Smart elliptic curve rational point search given Reg*#Sha Hi folks,
Let E be a global minimal model of an elliptic curve over QQ, with a
2-torsion point which generates the torsion subgroup, and with
Mordell-Weil rank 1 (under BSD). Let RegSha be equal to
Regulator(E)*#Sha(E) which has been computed using the BSD formula.
(Assume computing R the non-torsion rational point generator of the
Mordell-Weil group E(QQ) using descent by 2-isogeny is time
consuming.)
Does the RegSha information make it "easier" to compute R? If so
kindly supply search strategy.
Regards,
Ifti
 A: This paper
 called "Two $p$-adic $L$-functions and rational points on elliptic curves with supersingular reduction" by Kurihara and Pollack uses another idea than Heegner points, more in the spirit of the question. However rather than taking the real valued height and the real valued $L$-function, they take the $p$-adic version. If $p$ is a prime of supersingular reduction, then there are two of both, which gives more information. In fact one finds the $p$-adic logarithm of a point of infinite order to any desired accuracy by computing the leading terms of the two $p$-adic $L$-functions. To make thing fast, they use overcongergent modular symbols to compute it. 
A: I described an algorithm of this sort in the paper listed below. The algorithm is also good at finding (or ruling out) the existence of S-integral points. However, I will note that for finding a rational point of infinite order, it is probably more efficient to compute a Heegner point.


*

*MR1627825 (99i:11043)
Silverman, Joseph H.
Computing rational points on rank 1 elliptic curves via L-series and canonical heights.
Math. Comp. 68 (1999), no. 226, 835–858


@Dror Speiser: It really doesn't matter whether one knows that Sha is finite, one simply takes the value of Regulator(E)*#Sha(E) and does the computation as if Sha(E)=1. If that doesn't give a point, then one assumes that #Sha(E)=4 and repeats. It generally won't take very many iterations to strip away #Sha(E).
A: Since I cannot (yet) post comments, I'll post this as an answer, even though it is more a comment on Joe Silverman's answer.
Let $H$ be the bound for the multiplicative height (so $H = e^h$ if $h$ is a bound for the logarithmic height) of the point you want to find. Then (as is pointed out in Joe's paper) a naive search on the elliptic curve has a complexity of roughly $H^{3/2}$, whereas his algorithm gives you $H$. On the other hand, searching for the point on a 2-covering can be done in time essentially $H^{1/2}$ (the logarithmic height goes down by a factor of 4), and if I remember correctly, this is what mwrank does. Using $n$-coverings for arbitrary $n$, the complexity drops to $H^{1/((n-1)n)}$, but of course the implied constant grows (there are roughly $n^r$ covering curves to consider when the rank is $r$, and the complexity of the lattice computations also goes up, since one has to use rank-$n$ lattices). In addition, of course, you have to add the time you need to compute the $n$-coverings in the first place, which for $n = p$ a prime usually involves computing class and unit groups of fields of degree about $p^2$. For composite $n$, the complexity tends to be better, though.
The exponent $1/((n-1)n)$ comes from two ingredients: the first is that the logarithmic height of the preimage of your point on the covering curve goes down by a factor of $1/(2n)$. The second is the lattice-based point search method mentioned by Noam Elkies in his comment that gives you a complexity of $B^{2/(n-1)}$ for a search for points up to multiplicative height $B$ on a smooth curve in ${\mathbb P}^{n-1}$. Picking a good value of $n$ should lead to something subexponential; I assume this is what Noam was having in mind (his ANTS IV paper describes a version of the lattice-based point search).
The point I want to make is that descent and searching on coverings usually gives you a faster way of finding points than Joe's method, even without knowledge of the exact canonical height (the height is still useful since it gives you a bound for the search). Tom Fisher has demonstrated that this can be very effective already for relatively small values of $n$ like 6 or 12.
