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I'm working on an exercise in which I have to show that localising and completing are exact functors. More precisely I have a Dedekind domain $R$ and a prime ideal $\mathfrak{p}$ and I have to show that $M\mapsto R_\mathfrak{p} \otimes_R M$ and $M\mapsto \hat{R_\mathfrak{p}} \otimes_R M$ ($R_\mathfrak p$ the localisation at $\mathfrak p$, $\hat{R_\mathfrak p}$ the completion) preserve exact sequences.

My question in this context is the following: is it true that the localisation $R_\mathfrak{p}$ is contained in the completion $\hat{R_\mathfrak{p}}$ (using a suitable embedding)? I understand that this is true for the ring $\mathbb{Z}$, the localisation $\mathbb{Z}_{(p)}$ ($p$ a prime number) and the ring of $p$-adic integers, so in other words we have $\mathbb{Z}\subset \mathbb{Z}_{(p)} \subset \mathbb{Z}_p$. Is this correct in the case of an arbitrary Dedekind domain as well? Is it maybe also true in a more general setting, say if $R$ is "just" an integral domain?

I'm looking forward to your answers.

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Provided the statement is true and assuming I have already proved that localising is an exact functor, does anyone see a way I can use this to prove that completing is an exact functor? – Oliver Braun Oct 19 '11 at 22:14
up vote 1 down vote accepted

Ok, it turns out that in the lecture I'm attending, we only defined the completion for discrete valuation rings. Luckily, a Dedeking ring $R$ localised at a prime ideal $\mathfrak{p}$ is a discrete valuation ring with unique maximal ideal $\mathfrak{p}$. In that case $$\hat{R_\mathfrak{p}} = \lim\limits _{\longleftarrow} ~ R_\mathfrak{p} / \mathfrak p^n = \{ (a_0 , a_1 , ... ) ~|~ a_i \in R_\mathfrak{p}/\mathfrak{p}^{i+1} , ~ a_i + \mathfrak{p}^i = a_{i-1} \}$$ and $R_\mathfrak{p} \to \hat{R_\mathfrak{p}}, ~ a \mapsto (a+ \mathfrak{p} , ~ a+ \mathfrak{p^2} , ...)$ is an injective ring homomorphism. Thus in the given case $R\hookrightarrow R_\mathfrak{p} \hookrightarrow \hat{R_\mathfrak{p}}$.

In the more general case of an arbitrary ring $A$ and a prime ideal $\mathfrak{q}$ we don't necessarily have that $A_\mathfrak{q}$ is a discrete valuation ring. In that case I'm not even sure how to define a completion (or a valuation) with respect to $\mathfrak{q}$.

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Take a look at Liu's book "algebraic geometry and arithmetic curves". Especially Chapter 1.3 (or 1.4). This is probably what you're looking for. – Bana Oct 20 '11 at 21:57
Thanks, I'll look into it. – Oliver Braun Oct 22 '11 at 13:17

Most (or all?) Dedekind domains $R$ are faithfully flat over $\mathbf{Z}$. Therefore, the natural morphism from the localization at a prime $p \subset R$ to its completion is injective.

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Can you describe this morphism in more detail or, alternatively, direct me toward an online resource that describes it? Do you know if the statement is also true for arbitrary integral domains? – Oliver Braun Oct 19 '11 at 20:43

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