Jtdylktuy

$(k,v)$ i a local field, $mathfrak{p} = {x | x in k, v(x) > 0 }$, $mathfrak{o}={x | xin k, v(x) geq 0}$.
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 $omega(x) = lim_{ntoinfty}x^{q^n}$ 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 $ngeq 1$. 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
$$x^{q^{n+1}} = sum_{i=0}^{q}{qchoose i}x^{iq^{n-1}}y^{q-i}.$$
For $0<i<q$, the integer $${qchoose i}=frac{q}{i}{q-1choose i-1} $$ is divisble by $p$ so that ${qchoose i}y^{q-i}$ 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 ${qchoose i}$ by $p$ implies that ${qchoose i}y^{q-i}$ 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?

For 1) and 3)

Recall that a sequence $(x_n)$ of elements of $k$ is convergent iff there exist an element $lin k$ such that $v(l-x_n)underset{nto+infty}{longrightarrow}+infty$ and is Cauchy iff $v(x_{n+1}-x_n)underset{nto+infty}{longrightarrow}+infty$. Moreover, since the valution $v$ is discrete, we may assume that $v(k^*)=mathbb{Z}$ and so for every integer $n$, we have ${frak{p}}^n={xin kvert v(x)geqslant n}$. In particular, if $(x_n)$ is a sequence of $frak{o}$ such that for all $ngeqslant 1$, $x_{n+1}equiv x_n pmod{{frak{p}}^n}$ 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, $yin{frak{p}}^n$ and $q-igeqslant 1$, 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 $qngeqslant 2ngeqslant n+1$.