I have been using the Magma calculator recently, and while calculating ranks of elliptic curves with very big coefficients, there is a possibility to assume GRH is true, which signaficantly speeds up the calculation.
My question is, how is computation of the rank of an elliptic curve made faster by assuming the GRH. I am no expert in this field, so please keep your answers as simple as possible.
Computation of ranks of elliptic curves relies on descent. The first step of descent is the computation of a finite Selmer group, which in turn uses the computation of the class group of a potentially large number field. This is the step where GRH is used: it allows you to assume that the class group is generated by the set of prime ideals up to a relatively small norm bound, therefore speeding up the computation.

2$\begingroup$ I think you mean under GRH that the class group is generated by a small number of prime ideals of small norm (essentially bounded by a small power of $\log {\rm disc}(K)$, by Bach). $\endgroup$ – KConrad Jan 6 '18 at 14:32

$\begingroup$ Yes, I should probably have been more precise (I'll edit). But my goal was not to describe the full algorithm here anyway. $\endgroup$ – Aurel Jan 6 '18 at 16:26
Note: Here i present the method assuming GRH when the rank is large compared to the conductor .may this helping you .
Take $f(x)$ to be a function such that $f(0)=1$ and $f(x)\geq0$ for all real $x$
Then, assuming the Riemann hypothesis, the sum :$\sum f(\beta)$ where $1/2+i\beta$ runs over the nontrivial zeros of $L(s,E)$,will be an upper bound for the analytic rank of $E$ Moreover, for certain choices of $f(x)$ this sum may be efficiently evaluated using the explicit formula for the $L$function attached to $E$ , The method, is available as part of William Stein’s
(W. A. Stein et al., Purple PSAGE, The PSAGE Development Team, 2011,)

$\begingroup$ Well, this gives an upper bound of the rank, but how does this help compute the rank itself? $\endgroup$ – Alex M. Jan 6 '18 at 20:03

$\begingroup$ The link doesn't work. Let $\Lambda_E(p^k) = (p^k+1\#E(\mathbf{F}_{p^k})) \log p$ such that $\frac{L'}{L}(s,E) = \sum_{n=1}^\infty \Lambda_E(n) n^{s}$. Then the modularity theorem implies $\sum_\beta f(\beta)+C(f) = \sum_{n=1}^\infty \frac{\Lambda_E(n)}{n^{1/2}} (\hat{f}(\log n)+\hat{f}(\log n))$ where $C(f)$ is in term of $L(0,E)$, $\frac{\Gamma'}{\Gamma}$ and the analytic conductor. Then how do you relate $\text{rank}(E)$ to that ? @AlexM. $\endgroup$ – reuns Jan 6 '18 at 23:59

$\begingroup$ @reuns: Sorry to disappoint you, but I do not know. In fact, my comment was meant to tell Zeraoulia Rafik that I do not understand how to make use of his answer. $\endgroup$ – Alex M. Jan 7 '18 at 10:29

$\begingroup$ Concerning the software, @WilliamStein used to be active on MO, so maybe he can give a correct link. $\endgroup$ – Alex M. Jan 7 '18 at 10:37

$\begingroup$ @AlexM.I have provided the correct link , i think now it work $\endgroup$ – zeraoulia rafik Jan 7 '18 at 15:53