4
$\begingroup$

$\DeclareMathOperator\Sel{Sel}$Let $E$ be an elliptic curve defined over a number field $K$ with full $2$-torsion. The classical complete $2$-descent method tells that the $2$-Selmer group $\Sel_2(E/K)$ can be identified with the set of locally solvable everywhere homogeneous spaces. More precisely, we consider a quadratic twist $$E:y^2=x(x-na)(x+nb)$$ defined over $\mathbb Q$, where $n,a,b,\frac{a+b}{2}$ are positive odd integers. Then $\Sel_2(E/\mathbb Q)$ can be identified with $$\{\Lambda=(d_1,d_2)\in(\mathbb Q^\times/\mathbb Q^{\times2})^2: D_\Lambda(\mathbb{A}_K)\neq\emptyset\},$$ where $$D_\Lambda: d_1u_1^2-d_2u_2^2=na,\quad d_1u_1^2-d_1d_2u_3^2=-nb.$$

Assume that $n$ is a positive square-free integer prime to $2ab(a+b)$ and $\Lambda=(d_1,d_2)$ where $d_1,d_2$ are positive square-free odd divisors of $nab(a+b)$. Then we can show: if $D_\Lambda$ is locally solvable everywhere except $v=2$, then it is also locally solvable at $v=2$. This can be proved by the poduct formula on Hilbert symbols.

So my question is: when the following hold for any $\Lambda$? Does this relate the Selmer group or Shafarevich-Tate group of $E$?

Let $D_\Lambda$ be a homogeneous space in the form as above. If $D_\Lambda$ is locally solvable everywhere except a place $v$, then it is also locally solvable at $v$.

$\endgroup$

1 Answer 1

1
$\begingroup$

(Edit: I revise most of my question as my first answer overlooked that $d_1$ and $d_2$ are odd.) $\DeclareMathOperator{\res}{res}$

Let $E$ be an elliptic curve over $\mathbb{Q}$ with $E[2]\subset E(\mathbb{Q})$. Let $S$ be the normal $2$-Selmer group and let $S'$ be the relaxed Selmer group, that is the subset of $H^1\bigl(\mathbb{Q},E[2]\bigr)\cong {}^{\mathbb{Q}^\times}\!/{}_{\square} \times{}^{\mathbb{Q}^\times}\!/{}_{\square}$ that satisfy all local conditions away from $2$, but maybe not at $2$. Within $S'$ define $S''$ the subset of all $\xi=(d,e)$ where both $d$ and $e$ are odd. That is also a sort of a natural Selmer group, where the local condition at $2$ is that $\res_2(\xi)$ lies in the kernel of the valuation ${}^{\mathbb{Q}_2^\times}\!/{}_{\square} \times{}^{\mathbb{Q}_2^\times}\!/{}_{\square}\to {}^{\mathbb{Z}}\!/\!{}_{2\mathbb{Z}}\times {}^{\mathbb{Z}}\!/\!{}_{2\mathbb{Z}}$. The question is: When is $S''\subset S$?

This is not a local question as the elements in the kernel of the valuation in $H^1\bigl(\mathbb{Q}_2,E[2]\bigr)$ are not naturally compared with the image of the local Kummer map $\kappa_2\colon {}^{E(\mathbb{Q}_2)}\!/\!{}_{2E(\mathbb{Q}_2)} \to H^1\bigl(\mathbb{Q}_2,E[2]\bigr)$. Here I give the answer in one particular case; the methods generalise to other similar cases and one could determine the answer in general I imagine. Since the actual calculations with Hilbert symbols in very simple, it may help to understand better the original proof alluded to in the question.

Assume $E(\mathbb{Q}_2)[4]=E(\mathbb{Q}_2)[2]$. That is no point of order $2$ is divisible by $2$ in $E(\mathbb{Q}_2)$. Also assume that $E$ is given by $y^2=x(x-a)(x-b)$ with $a$ and $b$ odd integers. (I don't think $a\equiv b\pmod{4}$ is needed here.)

First consider the following exact sequence coming from global duality: $$ 0\to S\to S' \xrightarrow{\alpha} H^1\bigl(\mathbb{Q}_2, E\bigr)[2]\xrightarrow{\hat\beta} \hat{S} $$ where $\alpha$ and $\beta$ are restriction maps. First if $\beta\colon S\to {}^{E(\mathbb{Q}_2)}\!/\!{}_{2E(\mathbb{Q}_2)}$ is surjective, then $S'=S$ and hence $S''\subset S$. Therefore, suppose $\beta$ is not surjective. My assumption above imposes that the three $2$-torsion points (which are global points and hence in the image of $\beta$) are all distinct in ${}^{E(\mathbb{Q}_2)}\!/\!{}_{2E(\mathbb{Q}_2)}$. Therefore the image of $\beta$ is equal to group generated by the torsion points. It is $2$-dimensional in the $3$-dimensional target.

