I have some questions on formal group laws and their logarithms:

  1. Let $R$ be a graded commutative ring and $F \in R[[X, Y]]$ be a formal group law over $R$ that admits a logarithm. Can you tell me, why this logarithm is unique if $\mathbb{Q} \subset R$ and is unique for $R$ of characteristic $p$ if in addition we demand that the logarithm shall be a power series of the form $x+ a_1x+a_2x^2 +... $ with $a_{p^i-1}=0$ for all $ i \geq 1$? \
  2. If $R$ has characteristic $2$ and and $F$ is a graded formal group law fulfilling $F(x, x)=0$ why does there always exist a logarithm for $F$?

2 Answers 2


1a. If $\mathbb{Q} \subset R$, then you can construct a logarithm for any $F$ by explicit formula $l(x) = \int (1/F_y(x,0))dx$. Here, $$F_y(x,0) = \left. \frac{\partial F(x,y)}{\partial y} \right|_{y=0},$$ and by the fraction I mean the power series obtained from the relation $1/(1-t)=1+t+t^2+\cdots$. (Note that $F_y(x,0)=1+...$.) You need denominators to integrate, of course: $\int x^n dx = x^{n+1}/(n+1)$. Given the existence of a logarithm, you can check that it is unique by using the same technique as in my answer below: two distinct logarithms would give a nontrivial automorphism of $(\mathbb{G}_a)_R$, but any $\mathbb{Q}$-algebra has no (additive) torsion elements, so if $a_i(x+y)^i=a_i(x^i+y^i)$ (for $i>1$) then $a_i=0$.

1b. This should follow from the same sort of argument. These are good practice to work out yourself rather than watching somebody else do, so I'll leave it for you. Hint: Think about my counterexample below.

2) If you write $F(x,y)=\sum a_{ij}x^iy^j$, then the condition $F(x,x)=0$ is equivalent to saying that $\sum_{i+j=n}a_{ij}=0$. On the other hand, note that commutativity implies that $a_{ij}=a_{ji}$, so when $n$ is odd this is automatically satisfied and when $n=2m$ is even it implies that $a_{m,m}=0$. I'm not sure what you mean by a "graded formal group law", but hopefully it will imply that the above formula can be applied here (i.e., only even powers of $x$ show up in $F_y(x,0)$, so that we only need odd denominators, which necessarily exist: if $2=0$ then $2k+1=1$!).


below are answers to old questions; the question has been edited

  1. This is actually not true. If $l_1,l_2:F \rightarrow \mathbb{G}_a$ are two logarithms, then $l_1 \circ l_2^{-1}$ (compositional inverse) is an automorphism of $\mathbb{G}_a$. So the statement is equivalent to saying that these must be trivial. Suppose $l(z)=\sum_{i\geq 1}a_i z^i$ is such an automorphism. Then $l(x+y)=l(x)+l(y)$, so $\sum a_i(x+y)^i = \sum a_i (x^i+y^i)$. So if for example $\mbox{char }R=p$, then we can take $a_{p^j} \not= 0$.

  2. Sorry, your second question seems to have a typo...? If and when you edit it, comment on this question so I'll know to come back and answer.

  • $\begingroup$ Thank you very much! In fact I have already made a mistake in my first question, which I now have fixed. $\endgroup$
    – nick
    Commented May 20, 2012 at 13:24

Here is another answer to 1a, per a document of Neil Strickland.

Let $f(x),g(x) \in R[x]$ be strict isomorphisms $F_0 \to F_A$, and let $g^{-1}(x)$ denote the reverse of $g$. Define $k(x)=f(g^{-1}(x))$. Then $k(x+y) = fg^{-1}(x+y) = f(g^{-1}(x)+_{F_0} g^{-1}(y)) = fg^{-1}(x)+fg^{-1}(y) = k(x)+k(y)$.

Expand this series out and use the fact that binomial coefficients are invertible in $\mathbb{Q}$ and hence $R$. This will give $k(x)=\lambda x$, but since $f$ and $g$ are strict isomorphisms (i.e. $f'(0)=g'(0)=1)$ we can conclude that $\lambda=1$ and so $f$ and $g$ are isomorphisms.

  • $\begingroup$ I think this is just a compactified version of my argument (although I omitted any mention of "strictness"). $\endgroup$ Commented May 21, 2012 at 21:52
  • $\begingroup$ @Aaron: Dear Aaron, you may be correct! $\endgroup$
    – Juan S
    Commented May 27, 2012 at 6:27

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