# Show that $\mathcal C : Y^2=X(X^2-pX+p^2)$ has rank $0$, where $p$ is a prime $\equiv$ 2 (mod 3) and 5 (mod 8)

The question is from a masters' level elliptic curves exam (Oxford, 2017):

Let $$p$$ be a prime congruent to 2 (mod 3) and 5 (mod 8). Show that the elliptic curve $$\mathcal C : Y^2=X\left(X^2-pX+p^2\right)$$ has Mordell-Weil rank 0.

Because of the layout of the course, I am 99% certain that this question can be solved by a fairly standard 2-descent argument. The problem is, I get stuck at excluding a potential generator. I'll set up the 2-descent and then explain where I get stuck.

Let $$\mathcal D: V^2=U(U^2+2pU-3p^2)$$. Then $$\mathcal D$$ is an elliptic curve, and there are 2-isogenies $$\varphi\colon\mathcal C\to\mathcal D$$ and $$\hat{\varphi}\colon\mathcal D\to\mathcal C$$ such that $$\varphi\circ\hat{\varphi}$$ and $$\hat{\varphi}\circ\varphi$$ are the multiplication-by-2 maps. We calculate $$\mathcal C[2](\mathbb Q)$$ (the rational 2-torsion of $$\mathcal C$$), $$G:=\mathcal C(\mathbb Q)/\hat{\varphi}(\mathcal D(\mathbb Q))$$ and $$\hat{\varphi}(H):=\hat{\varphi}(\mathcal D(\mathbb Q)/\varphi(\mathcal C(\mathbb Q)))$$. From these last two sets we can calculate $$\mathcal C(\mathbb Q)/2\mathcal C(\mathbb Q)$$ and this, combined with the 2-torsion, allows us to determine the rank of $$\mathcal C(\mathbb Q)$$.

Obviously $$\mathcal C[2](\mathbb Q)=\{\mathcal O,(0,0)\}\cong C_2$$, so that's fine.

To calculate $$G$$, we define a map $$\hat{q}\colon G\to\mathbb Q^\times/\left(\mathbb Q^\times\right)^2$$ by $$\mathcal O\mapsto1,(0,y)\mapsto p^2=1$$ and $$(x,y)\mapsto x$$ when $$x\neq0$$. One can show that $$\hat{q}$$ is a well-defined injective group homomorphism, whose image consists of all squarefree integers $$r\mid p^2$$ for which there exist integers $$l,m$$ and $$n$$, not all 0, with $$\gcd(l,m)=1$$, such that $$\hat{W}_r : rl^4-pl^2m^2+p^2m^4/r = n^2.$$ Thus $$\mathrm{im}~\hat{q}\leq\{\pm1,\pm p\}$$. We have $$\hat{q}(\mathcal O)=\hat{q}(0,0)=1$$. The equations $$\hat{W}_{-1}$$ and $$\hat{W}_{-p}$$ have the form "nonpositive = nonnegative", so have no nontrivial solutions. On dividing $$\hat{W}_p$$ through by $$p$$ and completing the square, one can see that this has a solution mod $$p$$ only if 3 is a square mod $$p$$, or $$p\mid l$$ and $$m$$. The former is excluded by the given congruence info, the latter by $$\gcd(l,m)=1$$. Thus $$\mathrm{im}~\hat{q}=\{1\}$$ and $$G=\{\mathcal O\}$$.

To calculate $$H$$, in a similar vein we define a map $$q$$ by $$\mathcal O\mapsto1,(0,v)\mapsto -3p^2=-3$$ and $$(u,v)\mapsto u$$ when $$u\neq0$$. Now we're looking for squarefree integers $$r\mid-3p^2$$ satisfying $$W_r : rl^4+2pl^2m^2-3p^2m^4/r=n^2.$$ Thus $$\mathrm{im}~q\leq\{\pm1,\pm3,\pm p,\pm3p\}$$. Note that $$\mathcal D$$ factorises as $$V^2=U(U-p)(U+3p)$$, so we have $$q(\mathcal O)=1$$, $$q(0,0)=-3$$, $$q(p,0)=p$$ and $$q(-3p,0)=-3p$$. There's only one nontrivial coset left, so it suffices to exclude any one of $$-1, 3$$, $$-p$$ or $$3p$$ from $$\mathrm{im}~q$$ to demonstrate that $$\mathrm{im}~q=\{1,-3,p,-3p\}$$. The relevant equations are: $$W_{-1}:-l^4+2pl^2m^2+3p^2m^4=n^2,$$ $$W_{3}:3l^4+2pl^2m^2-p^2m^4=n^2,$$ $$W_{-p}:-pl^4+2pl^2m^2+3pm^4=n^2,$$ $$W_{3p}:3pl^4+2pl^2m^2-m^4=n^2.$$

The problem — It's in showing that one of the above equations has no solution that I get stuck. I've looked at them mod $$p$$, mod 3, mod 9, even mod 16, and I've tried completing the square, but I can't seem to arrive at a contradiction.

