# An elementary problem of proving congruences in a statement about absolute Galois group for $\mathbb{F}_{p}.$

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 $$n\in \mathbb{N}$$. 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 $$m\mid n$$ and $$m\in\mathbb{N}$$ and for any integer $$a$$, there exists $$l$$ s.t. $$a_{l}\not\equiv a \pmod n$$.

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 $$\langle F\rangle$$ generated by $$F$$ is a subgroup of $$\operatorname{Gal}(\overline{\mathbb{F}_{p}}/\mathbb{F}_{p}).$$ If the statement in Question is valid, then $$(F^{a_{i}})_{i}\nsubseteq \langle F\rangle$$.

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 $$\mathbb{Z}$$.
• Firstly I should really thank you for your answer for two questions above, by which I could recall important method learned in bachelor. Secondly, the second part is a beautiful insight which gives an answer for how does the element $(F^{a_{i}})$ of $\hat{\mathbb{Z}}$ constructed. Commented Dec 8, 2018 at 6:05
• That's right, Neukirch constructed a sequence in $\widehat{\mathbb{Z}}$ but not in $\mathbb{Z}$. Infact, $\mathbb{Z}$ is dense in $\widehat{\mathbb{Z}}$, you can try to show this! Commented Dec 8, 2018 at 11:28