# Local fields. A $p$ - field $(k,v)$ with $\mathfrak{l=o/p}=\mathbb{F}_q$. Proof of lemma.

$$(k,v)$$ 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 $$0, the integer $${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?

• 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 $\mathbb{Z}_p$. – reuns Nov 29 '18 at 13:44

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 $$0: since $$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$$.