Let $X$ be an elliptic curve over a complete local field.

The definition of semi-abelian reduction is: "the Neron model of $X$ is a semi-abelian scheme". On the other hand, the definition of semi-stable reduction is: "the minimal regular model of $X$ is semi-stable."

For elliptic curves, are the two definitions equivalent? How to prove it?

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    $\begingroup$ @kiseki I think a more precise title, as was the original one, would be more useful. $\endgroup$ Sep 10, 2012 at 14:47

1 Answer 1


The Néron model of $X$ is the smooth locus of the minimal regular model of $X$ (see Bosch-Lütkebohmert-Raynaud: Néron models, §1.5). The equivalence is then clear using the classification of Kodaira-Néron of the types of reduction of the minimal regular model.

Note that if $g(X)\ge 2$, then it is also true that $X$ has semi-stable reduction if and only if its Jacobian has semi-abelian reduction (Deligne-Mumford, based on Raynaud's description of Néron models of Jacobians).

If $g(X)=1$ but $X$ doesn't have a rational point, then the statement is no longer true. But one can show that the Jacobian of $X$ has semi-abelian reduction if and only if the type of the reduction of $X$ is a multiple of $I_n$, $n\ge 0$ (use the fact that $X$ covers its Jacobian. The additive reduction case is treated in a paper of Lorenzini, Raynaud and myself in 2004).

You don't need the completeness hypothesis on the base DVR.

  • $\begingroup$ @kiseki, I just check BLR's book. In Proposition 1.5/1, they state (and prove) that the smooth locus is the Néron model for elliptic curve, under the condition that the base field is strictly Henselian. In general, by Néron mapping property, the identity on the generic fiber extends to a morphism from the Néron model to the smooth locus of the minimal regular model. Base change this morphism to the strict henselization of the base ring. As both Néron and minimal regular models commute with this kind of base change, by BLR, we have an isomorphism upstair. As the base change is faithfully flat, $\endgroup$
    – Qing Liu
    Sep 11, 2012 at 8:16
  • $\begingroup$ continued: the morphism downstairs is also an isomorphism. If you don't want to use base change, the result can also be found in my book, Thereom 10.2.14. $\endgroup$
    – Qing Liu
    Sep 11, 2012 at 8:20
  • $\begingroup$ Thank you very much, Pof.Liu. I have found the result in your book. $\endgroup$
    – kiseki
    Sep 11, 2012 at 9:57

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