Rank of an elliptic curve

Question

How could we compute the rank of an elliptic curve? I looked for a methodology in my book, but I didn't find anything. Could you give me a hint?

I want to find the rank of the curve $Y^2=X^3+p^2X$ with $p \equiv 5 \pmod 8$.

EDIT:

$$E|_{\mathbb{Q}}: y^2=x^3+p^2x \ \ \ \ \ \ \ \ \ \ a=0, b=p^2$$ $$\overline{E}|_{\mathbb{Q}}: y^2=x^3-4p^2x \ \ \ \ \ \ -2a=0, a^2-4ab=-4p^2$$ $$\Gamma:E(\mathbb{Q})$$ $$\overline{\Gamma}:\overline{E}(\mathbb{Q})$$ $$r=rang(E(\mathbb{Q}))$$ $$2^r=\frac{|\alpha\Gamma||\overline{\alpha} \overline{\Gamma}|}{4}$$

From Tate theorem, we have: $$\alpha\Gamma=\{ \mathbb{Q}^{{\star}^{2}}, b\mathbb{Q}^{{\star}^{2}}\} \cup \{ b_1 \mathbb{Q}^{{\star}^{2}}, \text{ where } b1 \mid b, b=b_1b_2 \text{ and } z^2=b_1x^4+ax^2y^2+b_2y^4 \text{ has a solution in } \mathbb{Z} \text{ with } xy \neq 0\}$$

Since $b=p^2, b_1$ can be the following: $\pm 1, \pm p, \pm p^2$, but since $\pm p^2= \pm 1 \pmod { \mathbb{Q}^{\star^2}}$, $b_1=\pm 1, \pm p$, so we have to check if the following equations are solvable in $\mathbb{Z}$:

$$z^2=x^4+p^2y^4$$ $$z^2=-x^4-p^2y^4$$ $$z^2=px^4+py^4$$ $$z^2=-px^4-py^4$$

The equations are not solvable in $\mathbb{Z}$ so $\alpha \Gamma=\{ \mathbb{Q}^{{\star}^{2}}, p^2\mathbb{Q}^{{\star}^{2}} \}=\{ \mathbb{Q}^{{\star}^{2}} \} \Rightarrow |\alpha \Gamma|=1$

$$\overline{b_1} \mid \overline{b}$$

For $\overline{b_1}$ there are the following possible values $\pm 1, \pm 2, \pm 4, \pm p, \pm p^2, \pm 2p, \pm 4p, \pm 2p^2, \pm 4p^2$.

But since $\pm 4 = \pm 1 \pmod {\mathbb{Q}^{\star ^2}}, \\ p^2 = \pm 1 \pmod {\mathbb{Q}^{\star ^2}}, \\ 4p = \pm 2p \pmod {\mathbb{Q}^{\star ^2}}, \\ 2p^2 = \pm 2p \pmod {\mathbb{Q}^{\star ^2}}, \\ 4p^2 = \pm 1 \pmod {\mathbb{Q}^{\star ^2}}$

$b_1 \ : \ \pm 1, \pm 2, \pm p, \pm 2p$

So, we have to check if the following equations have a solution in $\mathbb{Z}$ with $xy \neq 0$:

$$\\z^2=x^4-4p^2y^4\\z^2=-x^4+4p^2y^4\\z^2=2x^4-2p^2y^4\\z^2=-2x^4+2p^2y^4\\z^2=px^4-4py^4\\z^2=-px^4+4py^4\\ z^2=2px^4-2py^4\\z^2=-2px^4+2py^4$$

I think that the only two equations that have a solution in $\mathbb{Z}$ with $x \cdot y \neq 0$ are these: $z^2=2px^4-2py^4$ and $z^2=-2px^4+2py^4$.

Am I right?

So, $\overline{\alpha} \overline{\Gamma}=\{ \mathbb{Q}^{{\star}^{2}} , -4p^2 \mathbb{Q}^{{\star}^{2}}, 2p \mathbb{Q}^{{\star}^{2}}, -2p \mathbb{Q}^{{\star}^{2}} \} = \{ \mathbb{Q}^{{\star}^{2}} , -\mathbb{Q}^{{\star}^{2}}, 2p \mathbb{Q}^{{\star}^{2}}, -2p \mathbb{Q}^{{\star}^{2}} \} \Rightarrow |\overline{\alpha} \overline{\Gamma}|=4$

Therefore:

$$2^r=\frac{1 \cdot 4}{4}=1 \Rightarrow r=0$$

Have I done something wrong?

Answer 1

As far as I know, there is no general algorithm to compute the rank of an elliptic curve. It is linked to the Conjecture of Swinnerton-Dyer, one of the ‘Problems of the Millenium’. If you can read french, you can look at this popularisation paper.

Answer 2

I assume that you consider the elliptic curve over $\mathbb{Q}$. There is no general algorithm known for computing the rank, which makes the problem difficult and interesting. However, there are several CAS which have algorithms implemented to compute the analytic rank. By the BSD-conjecture, this is equal to the geometric rank.
For the curve $y^2=x^3+p^2x$ I have computed for you the ranks, see here. It is known that if the analytic rank is zero, then so is the (geometric) rank.

And yes, Silverman's book is definitely the book you should have a look. It also helps for your other questions.