Let $K$ be a local field, complete with respect to a discrete valuation $v$ and let $E/K$ be an elliptic curve. We also let $E_0(K)$ is the set of points with nonsingular reduction.

It is known that $E(K)/E_0(K)$ is of order at most 4 if $E$ does not have split multiplicative reduction over $K$ (see Theorem VII.6.4 of Silverman's "The Arithmetic of Elliptic Curves").

My question concerns the case when the order of this group is exactly 4. How can we tell if this group is ${\mathbb Z} / {2\mathbb Z} \times {\mathbb Z} / {2\mathbb Z}$ or ${\mathbb Z} / {4\mathbb Z}$?

I looked at Tate's Algorithm, as described in Section IV.9 of Silverman's "Advanced Topics in the Arithmetic of Elliptic Curves", and although that tells us how to tell when the order is 4, and Table 4.1 states what the group is when the residue field, $k$, is algebraically closed, it does not tell us how to determine the actual group itself.

Any references, ideas, etc would be very welcome.

  • $\begingroup$ Let $\mathcal{E}$ be the special fiber of the Néron model of $E$. Then $E(K)/E_0(K) \cong (\mathcal{E}/\mathcal{E}^0)(k)$ injects into $(\mathcal{E}/\mathcal{E}^0)(\bar{k})$. So if you assume $E(K)/E_0(K)$ has order 4, then $(\mathcal{E}/\mathcal{E}^0)(k)=(\mathcal{E}/\mathcal{E}^0)(\bar{k})$ and everything can be read off from the Kodaira symbol of $E$. $\endgroup$ Apr 3 '17 at 19:30

This is not a complete answer, since there are some subtleties in residue characteristic $2$, but in all other cases there is a simple answer.

In residue characteristic $p>2$, you can simply look at the $2$-torsion of $E(K)$. If (a) it either has only one point of order $2$, or it has three, but one is contained in $E_0(K)$, then $E(K)$ is cyclic of order four. The reason is that, if $p>2$, then $E(K)/E_0(K)$ is a direct summand of $E(K)$. [Indeed, since the reduction is additive, this follows from the fact that $E_0(K)/E_1(K)$ is of order $p$, so the short exact sequence relating $E(K)/E_1(K)$ to the two smaller quotients splits; therefore any element of order $2$ in $E(K)/E_0(K)$ lifts to a unique element of order $2$ in $E(K)/E_1(K)$, which lifts to $E(K)$ by Hensel.]

Otherwise (b) we must have that $E(K)$ has three points of order $2$, none of which are in $E_0(K)$, so then $E(K)/E_0(K)$ must necessarily be isomorphic to the Klein four-group.

One brief thing that one can say about the case $p=2$ is that if $E(K)$ has three points of order $2$ which are all of bad reduction, or $E(K)$ has only one point of order $2$, then you're again immediately done. The tricky case remains the one where $p=2$, $\# E(K)[2] = 4$ and $\# E_0(K)[2] = 2$. In that case I can't exclude the possibility that $E(K)/E_0(K)$ could be the Klein-four group, but only one of its elements of order $2$ lifts to an element of order $2$ in $E(K)/E_1(K)$, the other two lifting to elements of order $4$. (But I have no idea whether of not this case actually occurs.)

  • $\begingroup$ That's a very useful, clear and helpfully detailed answer. Thanks very much! $\endgroup$
    – user106917
    Apr 4 '17 at 13:06

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