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The following is part of an exercise from Lenstra's Galois theory for schemes..

Let $a=\frac{b}{c}\in \mathbf{Q}^\times$, $n\in \widehat{\mathbf{Z}}^\times$.

Prove that there exists a sequence of positive integers $(n_i)_{i\geq 0}$ that satisfies $\operatorname{gcd}(n_i,2bc)=1$ for all $i$ and such that $n=\lim_{i\to\infty} n_i$ in $\widehat{\mathbf{Z}}$.

First of all, I have trouble understanding convergence of sequences in $\widehat{\mathbf{Z}}$. A similar exercise on convergence is treated here and here, but I don't seem to understand this.

Of course, taking the definition of convergence using the topology would give: $(a_n)$ converges (to $0$ for example) if and only if for every open set $U\ni 0$, $\exists N$ such that $n\geq N$ implies $a_n\in U$. We can know use that an open neighbourhood is given by finite intersections of $Ker \pi_k$ and so on, but this seems so complicated..

A claim in solutions I found on the internet says the following: a sequence $(a_k)$ converges in $\widehat{\mathbf{Z}}$ if and only if for all $n$, the sequence of reductions $(\pi_n(a_k))_{k\geq 1}$ in $\mathbf{Z}/n\mathbf{Z}$. I don't see why this claim is true. (One direction is trivial, since the projections are continuous.)

Could someone give and explain thoroughly how we should attack proving convergence of sequences in $\widehat{\mathbf{Z}}$ and give a hint for this exercise? I think we should use density of $\mathbf{Z}$ in $\widehat{\mathbf{Z}}$, but then I don't see how to incorporate the other conditions?

Dr. Heinz Doofenshmirtz
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1 Answers1

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"I don't see why this is true" : this is the definition of $\mathbf{\widehat{Z}=\varprojlim Z/mZ}$, the set of limits of sequences of integers that converge $\bmod m$ for all $m$. This proves that the $\gcd(n_i,bc)=1$ condition works iff $p | bc \implies n\not \equiv 0\bmod p$.

When restricting to $n\in \mathbf{\widehat{Z}}^\times$ this condition is satisfied.

Concretely let $ n_i$ such that $n \equiv n_i \bmod i!$ and $i > bc$.

reuns
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  • Thanks a lot for your answer! I am afraid I don't see why this is the definition. The definition I know is $\varprojlim \mathbb{Z}/n\mathbb{Z}={ (x_n){n\geq 1} \in \prod{m\geq 1} \mathbb{Z}/m\mathbb{Z} \mid x_k=x_\ell \bmod{k}\text{ for all }k\leq \ell }.$ – Dr. Heinz Doofenshmirtz May 14 '20 at 09:32
  • This is the same ! From a convergent sequence you get yours by reducing the limits $\bmod$ each $m$, and from yours you get a convergent sequence from the reduction $\bmod$ each $ i!$ – reuns May 14 '20 at 09:38
  • I'm sorry but I really don't understand :( It would be great if you could elaborate the most possible – Dr. Heinz Doofenshmirtz May 14 '20 at 10:02
  • What ? Take any integer $n_i$ representing $x_{i!}\bmod i!$ then $\lim_{i\to \infty} n_i$ converges to $x$ in the profinite integers. The next step is to construct some profinite integers, for example the one which is $\equiv 1\bmod 2^k$ for all $k$ and $\equiv 0\bmod p^k$ for all odd prime. – reuns May 14 '20 at 10:07
  • But my whole point is: what does convergence in the profinite integers mean? I understand that: "From a convergent sequence you get yours by reducing the limits mod each " because projection is continuous. But " from yours you get a convergent sequence from the reduction mod each !"? Where does $i!$ come from? How do we know that we get a convergent sequence if we don't know what convergence means? You say : "by definition", but this isn't the definition right? – Dr. Heinz Doofenshmirtz May 14 '20 at 11:14
  • $i! $ is divisible by $m$ once $i$ is large enough, it is a factorial, the reduction $\bmod i!$ contains the information $\bmod m$ whenever $m\ |\ i!$ – reuns May 14 '20 at 12:07