Surjectivity of map between Néron models $\mathcal{E} \to \mathcal{E}'$

Question

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?

Answer 1

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.