Let now $\xi\in S''$ with $\res_2(\xi) = (d,e)$. To say that $\xi\in S$ is equivalent to $\alpha(\xi)=0$. Since we know already that $\xi\in\ker(\hat\beta)$, all we need to check is that $\res_2(\xi)\cup \kappa(P)=0$ for one point $P\in E(\mathbb{Q}_2)$ which is not in the image of $\beta$. Here $\cup \colon H^1\bigl(\mathbb{Q}_2,E[2]\bigr)\times H^1\bigl(\mathbb{Q}_2,E[2]\bigr) \to {}^{\mathbb{Z}}\!/\!{}_{2\mathbb{Z}}$ is the local duality pairing. Under our identification it corresponds to the pairing $(d,e)\cup (d',e') = (d,e')_2 + (d',e)_2$ on ${}^{\mathbb{Q}_2^\times}\!/{}_{\square} \times{}^{\mathbb{Q}_2^\times}\!/{}_{\square}$ where $(,)_2$ is the Hilbert symbol with values in ${}^{\mathbb{Z}}\!/\!{}_{2\mathbb{Z}}$.

In our concrete case, there is a point $P$ with $x$-coordinate equal to $\tfrac{1}{4}$ on $E(\mathbb{Q}_2)$ because $a$ and $b$ are both odd. In fact the points $P$, $T_1=(0,0)$ and $T_2=(a,0)$ generate ${}^{E(\mathbb{Q}_2)}\!/\!{}_{2E(\mathbb{Q}_2)}$ because $T_2$ has bad reduction, $T_1$ has good non-trivial reduction, and $P$ has trivial reduction. We have $\kappa(P)=(1,1-4a)=(1,-3)$ since $a$ is odd. Now $$ (d,e)\cup(1,-3) = (d,-3)_2 + (1,e)_2 = 0 $$ because $d$ is odd and $-3\equiv 1 \pmod{4}$. Therefore $(d,e)\in S$.

Finally, I would like to point to Theorem 3 in Appendix 1 of Swinnerton-Dyer's 2-descent through the ages where a question of that nature is discussed, but for places away from 2.

Edit: This is the original answer.

I actually have troubles believing the statement you make in your question. Maybe I am wrong or I misunderstood something, but since it is too long for a comment, I include it as a possible "answer".

Take $y^2 = x^3 - x$. So $n=a=b=1$. I claim: The relaxed Selmer group, imposing all conditions but the one at $2$, is $\mathbb{F}^3$, while the rank of the curve is $0$ so the usual Selmer group is of dimension $2$.

Concretely, the local conditions away from $2$ and $\infty$ impose that $d_1$ and $d_2$ both belong to the group generated by $-1$ and $2$ modulo squares, since the curve has good reduction away from $2$. The condition at $\infty$ imposes that $d_1$ and $d_2$ have the same sign. Now we are down to the group generated by $(1,2)$, $(-1,-1)$ and $(2,1)$. The first two correspond to the torsion points $(1,0)$ and $(0,0)$ in $E(\mathbb{Q})[2]$. So the torsor $$\begin{align*} 2u^2 -v^2 & = 1 \\ 2u^2 -2w^2 &= -1 \end{align*}$$ is locally soluble at all places, except for $2$.

The reason, I was suspicious initially is that your statement, using global duality, would be equivalent to the surjectivity of the map from the Selmer group $S_2(E/\mathbb{Q})$ to $E(\mathbb{Q}_2)/2E(\mathbb{Q}_2)$. Your question would be equivalent to the surjectivity of the restriction from the Selmer group of some isogeny $\varphi$ to the group of points at this one place. And I don't see a reason why this should hold without conditions on $\varphi$.

$\endgroup$
2
  • $\begingroup$ You are right, but I require that $d_1,d_2$ are odd. Your answer makes me believe that this only holds in few special cases but is not true in general. Thanks for your answer. $\endgroup$ Jan 28, 2022 at 4:15
  • $\begingroup$ Oh, I am sorry that I overlooked the word "odd". That changes the question to an interesting one. But I need to think about it. It could be true and generalisable. $\endgroup$ Jan 28, 2022 at 10:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.