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.
Is the quotient of a complete ring, complete?
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0@DimaSustretov: Thank you. – 2012-03-08
1 Answers
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