Suppose $(R,\mathfrak{m})$ is a local ring and $I$ a proper ideal. If $R$ is $\mathfrak{m}$-adically complete is it also $I$-adically complete.
If a local ring is $\mathfrak{m}$-adically complete is it also $I$-adically complete
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0@QiL: I was not assuming that, but I would be interested in seeing a proof either way. – 2011-11-13
1 Answers
Let $(a_n)_n$ be an $I$-adic Cauchy sequence in $R$. It converges to some $a$ for the $\mathfrak m$-adic topology. Replacing $a_n$ with $a_n-a$, we can assume that $a=0$. We want to show that $a_n$ converges to $0$ in the $I$-adic topology.
Let $r$ be a positive integer. There exists $N=N(r)$ such that $a_n-a_m\in I^r$ for all $n, m\ge N$. Fix such an $n$. Then $a_n\in I^r+{\mathfrak m}^q$ for any $q> 0$. Denote by $\bar{\mathfrak m}$ the image of $\mathfrak m$ in $R/I^r$. Suppose $R$ is noetherian. As $R/I^r$ is a noetherian local ring, $\bar{a}_n\in \cap_{q>0} \bar{\mathfrak m}^q=0$ by Krull's intersection theorem. So $a_n\in I^r$. This being true for any $n\ge N$, we see that $a_n$ converges to $0$ $I$-adically.
In the non-noetherian case, does "complete" include separatedness in your definition ?
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0Yes, by a complete ring$I$do mean one which is $I$-adically separated. Thank you for this proof. – 2011-11-14