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Let $(R,\mathfrak{m})$ be a complete local ring, $a_{\lambda}$ be a decreasing net of ideals in $R$, indexed by a directed set. Consider the completion under $a_{\lambda}$-topology $A=\underleftarrow{\lim} R/\mathfrak{a}_{\lambda}$. Is $A$ still complete under the $\mathfrak{m}$-topology?

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Since inverse limits commute, I believe this is true. – the L Nov 30 '11 at 13:53
@Liran: This is a bit too easy - you forgot about the quotient rings. – Martin Brandenburg Nov 30 '11 at 20:11
Moreover (following Liran Shaul's comment), if we define $I=\cap_\lambda a_\lambda$, then isn't $A$ equal to $R/I$, so $\frak m$-complete? We can assume $I=0$, and then any sequence $(r_n)$ in $R$ that is Cauchy in the $a_\lambda$-topology is also Cauchy in the $\frak m$-topology, so has a limit in $R$. – inkspot Nov 30 '11 at 20:21
up vote 5 down vote accepted

I assume that $(R, \frak{m} )$ is a complete Noetherian local ring. Set $\frak{a} = \bigcap_\lambda \frak{a}_\lambda$. By passing to $R/\frak{a}$ we may assume that the $\frak{a}_\lambda$-topology is separated. Now, we use a Theorem of Chevalley (1946) which says that in the complete Noetherian local ring the $\frak{m}$-adic topology is weaker than every separated topology. Then every Cauchy sequence in the $\frak{a}_\lambda$-topology is also a Cauchy sequence in the $\frak{m}$-adic topology. Hence $R = \underleftarrow{\lim} R/\frak{a}_\lambda$ is complete.

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I'm not sure why there was some difficulty getting the latex to work out; it looks like it was choking on the \frak commands. I did the best I could. – Todd Trimble Dec 1 '11 at 14:43
Thanks a lot, Todd Trimble. – Pham Hung Quy Dec 2 '11 at 14:35
Can you please provide a reference for this Theorem of Chevalley? – the L Jan 8 '12 at 13:09
Lemma 7 - Chevalley: On theory of local ring, Annals of Math. 1946, 690-708. Theorem 2.1 - P. Schenzel, On use of local cohomology in algebra and geometry, in: six lecture of commutative algebre, 1998 (you can seacher it on internet) – Pham Hung Quy Jan 9 '12 at 12:16

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