Timeline for Automorphisms of k((X))
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9 events
| when toggle format | what | by | license | comment | |
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| Jan 13, 2015 at 0:15 | comment | added | KConrad | Sure, for any field $k$ every automorphism $\sigma$ of $k((X))$ satisfying $\sigma(k) = k$ must be $X$-adically continuous since all elements of $1 + Xk[[X]]$ are $n$-th powers for infinitely many $n > 0$ (if $k$ has characteristic $0$ use all $n > 0$ if $k$ has characteristic $p$ use all $n$ not divisible by $p$). When $k$ is perfect of positive characteristic then the condition $\sigma(k) = k$ can be proved rather than be taken as a hypothesis. | |
| Jan 12, 2015 at 22:52 | comment | added | YCor | If $k$ is not assumed perfect this still seems to prove that $k$-linear field automorphisms of $k((X))$ preserve $k[[X]]$ (as the $k$-subspace generated by elements that admits $n$-th roots for infinitely many $n$) and hence the continuity result. | |
| Jan 12, 2015 at 20:42 | comment | added | KConrad | Of course if we don't assume $\sigma$ is $k$-linear then the same argument works to prove $|\sigma(X)|_X < 1$ by writing $c = \sigma(b)$ for some $b \in k^\times$ (since $\sigma$ has to restrict to an automorphism of $k$) and looking at $\sigma(X/b) = 1 + Xg/c$. | |
| Jan 12, 2015 at 20:40 | comment | added | KConrad | Now we can show $X$-adic continuity. It suffices to prove $|\sigma(X)|_X < 1$. From the previous comment, $|\sigma(X)|_X \leq 1$. Suppose $|\sigma(X)|_X = 1$. Then $\sigma(X) = c + Xg$ where $c \in k^\times$ and $g \in k[[X]]$. Since $\sigma$ is assumed in the question to be the identity on $k$, $\sigma(X/c) = 1 + Xg/c$, so $\sigma(X/c)$ is an $n$th power for infinitely many $n > 0$, and because $\sigma$ is an automorphism we get that $X/c$ is an $n$th power for infinitely many $n > 0$, which is of course false (even for a single $n > 0$ there's a contradiction). Hence $|\sigma(X)|_X < 1$. | |
| Jan 12, 2015 at 20:35 | comment | added | KConrad | As a follow-up to user74230's comment, any abstract automorphism $\sigma$ of $k((X))$ must preserve $k[[X]]$. First suppose $f \in 1 + Xk[[X]]$. Then $f$ is an $n$th power in $k((X))$ for infinitely many $n > 0$ (in fact for all $n$ not divisible by $p$), so $\sigma(f)$ has the same property, which implies $|\sigma(f)|_X = 1$. Next, if $|f|_X < 1$ then $1 + f \in 1 + Xk[[X]]$, so $|\sigma(1+f)|_X = 1$, which implies $|\sigma(f)|_X \leq 1$. Finally, if $|f|_X = 1$ then $f = c + g$ where $c \in k^\times$ and $|g|_X < 1$, so from $\sigma(f) = \sigma(c) + \sigma(g)$ we have $|\sigma(f)|_X \leq 1$. | |
| Jan 12, 2015 at 18:20 | comment | added | user74230 | The hypothesis of restricting to the identity on $k$ is unnecessary since $k$ is the subset of elements that are $p^n$th-powers for all $n > 0$ (ensuring that $k$ has to be carried onto itself by any automorphism of $K$, so if $\sigma$ is that restriction to $k$ then composing with the continuous $\sum a_n X^n \mapsto \sum \sigma^{-1}(a_n) X^n$ brings us to the case $\sigma=1$). | |
| Jan 12, 2015 at 18:15 | comment | added | Yiftach Barnea | Kevin, you can try to check one of Benjamin Klopsch's papers. I think he worked on such groups in his phd thesis, but I am not sure what he proved and where he published it. | |
| Jan 12, 2015 at 18:07 | review | First posts | |||
| Jan 12, 2015 at 18:11 | |||||
| Jan 12, 2015 at 18:06 | history | asked | Kevin Keating | CC BY-SA 3.0 |