3
$\begingroup$

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 $(F^{a_{i}})_{i}\nsubseteq \langle F\rangle$.

$\endgroup$

1 Answer 1

3
$\begingroup$

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} $.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – user623904
    Commented Dec 8, 2018 at 6:05
  • $\begingroup$ Thirdly, I think Neukrich actually gave an answer to the following question where some people seems gave some nice talks : math.stackexchange.com/questions/256732/… $\endgroup$
    – user623904
    Commented Dec 8, 2018 at 6:06
  • $\begingroup$ 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! $\endgroup$
    – hellHound
    Commented Dec 8, 2018 at 11:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .