2
$\begingroup$

It is well known that the profinite completion of the integers, $\hat{\mathbb{Z}} = \varprojlim \mathbb{Z}/n\mathbb{Z}$ is, by the Chinese remainder theorem, isomorphic to $\prod_p \mathbb{Z}_p$.

I have two questions about this. 1) Is this ring isomorphism also topological? and 2) does it generalize to the ring of integers in a global field? Id est, if $\mathcal{O} \subset k$ is the ring of integers in a number field or global function field over a finite field, do we know that $\hat{\mathcal{O}}\cong \prod_\mathfrak{p} \mathcal{O}_\mathfrak{p}$, where $\mathfrak{p}$ varies over all (finite?) primes and $\mathcal{O}_\mathfrak{p}$ is the local completion with respect to $\mathfrak{p}$ ?

Since $\hat{\mathcal{O}} = \varprojlim \mathcal{O}/\mathfrak{n}$ and since we have a unique factorisation of ideals into prime ideals, I feel like the CRT would give us this same result. Is my intuition correct?

$\endgroup$
  • $\begingroup$ Yes, your intuition is correct... what fuller answer, if any, would be useful? $\endgroup$ – paul garrett Aug 3 '15 at 13:24
  • $\begingroup$ A "yes" to a yes/no question is sufficient. But the topological point is not clear to me. $\endgroup$ – mebassett Aug 3 '15 at 13:26
2
$\begingroup$

Certainly more discussion of various details could be warranted, depending on taste/interest, but the basic topological point is that those finite quotients $\mathfrak O/\mathfrak n$ have a unique locally compact Hausdorff topology, namely, the discrete topology. Projective limits of topological rings (with identities $1$ mapping to each other, for example) exist in a category of locally compact Hausdorff topological rings: the closed subset of the product defined by the transition maps in the limit.

Products and (projective) limits are both instances of "limit" in a somewhat larger sense, and "commute", so Sun-Ze's theorem (a.k.a. "Chinese Remainder Theorem") gives the indicated factorization.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.