Algorithm for computing rational points if the rank of Jacobian is 0

Question

Is there a general algorithm that can compute in finite time all rational points on any curve of genus $g\geq 2$ whose Jacobian has rank $0$?

If not, for what special cases such algorithm is known? For genus $g=2$, we have Chabauty0 command in Magma, is it always guaranteed to work? For genus $g=3$, natural families to consider are hyperelliptic curves $y^2=P(x)$ for $P$ of degree $7$ or $8$, or Picard curves $y^3=P(x)$ for $P$ quartic. Is there an algorithm for computing rational points on them if rank if the Jacobian is $0$? Is there a corresponding Magma code?

If yes, can you provide a reference? In Cohen's book, I found the following deep theorem Demyanenko-Manin.

Let C be a curve defined over a number field K. Assume that A is a K-simple Abelian variety such that $A^m$ occurs in the decomposition of the Jacobian J of C up to isogeny over K and that $$ m > \frac{rk(A(K))}{rk(End_K(A))}, $$ where as usual rk denotes the rank. Then C(K) is finite and can be determined explicitly.

Can we derive the result from here with, say, $K={\mathbb Q}$, $m=1$, and $A=J$? Even if yes, is there a better reference, ideally book/paper where the existence of an algorithm for the rank $0$ case is stated explicitly?

Answer 1

There is an algorithm due to Bjorn Poonen (Computing torsion points on curves, Experiment. Math. 10 (2001), no.3, 449–465) that, given a (not necessarily rational) base-point $P_0$ on the curve, finds all (geometric) points $P$ on the curve such that $[P-P_0]$ is torsion in the Jacobian. This solves an even harder problem! (Although, as far as I know, this has been implemented only for curves of genus 2 with a Weierstraß point as base-point.)

Answer 2

Here is a sketch of a method that could be able to do what you want. (I won't call it an algorithm, since I cannot prove that it will always work.) I will write $X$ for the curve and $J$ for its Jacobian.

1. For a number of odd primes $p$ of good reduction for $X$, determine $J(\mathbb F_p)$ (Magma can do this). Let $e$ be the gcd of the exponents of all these (finite abelian) groups.
2. Pick a base point $P_0 \in X(\mathbb Q)$ (or show that no rational point exists, in which case we are done).
3. For odd primes $q$ of good reduction for $X$, find the subset $S_q$ of $X(\mathbb F_q)$ consisting of points $\bar{P}$ such that $e \cdot [\bar{P} - \bar{P}_0] = 0$ in $J(\mathbb F_q)$. For each $\bar{P} \in S_q$, try to find a point $P \in X(\mathbb Q)$ reducing mod $q$ to it. If this can be done for each $\bar{P}$, the resulting set of points is $X(\mathbb Q)$. Otherwise, try the next prime $q$.

If this method terminates, it gives the correct answer (the reason for this is that the reduction maps $J(\mathbb Q) \to J(\mathbb F_p)$ and $X(\mathbb Q) \to X(\mathbb F_q)$ are injective). But I have no good intuition at this point how likely it is to be successful.