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I am reading Chapter 15 (page 319) of Cox's Book Primes of the Form $x^2+ny^2$. Given $K$ a quadratic field and $O$ an order of $K$, he defines the ring of $O$-adeles as $\hat O=O\otimes_{\mathbb Z}\hat {\mathbb Z} $ where $\hat{\mathbb Z}=\prod_p\mathbb Z_p$.

He says: since $O$ is a free $\mathbb Z$-module we have $\hat O=\prod_pO_p$ where $O_p=O\otimes_{\mathbb Z}\mathbb Z_p$.

I know the very basic stuff on tensor products, i.e. that tensor product preserves direct sums and that $\mathbb Z\otimes_{\mathbb Z} M=M$, but I can't explain his statement since I don't know how tensor product works with infinite products for example and how to use that $O$ is a free $\mathbb Z$-module.

user26857
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Blacksmith
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1 Answers1

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You need to know that $\mathcal{O}$ is a finite free module. In general, tensoring with a module doesn't infinite products unless the module is finitely presented; see this answer for a proof.

In this case we can be more explicit. There is a natural map

$$\mathcal{O} \otimes \widehat{\mathbb{Z}} \to \prod_p \mathcal{O}_p$$

and to show that it's bijective we describe the underlying map of abelian groups as

$$\mathcal{O} \otimes \widehat{\mathbb{Z}} \cong \mathbb{Z}^2 \otimes \widehat{\mathbb{Z}} \cong \widehat{\mathbb{Z}}^2 \cong \prod_p \mathbb{Z}_p^2 \cong \prod_p \mathbb{Z}^2 \otimes \mathbb{Z}_p \cong \prod_p \mathcal{O}_p.$$

This is a composition of isomorphisms (of abelian groups) hence is an isomorphism (of rings).

Qiaochu Yuan
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  • Thank you very much. Only one thing: how do you conclude that the isomorphism of abelian group is actually an isomorphism of rings? Is that because the forgetful functor from rings to abelian groups is fully faithful? This is quite enlightening for me as in the previous your answer. I mean, I often see canonical isomorphisms of modules treated as isomorphisms of rings and I never get why. I think your answer can clarify this to me. – Blacksmith Jan 15 '18 at 20:52
  • @Blacksmith: it's not fully faithful. What we need is that it's conservative, meaning that if $f : A \to B$ is a map of rings such that the underlying map of abelian groups is an isomorphism, then $f$ is an isomorphism. And this is true because in both rings and abelian groups a map is an isomorphism iff it's a bijection. – Qiaochu Yuan Jan 15 '18 at 20:54
  • So may I ask why it is conservative? Is there a general theory about conservative functors? I would like to know more about this since as I told you I often see isomorphisms of modules treated as isomorphisms of rings for example. – Blacksmith Jan 15 '18 at 20:57
  • There is a general theory but it's unnecessary in this case. All you need to observe is what I just said (which amounts to the observation that the forgetful functor all the way down to $\text{Set}$, in both cases, is conservative). – Qiaochu Yuan Jan 15 '18 at 20:59
  • Ok so if I want to apply what you said in your first comment I should confirm that the compositions of the isomorphisms of abelian groups that you wrote give together the underlying map of abelian groups of the map you call natural. Is this obvious? Then I'm done, thank you for your time. – Blacksmith Jan 15 '18 at 21:11
  • @Blacksmith: I prefer not to use the term "obvious" but I think it's straightforward; I'm not applying any big theorems or anything, just being careful. – Qiaochu Yuan Jan 15 '18 at 21:16