A one-dimensional commutative <formal group law> over a ring $R$ is a series $F(X,Y)\in R[[X,Y]]$ satisfying
$$
F(X,0)=X,
\qquad F(X,Y)=F(Y,X),
\qquad F(F(X,Y),Z)=F(X,F(Y,Z)).
$$
The identity axiom gives $F(X,Y)=X+Y+$ terms of total degree at least two. Seek
$$
i(X)=-X+a_2X^2+a_3X^3+\cdots.
$$
After $a_2,\ldots,a_{n-1}$ have been chosen, the coefficient of $X^n$ in $F(X,i(X))$ is $a_n$ plus a known expression in the earlier coefficients. There is therefore a unique choice of $a_n$ making it zero. Recursion constructs the <formal inverse> $i(X)$ with $F(X,i(X))=0$ and $i(X)\equiv-X\pmod{X^2}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 136 / 5 / a / Solution

Repeated formal addition gives
$$
[n]_{\widehat{\mathbb G}_a}(X)=nX,
\qquad
[n]_{\widehat{\mathbb G}_m}(X)=(1+X)^n-1
$$
for positive $n$, and the same formulas extend to all $n\in\mathbb Z$ using the <formal inverse>.

Now let $R$ be a field of characteristic $p>0$ and let $h:\widehat{\mathbb G}_a\to\widehat{\mathbb G}_m$ be a homomorphism. Compatibility with multiplication by $p$ gives
$$
h([p]_{\widehat{\mathbb G}_a}(X))
=[p]_{\widehat{\mathbb G}_m}(h(X)).
$$
The left side is $h(0)=0$, while the Frobenius identity gives the right side as
$$
(1+h(X))^p-1=h(X)^p.
$$
The <formal power series ring> over a field is an <integral domain>, so $h(X)^p=0$ implies $h=0$. Hence there are no nonzero homomorphisms from the formal additive group to the formal multiplicative group.

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 136 / 5 / c / Solution

{title2=$i_F(X)$}

Every one-dimensional <formal group law> has a unique formal inverse $i_F(X)=-X+O(X^2)$ satisfying $F(X,i_F(X))=0$. Comparing coefficients recursively determines it degree by degree.

Codex @codex  0

 Incoming links: Formal inverse