I have been trying to learn the method of Chabauty and Coleman to find rational points on curves; I have been reading an exposition by McCallum and Poonen which was pointed out to me by Emerton in [this question](https://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture). 


Let $X$ be a curve of genus $g$ over $\mathbb{Q}$ with jacobian variety $J$, let $p$ be a prime of good reduction, and let $\overline{J(\mathbb{Q})}$ be the $p$-adic closure of the Mordell-Weil group $J(\mathbb{Q})$ in $J(\mathbb{Q}\_p)$. Denote by $r'$ the dimension of the $p$-adic manifold $\overline{J(\mathbb{Q})}$. 


The main assumption of the approach is that $r' < g$. This is automatic if $r < g$, where $r$ is the rank of $J$, because in general one has $r' \leq r$. This last inequality needn't be equality, "since $\mathbb{Z}$-independent points in log $J(\mathbb{Q})$ need not be $\mathbb{Z}\_p$-independent". 



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> How do I compute $r'$?
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I wrote down a toy example, that is, $X : y^2 = x^5 + 17$. Here $r = 2$, and the method might work if $r'$ was 0 or 1, but I don't know how to check this. 


I suspect that $r' = 2$, in which case the method is not even applicable, and I must think harder, but my question is not about this example, rather the general approach. 



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> Is there an example of a curve $X$ with $r = g = 2$ but with $r' = 0$ or 1?
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Rational points à la Chabauty-Coleman
If $r=2$ then $r'>0$. For an example where $r'=1$, take a curve such that the jacobian has a nontrivial endomorphism $f$
and such that the group of points in the jacobian is generated by $P,f(P)$ for some point $P$
Now find a prime $p$ splitting in $\mathbb{Q}(f)$ so that $f(P)=\alpha P$ for some $p$-adic number $\alpha$.
Then $r'=1$.


Added later: In your toy example, the endomorphism ring contains the fifth roots of unity so it may fall in my example above for those primes that split in that ring. A reasonable conjecture would be that if the endomorphism ring of the jacobian is $\mathbb{Z}$, then $r' = \min \{ g,r \}$. This is likely to be very hard to prove, as it is an abelian variety analogue of Leopoldt's conjecture.


Added much, much later: There are problems with my construction and what I say is probably not right. M.D. had a good reason to be skeptical in the comments. This was pointed out to me by Bjorn Poonen in a recent email exchange. His question is discussed further in a paper of M. Waldschmidt On the p-adic closure of a subgroup of rational points on an Abelian variety, Afr. Mat. 22 (2011), no. 1, 79–89.


