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I'm looking for a good reference for the following fact:

Let $k$ be a perfect field of characteristic $p$ and let $K=k((X))$. Then every $k$-linear automorphism of $K$ is continuous with respect to the usual valuation on $K$.

R. Camina sketches a proof (New Horizons in pro-$p$ Groups, p. 206), but the statement applies only to the case where $k$ is finite, and the proof depends on the uniqueness of the standard valuation on $K$, (No reference is given for this last fact.)

Is there anything more definitive in the literature?

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  • $\begingroup$ 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. $\endgroup$
    – Yiftach Barnea
    Commented Jan 12, 2015 at 18:15
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    $\begingroup$ 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 $\sigma=1$). $\endgroup$
    – user74230
    Commented Jan 12, 2015 at 18:20
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    $\begingroup$ 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 $|\sigma(f)|_X \leq 1$. $\endgroup$
    – KConrad
    Commented Jan 12, 2015 at 20:35
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    $\begingroup$ 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 $|\sigma(X)|_X < 1$. $\endgroup$
    – KConrad
    Commented Jan 12, 2015 at 20:40
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    $\begingroup$ 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 $\sigma(X/b) = 1 + Xg/c$. $\endgroup$
    – KConrad
    Commented Jan 12, 2015 at 20:42

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