Timeline for When does the formal group of an abelian variety possess integral coefficients?
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| Jul 18 at 15:14 | comment | added | Learner | @KeerthiMadapusi, Well, thank you. So formal group is defined more as a functor rather than a formal power series representation, and for every abelian variety the associated formal group is unique but it's formal group laws are not unique, depends on the choice of coordinates and there are several formal group laws (isomorphic to each other) for a formal group. Does this sound okay? | |
| Jul 17 at 20:50 | comment | added | Keerthi Madapusi | @NT2024 Of course, studying the formal group is a very useful thing to do and has a long history in number theory. My point was simply that you should think about it in terms of the formal group rather than the formal group law. The latter does not have a canonical extension from the generic fiber because of choices made in defining it. | |
| Jul 16 at 18:17 | comment | added | Learner | @Lubin, Dear professor Lubin, thank you for the explanation. | |
| Jul 16 at 17:15 | comment | added | Lubin | Coming back to the logarithm issue, I point out that the logarithm of a formal group never has integral coefficients, unless that formal group is isomorphic to the additive f.g. (And thanks, Joe, for putting me right on the other issue.) | |
| Jul 16 at 15:17 | comment | added | Learner | @KeerthiMadapusi, thank you for the above helpful comments. To reply your last comment, I would like to study the torsion points of an abelian variety by studying those of the associated formal group law. My choice is those formal groups having integral coefficients. This was my motivation. | |
| Jul 16 at 14:38 | comment | added | Keerthi Madapusi | I'm not sure how you intend to apply this, but I would be careful about statements saying 'the' formal group law has any properties, since it is not well-defined, except up to isomorphism. | |
| Jul 16 at 14:37 | comment | added | Keerthi Madapusi | $A[[t_1,\ldots,t_n]]\to A[[t_1,\ldots,t_n]]\hat{\otimes}_AA[[t_1,\ldots,t_n]]$ which gives 'the' formal group law. However, since the choice of formal coordinates is not canonical, neither is this group law. On the other hand, it's clear from this that the group law has its coefficients in $A$. | |
| Jul 16 at 14:36 | comment | added | Keerthi Madapusi | The issue here is how one is writing down the coordinates for the formal group law. The formal completion of a smooth scheme $X\to \mathrm{Spec} A$ of dimension $n$ along a section $s:\mathrm{Spec} A \to X$ is always isomorphic to the formal spectrum of $A[[t_1,\ldots,t_n]]$, but this isomorphism is far from canonical. When you do this with a smooth group scheme, the multiplication on the group scheme induces a dual map on the power series ring | |
| Jul 16 at 14:27 | comment | added | Keerthi Madapusi | First, it's useful to make the following remark. In characteristic 0, every formal group law is isomorphic to the additive formal group via the logarithm. The additive formal group law obvious has integer coefficients: it has Z-coefficients! In that sense, your question has a trivial answer. | |
| Jul 16 at 9:46 | comment | added | Learner | @KeerthiMadapusi, can you sketch the proof as answer here or else any reference please? Also, in general, does the "good reduction" property ensure the coefficients of the associated formal group law to be integral? Or one needs more conditions? | |
| Jul 16 at 8:28 | comment | added | Keerthi Madapusi | The Neron model is a smooth group scheme and so its completion along the identity section is automatically a formal group law with integer coefficients. This is entirely formal (pun maybe intended). | |
| Jul 16 at 2:13 | comment | added | Learner | @JoeSilverman, Well, thank you for the helpful comments. | |
| Jul 16 at 1:57 | comment | added | Joe Silverman | I don't know a reference (see the last sentence of my comment), but possibly it's in Néron Models by Siegfried Bosch, Werner Lütkebohmert, Michel Raynaud. As for CM, I'm fairly certain that the endomorphism ring of $A$ injects into the endomorphism ring of its formal group (maybe assusming $A$ is princially polarized?), but again, I don't know a reference offhand. | |
| Jul 16 at 1:52 | comment | added | Learner | @JoeSilverman, thank you. Can you please give a reference to the fact that the formal group law of the Néron model has integral coefficients? Also, can you please confirm if the CM property of an abelian variety is inherited by its formal group law, which I discussed in the question. | |
| Jul 16 at 1:03 | comment | added | Joe Silverman | @Lubin Easy enough to create examples of abelian varieties with everywhere good reduction, just take one with CM and go to a finite extension where (via Neron-Ogg-Shafarevich) it has everywhere good reduction. Answering NT2024: In general, I think that if you take the Néron model of A over Spec$(\mathcal O_K)$, then the formal completion along the zero section will give you a scheme over $\mathcal O_K$ with a formal group law defined over $\mathcal O_K$. (Hopefully someone with more algebraic geometry knowledge than me can confirm or refute this.) | |
| Jul 16 at 0:44 | history | edited | Learner | CC BY-SA 4.0 |
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| Jul 16 at 0:43 | comment | added | Learner | @Lubin, Dear professor Lubin, I indeed meant logarithm with "integral coefficients', that is, if the logarithm $L$ of a formal group $F$ has integral coefficients, then $F(X,Y)=L^{-1}(L(x)+L(Y))$ will have integral coefficients. Next, assume that the abelian variety has good reduction at a prime $\mathfrak{p}$, then we can get a smooth extension of $A$ at the local ring $\mathcal{O}_{K, \mathfrak{p}}$. Can we have integral coefficients of the formal group law? or anyother conditions that ensures it? | |
| Jul 15 at 19:02 | comment | added | Lubin | What do you mean for the logarithm to have integral structure coefficients? And it’s not likely that you’ll find an Abelian variety with good reduction everywhere, I think. | |
| S Jul 15 at 16:56 | review | First questions | |||
| Jul 16 at 6:01 | |||||
| S Jul 15 at 16:56 | history | asked | Learner | CC BY-SA 4.0 |