Timeline for where do CM abelian varieties get good reduction?
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| Jan 25, 2014 at 23:18 | history | edited | conduc | CC BY-SA 3.0 |
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| Jan 25, 2014 at 23:16 | comment | added | conduc | @Joe, ah you're absolutely right! This is corollaire p. 517 of Fontaine's paper! | |
| Jan 25, 2014 at 23:16 | comment | added | Joe Silverman | @conduc Even for elliptic curves with CM by the maximal order, you generally can't take $F$ to be the Hilbert class field of $K$ and acquire everywhere good reduction. Also, you ask "how to determine $F$ in terms of $K$"? You do realize that there may not be a unique minimal $F$. Indeed, abelian varieties of dim $\ge2$ often have many different minimal fields of definition, since the field of moduli may not itself be a field of definition. | |
| Jan 25, 2014 at 23:10 | comment | added | Joe Silverman | @user76758 I agree that there are twisting subtleties, as well as the issue of whether there is a minimal field over which one attains good reduction. But neither, I think, applies to my comment. Regardless of which $\overline{\mathbb{Q}}$ model of $y^2=x^3+x$ that you take (i.e., no matter which twist), you won't find a Neron model over Spec($\mathbb{Z}[i]$) that has everywhere good reduction. (Indeed, my recollection is that there don't exist any elliptic schemes over Spec($\mathbb{Z}[i]$), although I may be misremembering. | |
| Jan 25, 2014 at 23:04 | history | edited | conduc | CC BY-SA 3.0 |
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| Jan 25, 2014 at 23:04 | comment | added | conduc | Thanks for your answers. I was tacitly assuming that the CM order was maximal. I'm going to edit the question. And I will have a look at Serre-Tate. | |
| Jan 25, 2014 at 22:54 | comment | added | user76758 | @Joe: There is a hidden subtlety in the formulation of the question which "rules out" twisting counterexamples (which is why I focused on non-maximal orders): the OP is beginning with an abelian variety over $\overline{\mathbf{Q}}$ and seems to allow the option to choose whatever descent to a number field we like best. So your (natural) viewpoint of first fixing a descent and then going up from there is sort of "opposite" to how the question is posed. Though I really have no idea if the OP recognizes the subtlety of this aspect of how the question was posed, and whether it is intended. | |
| Jan 25, 2014 at 22:31 | comment | added | Joe Silverman | Actually, the statement for elliptic curves isn't even true for maximal orders. Consider the case that $\text{End}(E)=\mathbb{Z}[i]$, so in the OP's notation, $F=K=\mathbb{Q}(i)$. But $E:y^2=x^3+x$ doesn't acquire good reduction over $\mathbb{Q}(i)$. As Damian has said, you can get an answer from Serre-Tate; roughly speaking, if you take a minimal field of definition for $A$ and then adjoin some torsion, you'll get $F$. | |
| Jan 25, 2014 at 20:41 | comment | added | user76758 | What you say in the elliptic curve case is false, since the CM order might not be maximal (so the $j$-invariant might generate over $K$ a much bigger class field). | |
| Jan 25, 2014 at 20:25 | comment | added | Damian Rössler | See Serre-Tate, "Good reduction of abelian varieties", (Annals of Math. 88, no. 3 (1968)), Th. 7, p. 505. | |
| Jan 25, 2014 at 19:58 | review | First posts | |||
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| Jan 25, 2014 at 19:43 | history | asked | conduc | CC BY-SA 3.0 |