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I am looking for a sufficient condition that ensures the formal group associated with an abelian variety has integral coefficients.

In precise, let $A$ be an abelian variety over a number field $K$ with ring of integers $\mathcal{O}_K$ and let $\hat{A}$ be the associated formal group of $A$. Here $\hat{A}$ is a formal power series of the form $$\hat{A}[[X,Y]]=X+Y+\text{higher degree terms}=\sum_{i,j \geq 0} a_{ij}X^iY^j \in K[[X,Y]].$$ Question: When does $a_{ij} \in \mathcal{O}_K$ for all $i,j$?

Note: A formal group will have integral coefficients if its logarithm has so.

I am assuming following two conditions:

  • $A$ has good reduction at all primes of the number field $K$

  • $A$ has CM property by the ring of integers of $K$.

Here is what I feel: (please check)

First case: Assume $A$ has good reduction at all primes of $K$, then $A$ can be extended to a smooth model over $\mathcal{O}_K$. That is, for every prime $\mathfrak{p}$ we get an abelian scheme $\mathcal{A}$ over the local ring $\mathcal{O}_{K, \mathfrak{p}}$, whose generic fiber is $A$.

Now the formal group $\hat{\mathcal{A}}$ of $A$ at the prime $\mathfrak{p}$ is obtained by the formal completion of $\mathcal{A}$. I think now the smoothness property ensures the coefficients of the formal group $\hat{\mathcal{A}}$ has coefficients in $\mathcal{O}_{K, \mathfrak{p}}$.

Second case: The endomorphisms of $A$ induces corresponding endomorphisms on its formal group $\hat{A}$. If $A$ has complex multiplication by the ring of integers $\mathcal{O}_K$, then $\mathcal{O}_K$ acts on $\hat{A}$ as well. So $\hat{A}$ inherits the CM property of $A$ as well. I am not sure if it give any information about the coefficients of $\hat{A}$. I don't know if the logarithm of $\hat{A}$ has integral coefficients.

Thanks

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  • $\begingroup$ 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. $\endgroup$
    – Lubin
    Commented Jul 15 at 19:02
  • $\begingroup$ @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? $\endgroup$
    – Learner
    Commented Jul 16 at 0:43
  • $\begingroup$ @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.) $\endgroup$
    – Joe Silverman
    Commented Jul 16 at 1:03
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    $\begingroup$ 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). $\endgroup$
    – Keerthi Madapusi
    Commented Jul 16 at 8:28
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    $\begingroup$ 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. $\endgroup$
    – Keerthi Madapusi
    Commented Jul 16 at 14:38

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