# Descent by 2-isogeny

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

Given two elliptic curves over $$\mathbb Q$$. $$$$E: y^2 = x(x^2-5) \quad E':y^2 = x(x^2+20)$$$$ These are related by a 2-isogeny: $$\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 $$$$E(\mathbb Q)\to^\phi E'(\mathbb Q)\to^q \mathbb Q^*/\mathbb Q ^{*^2}$$$$ 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 $$$$r^2l^4+20m^4=rn^2$$$$ has a integer solution. This can only happen if $$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 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.

(a)

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}

(b)

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 $$(-1,2)\in E(\Bbb Q)$$.