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May 7, 2015 at 17:43 history edited jmc CC BY-SA 3.0
Include a comment by René
May 7, 2015 at 12:13 comment added jmc @René — I thought that was true, but I wasn't 100% sure, so I left it out, to be on the safe side. I agree that if it is true, it should be included. Do you have a reference for your statements?
May 7, 2015 at 12:05 comment added R.P. In that case, +1. I do think it is worth emphasizing the fact that the middle term is a tangent space in its own right (and a free $R_{\mathcal{P}}$-module of rank $\operatorname{dim} A$), which naturally embeds into $T_e A$ and has a reduction map to $T_e A_0$.
May 7, 2015 at 11:54 history edited jmc CC BY-SA 3.0
Add some conditions and clarifications
May 7, 2015 at 11:52 comment added jmc @MatthiasWendt,JoeSilverman — Yes, I should have mentioned the Néron model.
May 7, 2015 at 11:52 comment added jmc @René — You are right, I forgot to specify that.
May 7, 2015 at 11:16 comment added R.P. With $\mathrm{Hom}_{S}(S[\varepsilon],\mathcal{A})$, do you mean the set of those $S$-morphisms $S[\varepsilon]\to\mathcal{A}$ that restrict to the identity section $e\colon S\to\mathcal{A}$? And in that case, wouldn't the middle term just be "the tangent space to $\mathcal{A}$ at the identity section" (i.e. something like $\operatorname{H}^0(S,e^\ast (\Omega^1_{\mathcal{A}})^{\vee})$)?
May 7, 2015 at 10:50 comment added Joe Silverman The definition of "good reduction" is that there is such an abelian scheme. And won't $\mathcal A$ be unique (as an abelian scheme over $R_{\mathcal P}$)? Actually, even if $A$ has bad reduction, one can specify that $\mathcal A$ be the Neron model, which will pin things down.
May 7, 2015 at 9:47 comment added Matthias Wendt Could you expand a bit more on the naturality? There seems to be a choice of abelian scheme.
May 7, 2015 at 9:31 history answered jmc CC BY-SA 3.0