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Let $p$ be a prime number, let $E$ be an elliptic curve defined over $\mathbb{Q}_p$. Let $\mathcal{E}_p$ be the special fiber of the Néron model of $E$ over $\mathbb{Z}_p$ and let $\mathcal{C}_p$ be the special fiber of the minimal regular model of $E$ over $\mathbb{Z}_p$

In their 1986 paper "K2 and $L$-functions of elliptic curves - computer calculations" Bloch and Grayson (p. 82, 83) state that the first $K$-group $K_1^'(\mathcal{E}_p)$ is non-torsion iff $E$ has split multiplicative reduction at $p$. In this case we have $K_1^'(\mathcal{E}_p)\cong \mathbb{Z}\oplus\mathrm{torsion}$ and their argument is that this follows because the first homology of the dual graph of $\mathcal{C}_p$ is isomorphic to $\mathbb{Z}$. But the only proof of this fact that I've seen (in the thesis of Rolshausen) does not use the dual graph at all. Note that the rank of the first homology of the dual graph is equal to the toric rank $t_p$ of (the connected component of) $\mathcal{E}_p$.

(1) How can you use the dual graph of $\mathcal{C}_p$ to show that the rank of $K_1^'(\mathcal{E}_p)$ is equal to $t_p$?

Presumably such a proof would use étale cohomology which unfortunately I don't know enough about.

Now suppose that $C$ is a (smooth, projective) curve of positive genus defined over $\mathbb{Q}_p$. Let $\mathcal{C}_p$ be the special fiber of the minimal regular model of $C$ over $\mathbb{Z}_p$. Let $t_p$ be the toric rank of (the connected component of the special fiber of) the Néron model of $\mathrm{Jac}(C)$ over $\mathbb{Z}_p$. As above, this is equal to the rank of the first homology of the dual graph of $\mathcal{C}_p$.

(2) Is it known that the rank of $K_1^'(\mathcal{C}_p)$ is equal to $t_p$?

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I don't understand (2). If $C^p$ has good reduction, then this would imply that $K_1(C_p)=0$. But $K_1(C_p)$ does not vanish because it contains at least the non-zero elements of ${\bf F}_p$. –  Damian Rössler Sep 17 '12 at 11:15
    
(sorry: "$C$ has good reduction", not "$C^p$ has good reduction") –  Damian Rössler Sep 17 '12 at 11:16
    
You're right, but if $C$ has good reduction, then $K_1(\mathcal{C}_p)$ is torsion. I'm not claiming that $t_p=0$ implies $K^'_1(\mathcal{C}_p)=0$. –  jsm Sep 19 '12 at 14:08

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