# Localisation contained in completion?

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?

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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

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 $\mathfrak{q}$.
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.