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The first lemma in Lubin-Tate theory says the following:

Let $K$ be a local field, $A$ its ring of integers, and $f\in A[[T]]$ be such that $f(0) = 0$, $f'(0)$ is a uniformizer, and $f$ induces Frobenius over the residue field. Then there exists a unique formal group law $F_f(X,Y)\in A[[X,Y]]$ that makes $f$ into a formal $A$-endomorphism.

If you go over the details of the lemma, you can (I think) generalize it as follows:

If $R$ is any ring, $f\in R[[T]]$ such that $f(0) = 0$ and $f'(0)\in R^\times$ (Edit: $u=f'(0)$ then $u^n - u\in R^\times$ for all $n$), then there exists a unique formal group law $F_f(X,Y)\in R[[X,Y]]$ that makes $f$ into a formal $R$-endomorphism.

The business about uniformizers and Frobenius in the Lubin-Tate lemma is just to ensure that everything converges on the maximal ideal of the ring of integers in the separable closure of $K$, so that you get an actual group.

So this is pretty cool---it says that you can take something purely analytic, $f$, and magically give it an algebraic structure. Specifically, the roots of the iterates $f^{(n)} = f\circ\cdots\circ f$ become a torsion $A$-module.

If the existence of $F_f$ generalizes like I think it does, a natural question is where does $F_f$ converge? I want to be able to answer the question for specific $f$, a simple example would be the following: if $R=\mathbb{C}$ and $f(z) = uz + z^2$, then what can you say about the convergence of $F_f$?

Edit: Okay, $\mathbb{C}$ was a bad choice, but suppose $R$ is a ring complete with respect to some $\mathfrak{a}$-adic topology. Would there be a reason not to study this case? Maybe the question I should be asking is, for what other $R$ and $f$ do people study these formal groups $F_f$?

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Fair enough, I added the extra condition on my claim. As far as the second claim, I only meant the roots when they make sense. The f I'm interested in are polynomials or rational functions. –  Sean Kelly Apr 19 '10 at 2:17
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As far as I know, the generalization of Lubin-Tate to even local fields where the residue field is not finite is still an open problem. The Frobenius condition is very crucial in the proof. –  Tran Chieu Minh Apr 19 '10 at 15:28
    
you CAN talk about roots without convergence matters since you may use Weierstrass preparation to reduce your question to a polynomial times some unit power series. Take the roots to be those of the polynomial. Similarly the Galois action sends units to units, so the Galois action is fine as well. –  olli_jvn Apr 23 '10 at 10:19
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2 Answers 2

I don't think you should be trying to interpret this stuff over the (real or) complex numbers. Discs do not have nice algebraic properties in the archimedean world as they do $p$-adically. Even $p$-adically, one never talks about an actual radius of convergence but just makes sure things converge on the maximal ideal and work there. For example, the Lubin--Tate series could even be a polynomial, with infinite radius of convergence, but still one just focuses on what it does inside the unit disc ($p$-adically).

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It does probably not work in the way you wish.

The crucial bit where convergence is used later in Lubin-Tate is to realize the Galois action via these series [a]. The first-level roots will lie in m minus m^2, so you really want convergence in this radius.

For finding roots itself I think convergence really is not the problem, but having some roots will not help you much if you cannot make the machine realizing the Galois action through power series work.

If you look at the proof (e.g. Yoshida's notes or Milne's notes), at some point one uses that

f(X^q) - f(X)^q

is zero modulo p, which is quite crucial for the convergence - as you say above. But this is not just a sufficient condition, I think if you try a series over some other ring where this fails, you'll really inavoidably get a series which doesn't converge in m. Still your Lubin-Tate polynomial (as proposed by KConrad as a good example above) has roots and all that, good, but there is no way to make the formal O_K-module series [a] act on them. I don't quite remember so well, but I am not even so sure whether it is certain that the extension made from the roots of a Lubin-Tate polynomial/powerseries is Galois anymore in general.

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