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My question is related to a previous question on the Mordell-Weil rank of the elliptic curve $E/\mathbf{Q} : y^2 = x^3- 2$ asked here. More precisely, I want to understand the following. Let $E'/\mathbf{Q}$ be the elliptic curve $y^2 = x^3 + 54$. There is a rational $3$-isogeny $\phi : E \to E'$ given by $$\phi(x,y) = \left( \frac{x^3 - 8}{x^2}, \frac{y(x^3 + 16)}{x^3}\right).$$ Let $\mathcal{E},\mathcal{E}'$ denote the Néron models of $E,E'$ respectively over $\mathbf{Z}_{(3)}$. Let $$ \Phi \colon \mathcal{E} \to \mathcal{E}' $$ denote the induced map on Néron models.

My goal: I would like to prove that $\Phi$ is étale surjective, so that the map of representable sheaves $\underline{\mathcal{E}} \to \underline{\mathcal{E}'}$ on $(\mathbf{Z}_{(3)})_{\text{ét}}$ is surjective.

It is enough to show this is true on special fibers. Let $\widetilde{\Phi} \colon \widetilde{\mathcal{E}} \to\widetilde{\mathcal{E}'}$ denote the map on reductions. Then I have shown by direct computation that on identity components, $$ \widetilde{\Phi}|_{\widetilde{\mathcal{E}^0}} \colon \widetilde{\mathcal{E}^0} \to\widetilde{\mathcal{E}'^0} $$ is étale surjective. Therefore $\widetilde{\Phi}$ is étale, because we can check it after base changing to $\overline{\mathbf{F}_3}$, in which case every component of $\widetilde{\mathcal{E}}$ is a translate of the identity component by a $\overline{\mathbf{F}_3}$ rational point.

Now for surjectivity, I have the following information (from a table in Silverman's ATEC). We define $k := \mathbf{F}_3$.

The component groups of the special fibers of the Néron models are: $$\widetilde{\mathcal{E}}(k)/\widetilde{\mathcal{E}^0}(k) = \widetilde{\mathcal{E}'}(k)/\widetilde{\mathcal{E}'^0}(k) = \mathbf{Z}/2\mathbf{Z}.$$

However, with this information I can't seem to conclude that $\widetilde{\Phi} \colon \widetilde{\mathcal{E}} \to \widetilde{\mathcal{E}'}$ is surjective. The problem seems to be that we don't know what the $\overline{k}$-points of the component groups are. Using things like Lang's Theorem to say that $\widetilde{\mathcal{E}}(k)/\widetilde{E^0}(k) = \widetilde{\mathcal{E}/\mathcal{E}^0}(k)$ don't seem to help too. I would appreciate any insight on how to get surjectivity of $\Phi$.

Edit: It seems that my description of component groups is wrong (see Chris's answer below). However, Chris has outlined reasons why $\Phi$ cannot be surjective. In any case, my ultimate goal is to show that as representable sheaves on the site ${\mathbf{Z}_{(3)}}_{\'{e}t}$, the map $\underline{\mathcal{E}} \to \underline{\mathcal{E}'}$ is surjective. Perhaps this can still be shown?

