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Dec 18, 2012 at 19:50 comment added David Corwin It seems that I wasn't reading carefully enough. My question is actually answered in the paragraph after Theorem 3.2.2 in Ribet's thesis.
Jul 26, 2012 at 21:35 comment added Xarles And observe that your last coment is directly related to what I did together with Siegfried Bosch in the paper Component groups of Néron models via rigid uniformization, Math. Ann. Volume 306, Number 1 (1996), 459-486.
Jul 26, 2012 at 21:32 comment added Xarles I don't understand exactly what is the question, but observe that the component group of the Neron model is isomorphic (as abstract group) to the cokernel of the map given by the valuation matrix (from $\mathbb{Z}^2$ to itself). This is shown, if I remember well, in the SGA7, Expose IX, and it is due to Raynaud.
Jul 16, 2012 at 21:39 comment added Will Sawin Since $GL_2(\mathbb Z)$ acts on your matrix on the right without changing the geometry of the abelian variety, only the image of the matrix should carry geometric meaning. In particular the quotient of $\mathbb Z^2$ by the image of the valuation matrix should be the group of components. This would follow from the fact that $\mathcal O_K^{\times} \times \mathcal O_K^{\times}$ is the group of sections which have good reduction at $0$, that is, lie in the identity component.
Jul 16, 2012 at 21:26 history edited Will Sawin CC BY-SA 3.0
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Jul 16, 2012 at 19:27 history asked David Corwin CC BY-SA 3.0