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If $(R,m)$ is a complete local ring (with respect to the $m-$adic topology) and $I$ a prime ideal in $R$, is $R/I$ complete (with respect to the $m/I-$adic topology)? It seems too strong, but I am unable to give a counterexample.

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complete for which topology? –  seporhau Mar 7 '12 at 17:00
    
@seporhau: Clarified. I thought, it was convention to assume "with respect to the maximal ideal" unless otherwise specified. –  Merlin Mar 7 '12 at 17:06
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What you'd like to be able to do is take a Cauchy sequence in $R/I$, then lift each element so that you have a Cauchy sequence in $R$, then take the limit and push it back down to the quotient. So the question becomes: under what conditions can you lift your Cauchy sequence mod $I$ to a Cauchy sequence in $R$?. –  Brett Frankel Mar 7 '12 at 21:58
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The completion functor from the category of $R$-modules to the catogory of $\hat R$-modules is exact. So the quotient by an ideal $I \subset R=\hat{R}$ is complete if and only if $I$ is complete as an $\hat R$-module. I tend to think that this is always the case at least for Noetherian $R$. –  Dima Sustretov Mar 7 '12 at 22:29
    
@DimaSustretov: Thank you. –  Merlin Mar 8 '12 at 16:29
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1 Answer

Proposition Let $(R, \mathfrak{m})$ be a complete Noetherian local ring. Let $I$ be a proper ideal of $R$. Then $(R/I, \mathfrak{m}/I)$ is complete.

Proof: The following sequence of $R$-modules is exact.

$0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0$

Since $R$ is Noethrian, the following sequence is exact, where $\hat I$ etc. are the completions of $I$ etc. with respect to $\mathfrak{m}$-adic toplogy.

$0 \rightarrow \hat I \rightarrow \hat R \rightarrow \hat (R/I) \rightarrow 0$

Since $R$ is complete, $\hat R = R$. It is well known that $I$ is a closed submodule of $R$ with respect to $\mathfrak{m}$-adic toplogy. Since $R$ is complete, so is $I$. Hence $\hat I = I$. Therefore we get the following exact sequence.

$0 \rightarrow I \rightarrow R \rightarrow \hat (R/I) \rightarrow 0$

Hence $R/I$ is complete. QED

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