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Over an algebra over a field whose scalar field is $\mathbb{Q}$, the series $\exp (X)-1$ defines an isomorphism from the formal additive group to the formal multiplicative group, with inverse $\log (1+X)$.

Work near the identity $O$ of a generalized Weierstrass equation of an elliptic curve. On the projective chart where $Y\ne 0$, use the local parameter $t=-x/y$ and the auxiliary coordinate $w=-1/y$. Substitution givesThe coefficient of $w$ in the linearized equation at $(0,0)$ is one. Consequently, successive coefficient comparison determines a unique formal power series $w(t)$, without dividing by any coefficient of the equation. Its initial terms areRecover the affine coordinates as Laurent series $x=t/w(t)$ and $y=-1/w(t)$; in particular $x=t^{-2}+O(t^{-1})$ and $y=-t^{-3}+O(t^{-2})$.

The elliptic curve group law is a regular morphism near $(O,O)$. Applying the local parameter to the sum and passing to completed local rings producesOne can construct the series without any nonunit division. PutThe divided difference is an integral formal power series, including its tangent line specialization at $T_1=T_2$. The line $w=\lambda t+\nu$ meets the cubic in the two chosen points and a third point $R$. Substitution in the equation shows that its cubic and quadratic coefficients in $t$ are $-A,-B$, whereSince $\lambda$ has total degree at least two and $\nu$ at least three, $A$ is a unit. The sum of the roots therefore givesThe last expression reflects the third intersection according to the chord-and-tangent group law. Both denominators have constant term one. This proves integrality of the universal construction and also provides an explicit coefficient algorithm. Expanding it, the resulting formal group of an elliptic curve beginsThe integral coefficient recursion and the unit-denominator addition calculation work universally over $\mathbb{Z}[a_1,a_2,a_3,a_4,a_6]$; thus these coefficients specialize in every characteristic of a field. This is the initial coefficients of the elliptic formal group at infinity construction, rather than a definition requiring division by the elliptic-curve discriminant.

The identity, commutativity and associativity of the elliptic curve group law give the defining identities of a one-dimensional commutative formal group law:Reflection in the generalized Weierstrass equation of an elliptic curve gives its formal inverse explicitly:Successive addition gives the multiplication series $[n{]}_F(T)=nT+O(T^2)$.

In characteristic zero, an especially useful description comes from the invariant differential on an elliptic curve:Integrating with zero constant term constructs the formal logarithmTranslation invariance of the invariant differential on an elliptic curve implies $L(F(X,Y))=L(X)+L(Y)$. Thus the formal logarithm converts the formal group law into the formal additive group in characteristic zero. Expanding this identity to degree three also recovers the displayed coefficients of $F$; in positive characteristic the integral addition construction remains valid although the logarithmic denominators may not exist.

Here a formal group law means a one-dimensional commutative law $F(X,Y)\in R[[X,Y]]$ with $F(X,0)=X$, $F(0,Y)=Y$, $F(X,Y)=F(Y,X)$ and $F(F(X,Y),Z)=F(X,F(Y,Z))$. It has leading part $X+Y$ and a recursively determined formal inverse $i_F(T)=-T+O(T^2)$. A homomorphism $h(T)\in TR[[T]]$ satisfies $h(F(X,Y))=G(h(X),h(Y))$. It is an isomorphism precisely when its linear coefficient is a unit of $R$: necessity follows by composing with an inverse, while sufficiency follows by recursive inversion of the series and then the homomorphism identity. Over a field the condition is $\boxed{h^{\prime }(0)\ne 0}$.

The characteristic-zero classification is that every such law over $K$ has a unique isomorphism to the formal additive group with linear coefficient one. This is its formal logarithm. To prove it, put $f(T)={\partial }_{Y}F(T,0)$, whose constant term is one, and defineIntegration is coefficientwise, which is possible over the characteristic-zero field. Differentiate associativity with respect to $Z$ at zero to obtainIt follows that ${\partial }_Y{\ell }_F(F(X,Y))=1/f(Y)={\ell }_F^{\prime }(Y)$. The difference ${\ell }_F(F(X,Y))-{\ell }_F(Y)$ is therefore independent of $Y$; evaluating at $Y=0$ givesIts linear coefficient one makes it invertible as a formal series. Conversely, a homomorphism from $F$ to the additive law with derivative $c$ satisfies $h^{\prime }(T)f(T)=c$, by differentiating its defining identity at the second variable zero. Thus $h=c{\ell }_F$. In particular the normalized logarithm is unique, and every homomorphism between two laws over $K$ is ${\ell }_G^{-1}(c{\ell }_F(T))$, with an isomorphism exactly when $c\ne 0$. This is a formal classification; analytic convergence is a separate issue.

Now suppose $F$ is defined over ${\mathcal{O}}_K$. Write $v_K(\pi )=1$, $e=v_K(p)$ and let the residue field have size $q$. Since $f(T)$ and its reciprocal have integral coefficients, writing ${\ell }_F(T)=T+{\sum }_{n\ge 2}{c}_n{T}^n$ gives the valuation bound for coefficients of an integral formal logarithmTherefore the logarithm converges on $\pi {\mathcal{O}}_K$, as $n{v}_K(t)-e{v}_p(n)$ tends to infinity. Choose an integer $s>e/(p-1)$ and put $g(t)={\ell }_F(t)-t$. For $t,u\in {\pi }^s{\mathcal{O}}_K$, factoring $t^n-u^n$ givesThe last inequality uses $v_p(n)\le (n-1)/(p-1)$, since $n\ge p^{v_p(n)}\ge 1+(p-1)v_p(n)$. It follows that $g$ maps this ball into ${\pi }^{s+1}{\mathcal{O}}_K$ and is strictly contracting, gaining at least one unit of valuation in differences.

