathbb{P}^2$ be the usual conic bundle with discriminant $\Delta$. Then does the section $\Delta\to X$ satisfy that $A_0(\Delta)→A_0(X)$ is surjective?

The motivation for this question i that by work of Colliot-Thelene, to show that $X$ is universally $Ch_0$-trivial, it suffices to show that $A_0(X) = 0$ and that there is a curve $C \to X$ so that $A_0(C_F) \to A_0(X_F)$ is surjective for all fields $F$ containing $k$.

Answer 1

If $C$ is the union of two lines defined over a field $K$, associated to each point of one of the two lines over $K$ we get a degree $2$ divisor on $C$ by taking the sum of that point and its Galois conjugate. The space parameterizing such divisors over $k$ is the Weil restriction from $K$ to $k$ of $\mathbb P^1$, which is a rational surface and in particular rationally connected, so any two divisors of this form are equivalent.

One divisor of this form is $2q$, taking the original point to be the intersection point, and another lies on $L$, taking the original point to be the intersection of one of the lines with $L$, and is thus rationally equivalent to $2p$.

So indeed $2p$ and $2q$ are rationally equivalent. This answers question 1, since as you point out this is the only remaining case.

I am not sure about your other questions.