# Rank of an elliptic curve

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?

• Which book is "my book"? – KCd Dec 30 '14 at 16:09
• @KCd My prof wrote a book and gave it to use... Do you have an idea? – evinda Dec 30 '14 at 16:12
• Later parts of Silverman's "Arithmetic of a Elliptic Curves" book discuss computing the rank. It is not clear from your question if this book is at your level or not. – KCd Dec 30 '14 at 16:15

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.
• So even if $p$ is a prime that is equivalent to $5$ modulo $8$, it does not hold that the rank is equal to $0$ ? – evinda Dec 30 '14 at 20:06
• I tried for some primes $p\equiv 5 \bmod 8$, but did not find a counterexample so far. In Silverman's book, such a result is not explicitly mentioned. – Dietrich Burde Dec 30 '14 at 20:08
• I edited my post... Does this set: $\{ \mathbb{Q}^{{\star}^{2}} , -4p^2 \mathbb{Q}^{{\star}^{2}}, 2p \mathbb{Q}^{{\star}^{2}}, -2p \mathbb{Q}^{{\star}^{2}} \}$ maybe contain two same elements modulo $\mathbb{Q}^{{\star}^{2}}$? – evinda Dec 30 '14 at 20:36