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Suppose $a \in \widehat{\Bbb{Z}}$, say $a = (\overline{a_1}, \overline{a_2}, \overline{a_3}, \overline{a_4}, \dots)$.

I'm wondering if for any $b \in \widehat{\Bbb{Z}}$ whether there exists a number $q \in \Bbb{Z}$ such that:

$$ (\overline{q a_i})_{i \geq 1} = ab = (\overline{a_i b_i})_{i \geq 1} $$

  • We can't simply suppose $q = b_i \pmod i$ because the $i$ are not necessarily coprime (so usual CRT doesn't apply).
  • On the other hand we can! Reason: the CRT system is consistent by definition of inverse system.

See this answer for more information about that.

However, that only applies to the case when the tuple of moduli is finite!

Question. So what about in this infinite case of the moduli $(1,2,3,4, \dots)$?

I am asking this in an attempt to answer this bountied question.

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  • $\begingroup$ I have a hunch that it probably won't work, i.e. there in general does not always exist such an integer. $\endgroup$ Commented Oct 7, 2023 at 21:35
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    $\begingroup$ What do you mean by $\widehat{\mathbb Z}$? Often, it turns out to be (either by definition or provably) to be the product of $\mathbb Z_p$, $p$-adic integers, over all (finite) primes $p$. Your "mod $i$" condition seems to be something else. Can you clarify? :) $\endgroup$ Commented Oct 7, 2023 at 21:45
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    $\begingroup$ @paul: I assume the OP is thinking of the profinite integers as the cofiltered limit of the quotients $\mathbb{Z}/n\mathbb{Z}$. So one can specify a profinite integer by specifying its images $\bmod n$ for every $n$. $\endgroup$ Commented Oct 7, 2023 at 22:03
  • $\begingroup$ @QiaochuYuan, ah, ok, sure, and/but then of_course there are compatibility relations... Oh, well, :) Thanks. :) (Also, then, ordering the moduli by magnitude rather than by divisibility is maybe not so good...) $\endgroup$ Commented Oct 7, 2023 at 22:21

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Edit: Whoops, sorry about that, I didn't see that $a$ was fixed. The answer is still no but we need to do more work.

Recall that by CRT we have an isomorphism $\widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$. If $a = 0$ then the answer is trivially yes so I assume you meant to assume that $a \neq 0$. Then there is at least one prime $p$ such that the image $a_p$ of $a$ in $\mathbb{Z}_p$ is nonzero. Now we can take $b$ to be any element of $\widehat{\mathbb{Z}}$ such that $b_p \in \mathbb{Z}_p \setminus \mathbb{Z}$, which always exists because $\mathbb{Z}_p$ is uncountable, and then if $ba = qa$ then $b_p a_p = q a_p$ in $\mathbb{Z}_p$.

But since $\mathbb{Z}_p$ is an integral domain and $a_p \neq 0$ this gives $b_p = q$; contradiction.

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