I believe that we need to exclude the above values of $$r$$ to get rank 0, which I will now justify. Let's suppose on the contrary that $$\mathrm{im}~q=\{\pm1,\pm3,\pm p,\pm3p\}$$. Then $$H=\langle (0,0),(0,p),R \rangle$$, where $$R$$ is some other point on $$\mathcal D(\mathbb Q)$$ (say $$q(R)=-1$$). Now $$\hat{\varphi}(\mathcal O)=\hat{\varphi}(0,0)=\mathcal O$$ and $$\hat{\varphi}(0,p)=\hat{\varphi}(0,-3p)=(0,0)$$, so since $$\hat{\varphi}$$ is 2-to-1, $$\hat{\varphi}(R)$$ is independent of $$(0,0)$$ in $$\mathcal C(\mathbb Q)/2\mathcal C(\mathbb Q)$$. Thus $$\mathcal C(\mathbb Q)/2\mathcal C(\mathbb Q)=\langle (0,0),R \rangle\cong C_2\times C_2$$. So on removing the 2-torsion we get that the rank of $$\mathcal C(\mathbb Q)$$ is 1. Which is not 0. So we need to exclude this point $$R$$.

Edit: Simplicity of Solution — I'm not entirely sure whether this contravenes the rules of this site, but since this question comes from an exam, I require a solution that could reasonably be believed to be arrived at by a late undergraduate/early graduate who has only been acquainted with elliptic curves for one term, has approximately 30 minutes, and no access to a computer.

• Since your $2$-covering $W_{-p}$ has a $3$-adic and $p$-adic point (looking modulo $9$ was always going to be fruitless because Hensel) you should definitely be looking for a contradiction to a $2$-adic point. I would look at $W_{-p}$ but divide both sides by $p$ to get $-l^4 + 2l^2m^2 + 3m^4 = pn^2$ and try looking for a contradiction there. Commented Jun 10, 2021 at 21:25
• $W_{-p}$ has a point mod 16, namely $(l,m,n)=(1,1,2)$ (in the simplified form you give). About an hour's fiddling leads me no further than $l,m\equiv1$ mod 2 and $n\equiv2$ mod 4. I noticed that you can factorise $W_{-p}$ as $(l^2+m^2)(l^2-3m^2)=-pn^2$, but I don't know if that helps... Commented Jun 11, 2021 at 11:08

## 1 Answer

Let's consider your $$2$$-covering $$W_{-p} : -pl^4+2pl^2m^2+3pm^4=n^2$$

Clearing denominators we can take $$l,m,n$$ as integers. I clain there is no $$2$$-adic point. I am going to do this the only way I could figure out, and it may seem like magic - so I apologise for that.

Consider the transformation $$n = n' + 4l'^2 - 4l'm' + 2m'^2$$ $$l = (2l' - m')/2$$ $$m = m'/2$$ which leaves us with (after dropping the ') $$n^2 + (2l^2 - 2lm + m^2)n = -(p+1)l^4 + 2(p+1)l^3m - (p+2)l^2 m^2 + lm^3 + \frac{p-1}{4}m^4$$ Now, because of the mod $$8$$ condition, the last term is an integer and modulo $$2$$ we have $$n^2 + m^2n = l^2 m^2 + lm^3 + m^4 \pmod{2}$$ then $$n + mn = m \pmod{2}$$ so that $$m=n=0 \pmod{2}$$. But then looking mod $$4$$ we see find $$0 = -(p+1)l^4 = 2l^4 \pmod{4}$$
so $$l \equiv 0 \pmod{2}$$ and we reach a contradiction.

Discussion of Magic: To find the magical transformation above what I did is "minimise the $$2$$-covering $$W_{-p}$$ at the prime $$2$$". If you search for something like "Local Solubility On 2-Coverings" and you'll find such papers of Stoll, Cremona, and Fisher for general algorithms to do this. Implement them by hand will lead you to this change of variables. If you have a $$2$$-covering defined over $$\mathbb{Q}$$ the Magma function Minimise(C : CrossTerms:=true) will minimise your covering everywhere. The point of this exercise is to make sure you can either (1) run into a local obstruction or (2) can use Hensel to lift a solution - hence why we get weird cross terms when working with $$\mathbb{Z}_2$$

I would also appreciate any answer which does not invoke witchcraft.

• Thanks for the effort you've put into this answer! However, as you yourself point out, the transformation required for it is completely magical, and I can hardly believe that this is the simplest way it can be solved, since it was set on a "first course in elliptic curves" exam. Commented Jun 11, 2021 at 13:06
• @TomSharpe Yes I completely agree. I think there must be some elementary trick which can be spotted. Even more - I would love to know what it is! Commented Jun 13, 2021 at 20:52