Vrftsjtryk

I know little bit basic algebraic number theory but do not major in it.
This question might be trivial for the experts. I should be ashamed of my failure in proving it. If it is too easy, please devote me.

Let $p$ be a prime.
Question is the proof of the following statement:
For $nin mathbb{N}$. Let $a_{n}:=n'x_{n}$ where $n'$ and $v_{p}(n)$ is defined s.t. $n'p^{v_{p}(n)}=n$, $(n',p)=1$ and $x_{n}$ is defined s.t. $1=n'x_{n}+p^{v_{p}(n)}y_{n}$.
Then $a_{n}equiv a_{m} pmod m$ for $mmid n$ and $minmathbb{N}$ and for any integer $a$, there exists $l$ s.t. $a_{l}notequiv a pmod n$.

The author failed in dealing with the explicit expressions for $x_{n}$ and $y_{n}$ although there is an algorithm for that and he failed in working out another approach.

Background:
This is one of the steps in a proof from Neukirch's Algebraic Number Theory.
His aim is proving that, if $F$ is Frobenius element, then cyclic group $langle Frangle$ generated by $F$ is a subgroup of $operatorname{Gal}(overline{mathbb{F}_{p}}/mathbb{F}_{p}).$
If the statement in Question is valid, then $(F^{a_{i}})_{i}nsubseteq langle Frangle$.

As far as the proof is concerned, you do not really need explicit expressions for $ x_n, y_n $.

For $ m = m' p^{v_p(m)} $ and $ n = n'p^{v_p(n)} $, if $ m | n $, then we have $ m' | n' $ and $ v_p(m) le v_p(n) $. Now, $$ a_n - a_m = n'x_n - m'x_m $$ and also $$ a_n - a_m = n'x_n - m'x_m = (1-p^{v_p(n)}y_n) - (1-p^{v_p(m)}y_m) = p^{v_p(m)}y_m - p^{v_p(n)}y_n $$ The first equation shows that $ m' | a_n - a_m $ and the second shows that $ p^{v_p(m)} | a_n - a_m $. Together, we get $ m | a_n - a_m $.

Now suppose that $ a $ is an integer such that $ a equiv a_l pmod l $ for all $ l ge 1 $. Note that by the definitions, for all $ k ge 1 $ we have $ a_{p^k} equiv 1 pmod {p^k} $ and whenever $ q $ is a prime not equal to $ p $, $ a_{q^k} equiv 0 pmod {q^k} $ . Then setting $ l = p^k $ for $ k ge 1 $, we get that $ a equiv 1 pmod {p^k} $ for all $ k ge 1 $ which implies $ a = 1 $. But this is impossible as otherwise we'd have as $ a_q equiv 1 pmod q $ for $ q neq p $.

The above might sound uninsightful at first but what is really happening is this: For a finite field $ F $, the absolute Galois group $ text{Gal} (overline{F} / F) $ is isomorphic to $ widehat{mathbb{Z}} $, the profinite completion of $ mathbb{Z} $ given by the inverse limit $ varprojlim mathbb{Z} / n mathbb{Z} $. This is isomorphic to the direct product $ prod_l mathbb{Z}_l $ of the rings of $ l $-adic integers and $ mathbb{Z} $ embeds inside this product diagonally (corresponding to the infinite cyclic group generated by the Frobenius in $ text{Gal} (overline{F} / F) $). In the above example, Neukirch is really choosing an element $ (c_l) in prod_l mathbb{Z}_l $ given by $ c_l = 1 $ when $ l = p $ and $ c_l = 0 $ otherwise, and hence such an element is not in the diagonally embedded $ mathbb{Z} $.