For any $z\in {\pi }^s{\mathcal{O}}_K$, iterate $t_{j+1}=z-g(t_j)$ from $t_0=z$. The iterates remain in the ball, their differences gain valuation at each step, and completeness gives a limit solving ${\ell }_F(t)=z$. The same difference estimate makes this solution unique. Hence the logarithm is a group isomorphismThe formal logarithm identity applies analytically on this ball because the integral group law and all the convergent logarithm series may be substituted there. Finally reduction of the formal parameter modulo ${\pi }^s$ is a group homomorphism from $F(\pi {\mathcal{O}}_K)$ onto the finite group on $\pi {\mathcal{O}}_K/{\pi }^s{\mathcal{O}}_K$, with kernel $F({\pi }^s{\mathcal{O}}_K)$. Its index is $q^{s-1}$. Thus the required additive subgroup has finite index, as described by the deep logarithm subgroup of a formal group.

In this setting a formal group law means a one-dimensional commutative series $F(X,Y)\in R[[X,Y]]$ with identity zero andIt starts with $X+Y$; there is a unique inverse series $\iota (T)=-T+\cdots$ satisfying $F(T,\iota (T))=0$. For coefficients in $\mathbb{Z}$, the series and its inverse converge on $p{\mathbb{Z}}_p$, making this ideal a topological abelian group under $F$.

Set $a(T)={\partial }_{Y}F(T,0)\in 1+T\mathbb{Z}[[T]]$, and write $a(T{)}^{-1}={\sum }_{j\ge 0}{b}_j{T}^j$. The formal logarithm isDifferentiate associativity in the third variable at zero. It gives $a(F(X,Y))={\partial }_{Y}F(X,Y)a(Y)$, so ${\partial }_{Y}L(F(X,Y))=L^{\prime }(Y)$. Subtracting $L(Y)$ leaves a series independent of $Y$, whose value at zero is $L(X)$. Thus $L(F(X,Y))=L(X)+L(Y)$.

For odd $p$, the logarithm converges on $I=p{\mathbb{Z}}_p$: the valuation of its degree-$n$ term is at least $n-v_p(n)$, which tends to infinity. Put $u(T)=L(T)-T$. For $x,y\in I$, factor $x^n-y^n$ to obtainfor every $n\ge 2$ and odd $p$. Hence $u$ is a contraction with constant at most $1/p$ and maps $I$ into $pI$. The equation $L(x)=z$ is equivalent to $x=z-u(x)$; the contraction mapping theorem gives a unique solution in $I$ for every $z\in I$. The same estimate shows $|L(x)-L(y){|}_p=|x-y{|}_p$, so the logarithm is a topological group isomorphism. Dividing its values by $p$ givesThis is the deep logarithm subgroup of a formal group argument, here with depth one.

For $p=2$, the degree-two estimate at depth one need not be strict. At depth two, however,so the same proof gives $F(4{\mathbb{Z}}_2)\cong (4{\mathbb{Z}}_2,+)\cong {\mathbb{Z}}_2$. This subgroup has index two in $F(2{\mathbb{Z}}_2)$, because $F(x,y)\equiv x+y(\mathrm{mod}4)$ when $x,y\in 2{\mathbb{Z}}_2$.

Indeed the topological structure of a formal group on twice the 2-adic integers has just two possibilities. Let $G=F(2{\mathbb{Z}}_2)$, let its index-two subgroup $H$ have topological generator $h$, and choose $v\notin H$. Write $2v=ch$, with $c\in {\mathbb{Z}}_2$, using group notation. If $c$ is odd, $2v$ generates $H$ and $v$ generates $G$, giving $G\cong {\mathbb{Z}}_2$. If $c$ is even, $v-(c/2)h$ has order two and lies outside $H$, giving $G\cong H\times \mathbb{Z}/2\mathbb{Z}$. The ${\mathbb{Z}}_2$ multiples are defined by continuity in this compact pro-two group.

For explicit contrasting examples, the formal additive group $F_a(X,Y)=X+Y$ gives $F_a(2{\mathbb{Z}}_2)\cong {\mathbb{Z}}_2$, which is a torsion-free group. The formal multiplicative group $F_m(X,Y)=X+Y+XY$ is identified by $x\mapsto 1+x$ with $1+2{\mathbb{Z}}_2$. Its element $x=-2$ corresponds to $-1$ and has order two. Moreoverand the logarithm identifies the second factor with $4{\mathbb{Z}}_2$. The two resulting groups are ${\mathbb{Z}}_2$ and $\mathbb{Z}/2\mathbb{Z}\times {\mathbb{Z}}_2$, so they are not isomorphic.

The formal additive group and formal multiplicative group have lawsIf $R$ is an algebra over a field whose scalar field is $\mathbb{Q}$, the serieshas linear coefficient one and satisfiesIt is therefore the exponential isomorphism between the additive and multiplicative formal groups, with inverse $\log (1+X)$.

For a one-dimensional commutative integral formal group law, the formal logarithm identifies $F(4{\mathbb{Z}}_2)$ with $4{\mathbb{Z}}_2$. This is an index-two subgroup of $F(2{\mathbb{Z}}_2)$. Choose a topological generator $h$ of the subgroup and an element $v$ outside it, and write $2v=ch$. If $c$ is odd, $v$ generates the whole group. If $c$ is even, $v-(c/2)h$ has order two and splits off a cyclic factor. The formal additive group and formal multiplicative group realize the two cases.