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Let $k$ be a field. I have two questions about the formal power series $k[[x]]$.

  1. How to determine elements in $\mathrm{Aut}(k[[x]])$?

  2. How to determine elements in this set:$$\{f\in k[[x]]: \exists g\in k[[x]] \text{ s.t. } f(g(x))=g(f(x))=x\}$$

It seems these two questions are related. Could you provide some help? Thanks!

user26857
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Richard
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    For question two: https://en.wikipedia.org/wiki/Formal_power_series#Composition_inverse – Chris Sanders Feb 19 '22 at 06:42
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    For question one: therefore the automorphisms of $(k[[x]])$ are precisely those ring endomorphisms that send $x$ to one of these compositionally invertible elements – Chris Sanders Feb 19 '22 at 06:46
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    Abstract duplicate of [The group of $k$-automorphisms of $k[[x,y]]$, $k$ is a field](https://math.stackexchange.com/questions/2225740/the-group-of-k-automorphisms-of-kx-y-k-is-a-field) – KReiser Feb 19 '22 at 09:13
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    So the 3 steps are: any $k$-automorphism sends $x$ to a non-unit, in $xk[[x]]$, so it is continuous determined by where it sends $x$, for being surjective it must not be in $x^2k[[x]]$, and any such element of $xk[[x]]-x^2k[[x]]$ has a compositional inverse. – reuns Feb 19 '22 at 09:25

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