# What is the classification of 1-dimensional commutative formal group laws over $\mathbb{Z}$ up to isomorphism?

All 1-dimensional commutative formal group laws over a field $k$ of characteristic $\geq$ 0 are classified up to isomorphism by the characteristic of $k$ and their height. This is result of Michel Lazard.

Additionally, all 1-d commutative formal group laws over $\mathbb{Q}$ are isomorphic.

What is known about the classification of 1-dimensional commutative formal group laws over rings of characteristic 0?

In general, classifying 1-dimensional commutative formal group laws over a general $$\mathbb{Z}$$-flat characteristic $$0$$ ring $$R$$ can be complicated. If $$R$$ contains a copy of $$\mathbb{Q}$$, then it is easy, as the logarithm, $$f : F(X,Y) \rightarrow \widehat{\mathbb{G}}_a,$$ provides an isomorphism with the additive formal group law. However, if not, then there can be many non-isomorphic formal group laws, and for many rings a classification is not known.
As an example where the classification is known, but is more complicated, consider $$R = \mathbb{Z}_p$$.
Here it is a result of Hazewinkel that two formal group laws over $$\mathbb{Z}_p$$ are isomorphic if and only if their reductions to $$\mathbb{F}_p$$ are isomorphic. So the classification is the same as the classification of formal group laws over $$\mathbb{F}_p$$.
Isomorphism classes of $$1$$-dimensional formal group laws over $$\mathbb{F}_p$$ are in bijection with Eisenstein polynomials with coefficients in $$\mathbb{Z}_p$$. The height of the formal group law is equal to the degree of the corresponding Eisenstein polynomial. This can be found in Hazewinkel's book.
Note that, contrary to the OP, whilst isomorphic formal groups over a characteristic $$p$$ field always have the same height, the height only classifies formal group laws over a separably closed field.