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The problem

Let $ν$ be a normalised discrete order function of a field $F$. Suppose that $F$ is complete. Put $$R=\left\{x\in F:v\left(x\right)\geqslant0\right\}$$ and $$ M=\left\{x\in F:v\left(x\right)>0\right\}$$ then $R/M$ is a field. Suppose that $R/M$ is finite with $q$ elements, choose $c\in F$ such that $v\left(c\right)=0$. Proof that: the sequence $\left\{c^{q^k}\right\}$ is convergent.

My try

I try to prove that the sequence is Cauchy. I want to prove that for any positive $m$, $$v\left(c^{q^n}-c^{q^{n+m}}\right)\to \infty$$ as $n\to \infty$, with speed irrelevant to $m$. Note that $$c\equiv c^{q^m} \mod M,$$ there will be a $b\in M$ such that $c-c^{q^m}=b$. Therefore for any positive $n$ $$\left(c-c^{q^m}\right)^{q^n}=b^{q^n}$$ and therefore $$v\left ( \sum_{l=0}^{q^n} \binom{q^n}{l} c^{q-l+lq^m} \right ) \geqslant q^n.$$ Put $$ \sum_{l=1}^{q^n-1} \binom{q^n}{l} c^{q-l+lq^m}=A,$$ then if $v\left(A\right)>v\left(c^{q^n}-c^{q^{n+m}}\right)$, $$ v\left(c^{q^n}-c^{q^{n+m}}\right)>v\left(A\right)= v\left ( \sum_{l=0}^{q^n} \binom{q^n}{l} c^{q-l+lq^m} \right ) \geqslant q^n;$$ and if $v\left(A\right)<v\left(c^{q^n}-c^{q^{n+m}}\right)$, then $$ v\left(c^{q^n}-c^{q^{n+m}}\right)=q^n.$$

Now my problem is: I cannot deal with the case $v\left(A\right)=v\left(c^{q^n}-c^{q^{n+m}}\right)$. Any alternative solutions or hints would be greatly appreciated. Thank you very much!

Zhang
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  • Some remarks: (1) It should be $c^{q^n-l+lq^m}$. (2) For even $q$, the power expands as $c^{q^n} + \ldots + c^{q^{n+m}}$. (3) If $F = \Bbb Q_p$, the limit is called the Teichmüller representative of $c$, and the claim can be proved by taking $c \in \Bbb Z$ and noting $\overline{\Bbb Z} = \Bbb Z_p$. Maybe expressing $c$ as a power series in the uniformizer and trimming it after a large enough power would do the trick? – Amateur_Algebraist Feb 17 '24 at 08:27
  • Search for the "Lifting the Exponent" (LTE) lemma. – Torsten Schoeneberg Mar 02 '24 at 00:29

1 Answers1

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This morning I solved it in a different way, so I decide to post it here. First, we shall prove the following lemma.

Lemma For any positive $n$, $$v\left(c^{q^n}-c^{q^{n+1}}\right)\geqslant n.$$

Proof. We shall use induction. For $n=0$, this holds since $c$ and $c^{q}$ belongs to the same equivalence class in $R/M$. Suppose that the proposition is true for $n\leqslant m$, then for $n=m+1$, note that $$v\left(c^{q^n})-c^{q^{n+1}}\right)=v\left(1-c^{q^{n+1}-q^n}\right),$$ and $$c^{q-1}\equiv 1\mod M,$$ write $c^{q-1}=1+d$ for some $d\in M$. Then $$c^{q^{n+1}-q^n}=\left(1+d\right)^{q^n},$$ and hence $$1-c^{q^{n+1}-q^n}=\left(1-\left(1+d\right)^{q^{m-1}}\right)\sum^{q-1}_{k=0}\left(1+d\right)^{kq^{m-1}}. $$ Note that $$\sum^{q-1}_{k=0}\left(1+d\right)^{kq^{m-1}}=q\cdot 1+dA$$ for some $A\in R$, while $q\cdot1=q\in M$, we have $$v\left(\sum^{q-1}_{k=0}\left(1+d\right)^{kq^{m-1}}\right)\geqslant 1.$$ On the other hand, $$v\left(1-\left(1+d\right)^{q^{m-1}}\right)=v\left(c^{q^n}-c^{q^{n-1}}\right).$$ Hence the proposition holds by induction.Q.E.D.

Note that for any positive $k$, $$v\left(c^{q^n}-c^{q^{n+k}}\right)\geqslant \min_{0\leqslant l\leqslant k-1}v\left(c^{q^{n+l}}-c^{q^{n+l+1}}\right)\geqslant n\to \infty,$$ as $n\to \infty$, the sequence is Cauchy and hence converges.

Zhang
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