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  • $\begingroup$ If $\Phi$ is a $3$-isogeny then there exists $\Psi:{\cal E}'\to{\cal E}$ such that $\Psi\circ\Phi=[3]_{\cal E}$ (where $[3]$ is the multiplication by $3$ morphism). So $\widetilde{\Psi}\circ\widetilde{\Phi}=[3]_{\widetilde{\cal E}}$. If $\widetilde{\Phi}$ were not surjective then its image would be the id. component of $\widetilde{\cal E}'$, according to you computation. But then $[3]_{\widetilde{\cal E}}$ would not be surjective, which is false. $\endgroup$ – Damian Rössler Dec 16 '16 at 9:03
  • $\begingroup$ The answer below suggests that the component group of ${\cal E}'$ is not ${\bf Z}/2{\bf Z}$, in which case the argument I give does not work (I only based myself on your question). $\endgroup$ – Damian Rössler Dec 16 '16 at 11:27
  • $\begingroup$ @DamianRössler Hi Damian, I've looked at the tables in Silverman again and I'm pretty sure they're both $\mathbf{Z}/2\mathbf{Z}$, unless Silverman is wrong. $\endgroup$ – Ben Lim Dec 16 '16 at 22:25
  • $\begingroup$ @DamianRössler Also, granted that it's true that both component groups are $\mathbf{Z}/2\mathbf{Z}$, I'm trying to understand your argument. If $\widetilde{\Phi}$ were not surjective then why must the image lie in the identity component? It seems to me that your argument is using the fact that $\widetilde{\mathcal{E}'/\mathcal{E}'^0}(\overline{k}) = \mathbf{Z}/2\mathbf{Z}$, which is something we don't know. $\endgroup$ – Ben Lim Dec 17 '16 at 0:05
  • $\begingroup$ Hello Ben. You may assume that we are working over the maximal unramified extension of ${\bf Z}_3$ to check surjectivity, so that w.r.o.g. $k=\bar k$. It now comes down to checking that on the special fibre of $\mathcal E$, the mult. by $3$ morphism is surjective. But the component group of the special fibre is ${\bf Z}/2{\bf Z}$, according to you (if I got it right) so the mult. by $3$ morphism on the geometric component group is the identity, which implies the above mentioned surjectivity. But this works only if the component group has order prime to $3$. $\endgroup$ – Damian Rössler Dec 17 '16 at 12:43
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I get that the Kodaira types of $E$ is II and for $E'$ it is IV${}^{*}$. This means that $\mathcal{E}$ is connected (or in simple terms no point over $\mathbb{Q}_3^{\text{unr}}$ reduces to the singular point $(-1,0)$ of the reduced Weierstrass equation) and $\mathcal{E}'$ has component group $\Phi' = \mathbb{Z}/3\mathbb{Z}$ (for instance the point $(3,9)\in E'(\mathbb{Q})$ reduces to the singular point $(0,0)$). It follows that $\hat\phi$ is surjective. The kernel of $\phi$ is étale: the point $(0,1+3+2\cdot 3^2+\cdots)\in E(\mathbb{Q}_3)[3]$ generates the kernel. Instead the kernel of $\hat\phi$ is not even quasi-finite.

One can also see directly that $\phi$ is not surjective: If $(x,y)$ in $E$ is mapped to $(3,9)$ then $x^3-3x^2-8=0$, but that has no solution in $\mathbb{Q}^{\text{unr}}_3$.

Edit: I corrected my statement about the kernel of $\hat\phi$. Yes, subgroups of order $p$ when the elliptic curve has additive reduction at $p$ are nasty.

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  • $\begingroup$ Thanks for your answer. I realized that thereally was a typo in the definition of \phi that I have written above. It is now corrected. $\endgroup$ – Ben Lim Dec 16 '16 at 12:49
  • $\begingroup$ Yes, I noted that. But more importantly you have got the group of components wrong (I believe), $\endgroup$ – Chris Wuthrich Dec 16 '16 at 13:11
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    $\begingroup$ Ok. But I'm looking at pg 365 of Silverman's ATEC. The entry for type II says that char $k \neq 2,3$ and $\nu_3(\Delta_E) = 2$. This is not true for E as the residue characteristic is 3 and the valuation of the minimal discriminant is $3$. This agrees with the type III entry of the table. So if the table is right, the component group of $\mathcal{E}$ is $\mathbf{Z}/2\mathbf{Z}$. As for $E'$, we have $\nu_3(\Delta_{E'}) = 9$ and this agrees with the III* entry. So the component group for $\mathcal{E}'$ is also $\mathbf{Z}/2\mathbf{Z}$. $\endgroup$ – Ben Lim Dec 16 '16 at 22:19
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    $\begingroup$ @BenLim: Instead of repeatedly referring to that table (which you'll see in the first edition rather than the 2nd actually did not allow $p=3$ in that case), you should do as Chris advises and actually work out the minimal regular proper model to compute the component group yourself. Chris has pointed out specific numerical reasons why the component group cannot possibly be what is being asserted from the table, so the table must be wrong (or rather, permitting $p=3$ there in the 2nd edition is incorrect). $\endgroup$ – nfdc23 Dec 17 '16 at 1:55
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    $\begingroup$ @BenLim: Since the composition of $\phi$ and its dual is $[3]$ that vanishes on the identity components of the mod-3 fibers, we know that at least one of the two must vanish between identity components of the mod-3 fibers due to additive reduction. Hence, Chris' assertion that both are etale over $\mathbf{Z}_3$ (or equivalently between identity components of mod-3 fibers) is incorrect; specific subgroups of residual kernels were identified, but no argument was given for why those kernels cannot be bigger. Regardless, he showed your description of the (geometric) component groups is incorrect. $\endgroup$ – nfdc23 Dec 17 '16 at 18:43

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