In my old number theory notebook this is stated as a fact. However, I ran into problems when I tried to prove it. First let me state the (supposed) theorem accurately:

Theorem (?)

Let $K$ be a number field with ring of integers $O_K$. Let $p$ be a prime of $\mathbb{Z}$. Then $O_K\otimes \mathbb{Z}_p\cong \oplus_{\mathfrak{p}|p}O_{K,\mathfrak{p}}$, where $\mathfrak{p}$ runs over those primes of $O_K$ that lie over $p$, and where $O_{K,\mathfrak{p}}$ is the formal local ring of $O_K$ at $\mathfrak{p}$ (i.e., $\varprojlim O_K/\mathfrak{p}^n$).


The way I was thinking of proving this is through the Chinese Remainder Theorem. The following is true for every natural number $n$: $$O_K\otimes \mathbb{Z}/p^n\mathbb{Z}\cong O_K/p^nO_K\cong O_K/\mathfrak{p}_1^{ne_1}\cdots \mathfrak{p}_m^{ne_m}\cong O_K/\mathfrak{p}_1^{ne_1}\oplus...\oplus O_K/\mathfrak{p}_m^{ne_m}.$$ where $pO_K=\mathfrak{p}_1^{e_1}\cdots \mathfrak{p}_m^{e_m}$.

The next natural step is to take inverse limits of both sides with respect to n. As inverse limits commute with direct sums (as both are limits), $\varprojlim O_K/\mathfrak{p}_1^{ne_1}\oplus...\oplus O_K/\mathfrak{p}_m^{ne_m}\cong O_{K,\mathfrak{p}_1}\oplus...\oplus O_{K,\mathfrak{p}_m}$.

This is where the wheels come off the argument: inverse limits don't generally commute with tensor products. This is because inverse limits are categorical limits, whereas tensor products are categorical colimits (indeed they are the coproduct of the category of $\mathbb{Z}$-algebras). So there is no reason, a priori, that $\varprojlim O_K\otimes \mathbb{Z}/p^n\mathbb{Z}$ would be isomorphic to $O_K\otimes \mathbb{Z}_p$.

So this leads me to the


Why is the theorem true? (if it even is.)

  • 8
    $\begingroup$ Why not consider the map ${\mathbf Z}_p \otimes_{\mathbf Z} {\mathcal O}_K \rightarrow \bigoplus_{{\mathfrak p}|p} {\mathcal O}_{K,{\mathfrak p}}$ where $x \otimes \alpha \mapsto (x\sigma_{\mathfrak p}(\alpha))_{{\mathfrak p}|p}$, where $\sigma_{\mathfrak p}$ is the embedding of ${\mathcal O}_K$ into its $\mathfrak p$-adic completion, and show this is a ${\mathbf Z}_p$-algebra isomorphism? (Hint: I think it's easier to show it is surjective first, using continuity and compactness, and then you can get injectivity by looking at both sides just as ${\mathbf Z}_p$-modules.) $\endgroup$ – KCd Apr 28 '12 at 2:54
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    $\begingroup$ Try to work through a proof first in a concrete case, such as $K = {\mathbf Q}(i)$ and $p=5$ (thus showing ${\mathbf Z}_5 \otimes_{\mathbf Z} {\mathbf Z}[i] \cong {\mathbf Z}[i]_{(1+2i)} \oplus {\mathbf Z}[i]_{(1-2i)}$, where the rings on the right are completions, not localizations), before doing the general case. I also think it's better to denote the right side as a direct product instead of a direct sum, since a finite product of rings feels more like a ring than a finite direct sum of rings, even though it's the same thing. $\endgroup$ – KCd Apr 28 '12 at 3:00
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    $\begingroup$ (Some would say that this is the problem with my generation.) $\endgroup$ – Nicole Apr 28 '12 at 3:02
  • 1
    $\begingroup$ $\mathcal{O}_K$ is a finite generated free ablian group. $\endgroup$ – wxu Apr 28 '12 at 5:55
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    $\begingroup$ Dear @Nicole, I just want to say that I always enjoy reading your questions. Regards, $\endgroup$ – Bruno Joyal May 3 '12 at 3:35

As noticed wxu in the comments, $O_K$ is finite free over $\mathbb Z$. Write $O_K=\oplus_{1\le i\le d} e_i\mathbb Z$. Then we have $$ O_K\otimes_\mathbb Z \mathbb Z/p^n \mathbb Z =\oplus_{1\le i\le d} e_i(\mathbb Z/p^n\mathbb Z).$$ Passing to the inverse limit, we get $$\varprojlim_n O_K\otimes \mathbb Z/p^n \mathbb Z =\oplus_{1\le i\le d} e_i\mathbb Z_p=O_K\otimes_\mathbb Z \mathbb Z_p.$$


Your problem can be solved under more general conditions. Let $A$ be a Noetherian commutative ring. Let $I$ be an ideal of $A$. Let $M$ be a finitely generated $A$-module. Let $A_I$ = $\varprojlim A/I^n$. Then $M\otimes A_I$ $\cong$ $\varprojlim M\otimes (A/I^n)$. The proof can be found, for example, in Matsumura's "Commutative ring theory", Theorem 8.7 in page 60.


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