Im practicing with a exercise about 2-isogenies, but struggling a bit. Im doing the following exercise:

Given two elliptic curves over $\mathbb Q$. \begin{equation}E: y^2 = x(x^2-5) \quad E':y^2 = x(x^2+20)\end{equation} These are related by a 2-isogeny: $\phi((x,y)) = (x-\frac 5x, y + \frac{5y}{x^2})$ if $x\neq 0$, $\phi((0,0)) = O$.

(a) Show that the group $E'(\mathbb Q)/\phi(E(\mathbb Q))$ has order 2.

(b) Compute $E(\mathbb Q)/\hat\phi(E'(\mathbb Q))$, and hence calculate the rank of $E(\mathbb Q)$.

This is what I have found so far:

Define a function (which is a homomorphism of groups) $q:E'(\mathbb Q) \to \mathbb Q^*/\mathbb Q ^{*^2}$ $q((u,v)) = [u]$, if $u\neq 0$, $q((0,0)) = [20], q(O) = [1]$. Then the sequence \begin{equation} E(\mathbb Q)\to^\phi E'(\mathbb Q)\to^q \mathbb Q^*/\mathbb Q ^{*^2}\end{equation} is exact.

$\textbf{Lemma.}$ Let $r$ be a squarefree integer. $[r]\in\mathbb Q^*/\mathbb Q ^{*^2}$ is in the image of $q$ if and only if \begin{equation}r^2l^4+20m^4=rn^2\end{equation} has a integer solution. This can only happen if $r|20$.

This is what i have done so far:

The square free integers dividing 20 are $r = \pm1, \pm2,\pm5,\pm10,\pm20$. One easily verifies that $r = 1$ gives rise to the solution $(1,0,1)$ and $r = 20$ to $(0,1,1)$. Easily reasoning gives that negative $r$ cannot work, as it gives a positive right hand side and negative left hand side. So we are left with $r = 1,2,5,10,20$. Do I need to prove for $r = 2,5,10$ that solutions doesn't exist? Or are there easier methods for showing only 1 and 20 satisfy this condition?

For (b), the squarefree integers dividing 5 are $r = \pm1,\pm5$. Again, $r=1,5$ give rise to solutions. For $r = -1$, we have [begin{equation} -l^4+5m^4=n^2\end{equation} which has solution $(1,1,2)$. Hence $\text{im}(q) = \{\pm1,\pm5\}$ and $E(\mathbb Q)/\hat\phi(E'(\mathbb Q))$ has order 4. Finding generators gives that it is generated by $(0,0)$ and $(-1,2)$. Is this correct? please correct me if im wrong.

And from this point, how to calculate the rank of $E(\mathbb Q)$.

Thanks a lot in advance :)


I will take for granted that $$ E (\mathbb Q)\overset\phi\longrightarrow E'(\mathbb Q)\overset q\longrightarrow \mathbb Q^\times/(\mathbb Q ^\times)^2 $$ is an exact sequence.

I will also take the lemma for granted.


Then the image in $\mathbb Q^\times/(\mathbb Q ^\times)^2 $ of the positive divisors $1,2,4,5,10,20$ of $20$ has representatives $1,2,5,10$.

We already know that $1$ and $20\equiv 5$ (modulo rational squares) are in the image.

Because of the group structure it is enough to show that $2$ is not in the image, by the lemma we consider the equation in integers $$ 4l^4+20m^4=2n^2 $$ and show it has no solutions. If $l\ne 0$ modulo $5$, then also $n\ne 0$ modulo $5$, and it follows that $2$ is a quadratic residue modulo $5$, contradiction. So $l=5l_1$, so $n=5n_1$ and we get $$ 4\cdot 5^3l_1^4+4m^4=2\cdot 5n_1^2\ . $$ So $m=5m_1$, and we can further write $$ 4\cdot 5^3l_1^4+4\cdot 5^4m_1^4=2\cdot 5n_1^2\ . $$ So $n_1=5n_2$, and we can further write $$ 4\cdot 5^3l_1^4+4\cdot 5^4m_1^4=2\cdot 5^3n_2^2\ . $$ Simplifying with $5^3$ delivers a smaller solution. We could have started with a minimal solution, or here we can invoke an infinite descent, thus getting a contradiction.

We have so far: $$ \begin{aligned} E'(\Bbb Q)/\phi E(\Bbb Q) &\overset\cong{\underset q\longrightarrow} \operatorname{Image} (q) = \langle 5\rangle\subseteq \mathbb Q^\times/(\mathbb Q ^\times)^2 \\\\ &\text{has one generator and two elements.} \end{aligned} $$


It is ok, we have on $E$ the $\Bbb Q$-rational points $(0,0)$, a torsion point, mapping to $5$, and $(-1,2)$, mapping to $-1$, so if the Lemma also aplies for the dual map we have: $$ \begin{aligned} E(\Bbb Q)/\hat\phi E'(\Bbb Q) &\overset\cong{\underset {q'}\longrightarrow} \operatorname{Image} (q') = \langle -1,5\rangle\subseteq \mathbb Q^\times/(\mathbb Q ^\times)^2 \\\\ &\text{has two generators and four elements.} \end{aligned} $$ Consider now the composition of the two isogenies, $[2]=[2]_E=\hat\phi\circ\phi$, which leads to the following exact diagram: $\require{AMScd}$ \begin{CD} @. 0 @. 0 @. 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> \hat\phi\phi E(\Bbb Q) @>>> \hat \phi E'(\Bbb Q) @>>> \boxed{0} @>>> 0\\ @. @| @VVV @VVV \\ 0 @>>> [2] E(\Bbb Q) @>>> E (\Bbb Q) @>>> (?) @>>> 0\\ @. @VVV @VVV @VV{\cong}V \\ 0 @>>> 0 @>>> \langle 5,-1\rangle @= \langle 5,-1 \rangle @>>> 0\\ @. @VVV @VVV @VVV \\ @. 0 @. 0 @. 0 \end{CD} (The left upper square is made of inclusions.)

The entry $\boxed{0}$ in the first horizontal short exact row is indeed vanishing, because we know that $(0,0)$ generates $E'(\Bbb Q)$ modulo the image of $\phi$ (because it is mapped by $\phi$ to a non-trivial element), but we pass to the image via $\hat \phi$, and this torsion point is mapped to the zero element $\infty\in E(\Bbb Q)$.

It follows that the $(?)$ entry is also a vector space of dimension two over $\Bbb F_2=\Bbb Z/2$. One part of it comes from the $2$-torsion element $(0,0)$ of $E(\Bbb Q)$, and this is the only $2$-torsion part. So the rest comes from the generator of $E(\Bbb Q)$ modulo torsion, the preimage of $-1$ is such a generator, so $(-1,2)\in E(\Bbb Q)$.


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