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