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I'm studying the chapter on Neron Models in Silverman's book "Advanced Topics in the Arithmetic of Elliptic Curves" at the moment, and I do not quite understand why in the split multiplicative case, we have that the component group is cyclic.

So, the set-up is as follows: we have a Dedekind domain $R$ with fraction field $K$. We have an elliptic curve $E/K$ with minimal proper regular model $\mathcal{C}/R$. Theorem 6.1 then states that the largest subscheme $\mathcal{E}/R$ which is smooth over $R$ will be the Neron Model for $E/K$ over $R$.

I will now assume that $R$ is in fact local. Also, my elliptic curve will be assumed to have split multiplicative reduction. In corollary 9.2 we find the identification $E(K)/E^{0}(K)\cong{\mathcal{E}(R)/\mathcal{E}^{0}(R)}$, with an inclusion $\mathcal{E}(R)/\mathcal{E}^{0}(R)\hookrightarrow{\tilde{\mathcal{E}}(k)/\tilde{\mathcal{E}}^{0}(k)}$ where $k$ is the residue field of $R$. This last group is the component group of $\tilde{\mathcal{E}}$. In Tate's algorithm we find that this group is finite of order $n=v(\Delta)$, the valuation of the minimal discriminant (I've convinced myself of this fact by calculating the blow-ups in question).

This is the point where I get stuck. I see no reason why this group should be cyclic. For instance, let $n=4$. Why is the group of components not $\mathbb{Z}/2\mathbb{Z}\times{\mathbb{Z}/2\mathbb{Z}}$? It feels like I'm missing some small step here.

At any rate, using p-adic uniformization we can get this result from chapter 5 by considering the corresponding Tate curve (for $K$ complete). I'd like to know whether we can obtain the result without referring to this chapter.

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  • $\begingroup$ But the group can be non-cyclic, did you take a look at the table of reduction types? (page 365) $\endgroup$ – Vinicius M. Jan 9 '15 at 12:40
  • $\begingroup$ I'm talking about the split multiplicative case, where the group is cyclic of order $n$ (In the table the special fibre is an $n$-gon). $\endgroup$ – Paul Helminck Jan 9 '15 at 14:14

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