$(k,v)$ i a local field, $\mathfrak{p} = \{x | x \in k, v(x) > 0 \}$, $\mathfrak{o}=\{x | x\in k, v(x) \geq 0\}$. I'm working on Local Fields and I don't understand few things in proof of this Lemma:

Let $\mathfrak{l=o/p}=\mathbb{F}_q$ for a $p$-field $(k,v)$. Then, for each $x \in \mathfrak{o}$, the limit $\omega(x) = \lim_{n\to\infty}x^{q^n}$ exists in $\mathfrak{o}$, and the map $\omega :\mathfrak{o} \to \mathfrak{o} $ has the following properties: $\omega(x)\equiv x \ mod \ \mathfrak{p}$, $ \ $ $\omega(x)^q = \omega(x) \ $, $\omega(xy)=\omega(x)\omega(y)$.

Proof: By induction, we shall prove the congruences $x^{q^n}\equiv \ x^{q^{n-1}} \ mod \ \mathfrak{p}^n$ for all $n\geq 1$. For $n=1$,$x^q \equiv x \ mod \ \mathfrak{p}$ follows from th fact that $\mathfrak{l=o/p}$ is a finite field with $q$ elements. Assume the congruence for $n \geq 1$ so that $x^{q^{n}}=x^{q^{n-1}}+y$ with $y \in \mathfrak{p}^n$. Then $$x^{q^{n+1}} = \sum_{i=0}^{q}{q\choose i}x^{iq^{n-1}}y^{q-i}.$$ For $0<i<q$, the integer $${q\choose i}=\frac{q}{i}{q-1\choose i-1} $$ is divisble by $p$ so that ${q\choose i}y^{q-i}$ is contained in $\mathfrak{p}^{n+1}$. Since the same is obviously true for $i=0$, we obtain $$ x^{q^{n+1}} \equiv x^{q^{n}}mod \ \mathfrak{p}^{n+1}.$$ Now, we see from these congurences that $\{x^{q^{n}} \}_{n \geq 1 }$ is a Cauchy sequence in $\mathfrak{o}$ in the $v$ - topology. As $v$ is complete and $\mathfrak{o}$ is closed in $k$, the sequence converges to an element $\omega(x)$ in $\mathfrak{o}$. It is clear that congruences yield $x^{q^{n}} \equiv x \ mod \ \mathfrak{p}$ for all $n \geq 1$. Hence $\omega(x) \equiv x \ mod \ \mathfrak{p} $. We also see $$\omega(x)^q = lim_{n \to \infty}x^{q^{n+1}} = \omega(x), \ \omega(x)(y) = lim_{n \to \infty}x^{q^{n}}y^{q^{n}} = \omega(x)\omega(y).$$ QED

1) Why we are interested in congruences modulo $\mathfrak{p}^n$ ?

2) Why divisibility of ${q\choose i}$ by $p$ implies that ${q\choose i}y^{q-i}$ is contained in $\mathfrak{p}^{n+1}$?

3) Why we see from these congruences that $\{x^{q^{n}} \}_{n \geq 1 }$ is a Cauchy sequence in $\mathfrak{o}$ in the $v$ - topology?

  • $\begingroup$ To make things concrete pick an element $\pi \in \mathfrak{p}, \pi \not \in \mathfrak{p}$ then $ \mathfrak{p} = (\pi)$ and everything works exactly as in $\mathbb{Z}_p$. $\endgroup$ – reuns Nov 29 '18 at 13:44

For 1) and 3)

Recall that a sequence $(x_n)$ of elements of $k$ is convergent iff there exist an element $l\in k$ such that $v(l-x_n)\underset{n\to+\infty}{\longrightarrow}+\infty$ and is Cauchy iff $v(x_{n+1}-x_n)\underset{n\to+\infty}{\longrightarrow}+\infty$. Moreover, since the valution $v$ is discrete, we may assume that $v(k^*)=\mathbb{Z}$ and so for every integer $n$, we have ${\frak{p}}^n=\{x\in k\vert v(x)\geqslant n\}$. In particular, if $(x_n)$ is a sequence of $\frak{o}$ such that for all $n\geqslant 1$, $x_{n+1}\equiv x_n \pmod{{\frak{p}}^n}$ then $x_{n+1}-x_n \in {\frak{p}}^n$ ie $v(x_{n+1}-x_n)\geqslant n$ so $(x_n)$ is a Cauchy sequence.

For 2)

For $0<i<q$: since $p$ is in $\frak{p}$ and $\frak{p}$ is an ideal of $\frak{o}$, the divisibility of $\binom{q}{i}$ by $p$ implies that $\binom{q}{i}$ is in $\frak{p}$. Likewise, ${\frak{p}}^n$ is an ideal, $y\in{\frak{p}}^n$ and $q-i\geqslant 1$, so $y^{q-i}\in{\frak{p}}^n$ hence $\binom{q}{i}y^{q-i}\in {\frak{p}}{\frak{p}}^n={\frak{p}}^{n+1}$. For $i=0$: $y^q \in {\frak{p}}^{qn}\subseteq {\frak{p}}^{n+1}$ since $qn\geqslant 2n\geqslant n+1$.


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