# Automorphism group of a formal power series ring

Let $A$ be a commutative ring. Let $A[[x]]$ be the ring of formal power series in one variable. Can we determine the structure of the automorphism group of $A[[x]]$ over $A$?

This is a related question.

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Can you do the case in which $A$ is a field? – Mariano Suárez-Alvarez Jul 21 '12 at 2:52
If you know the answer, please post it. I prefer not to answer this question since I'm the OP. This does not necessarily mean that I know the answer. – Makoto Kato Jul 21 '12 at 3:46
If you browse a bit the site, you will notice that it is usually a good idea (and other users appreciate it) to explain what you already know about what you ask in the question itself. – Mariano Suárez-Alvarez Jul 21 '12 at 4:28
As it should have been made clear to you by now, you can answer your own question. in any case, please ask meta questions in the meta site, not here. – Mariano Suárez-Alvarez Jul 21 '12 at 4:34
Then don't... Is there a point to all this which should not be better treated in the meta site? – Mariano Suárez-Alvarez Jul 21 '12 at 4:36

## 1 Answer

A continuous automorphism $\psi$ of $A[[x]]$ is determined by what it does to $x$ and what it does to the elements of $A$. If $\psi$ leaves elements of $A$ fixed, as I believe you are specifying, then the only question is which elements of $A$ may occur as $x^\psi$ (I write the image of $x$ under $\psi$ this way to avoid later confusion). Since $x$ itself is “analytically nilpotent”, by which I mean that its powers converge to $0$, continuity implies that $x^\psi$ must be a.n. as well. As you know from the comments to your recent post, any such series $u(x)=\sum\alpha_ix^i\in A[[x]]$ will definitely determine a homomorphism from $A[[x]]$ to itself.

Then two questions remain: precisely which are the analytically nilpotent elements of $A[[x]]$, and precisely what characterizes invertible series (to make $\psi$ bijective). If $A$ has discrete topology, then the set of a.n. elements of $A[[x]]$ is just $xA[[x]]$, but for a general topological ring, I’m not at all sure what the answer might be. But suppose $A$ is complete under a topology given by the powers of some ideal $I$ for which $\bigcap_n I^n=(0)$. Then when we call $\sqrt I$ the radical of $I$, that is the set of all elements of $A$ for which some positive power lands in $I$ (so that $\sqrt I$ is an ideal of $A$), then $\sqrt{I}+xA[[x]]$ is the set of all a.n. elements of $A[[x]]$.

What about invertibility of a series $u(x)=\sum_0^\infty\alpha_ix^i$? You’ve already seen that if $A$ has discrete topology and $\alpha_0=0$, $u$ is invertible if and only if $\alpha_1$ has a reciprocal in $A$. Again, in the general case of a topological ring, I don’t know the answer, but in the interesting case that $A$ is complete under the $I$-adic topology mentioned before, all that’s necessary again is that $\alpha_1$ should have reciprocal in $A$. The only proof of this that I know involves Weierstrass Preparation, but I don’t doubt that others have more direct arguments.

As an example of a nontrivial, maybe even nonobvious automorphism of the most general type, consider the $\psi$ for which $x^\psi=u(x)=p -x + x^p\in{\mathbb{Z}}_p[[x]]$. The constant ring is complete under the $(p)$-adic topology, so everything I said before applies. The issue of what kind of noncontinuous automorphisms of $A[[x]]$ there might be, I leave to you.

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Whoops! You don’t need Weierstrass for that: Newton’s method works fine. (Sorry) – Lubin Jul 21 '12 at 20:48