How does one study Krull dimension of some I-adic completion of a ring or, more generally, a module? I know that in case of Noetherian local ring Krull dimension of its completion equals Krull dimension of the ring, but what can we say in general case?
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2$\begingroup$ How general? Are you concerned primarily about the local hypothesis, or do you also want something that works in the non-Noetherian case? $\endgroup$– Charles StaatsCommented Apr 17, 2010 at 23:54
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$\begingroup$ @Charles I'm mainly interested in Noetherian rings. If I'm not mistaken, Zariski rings also preserve Krull dimension under completion. Are there other well-behaved types of rings? Is it true for modules over Zariski ring? Or, maybe, there is a general technique to find it out... $\endgroup$– Andrew PenningCommented Apr 18, 2010 at 7:04
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1 Answer
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For a Noetherian ring R, the Krull dimension of its $I$-adic completion, $\hat{R}$ is given by $\sup h(J)$, where $J$ ranges over all maximal ideals of $R$ containing $I$ and $h(J)$ is the height of $J$. Therefore $\dim \hat R\le \dim R$ with equality only when $I\subset \operatorname{rad} R$. A reference is "Topics in $\mathfrak m$-adic topologies" by S.Greco, P.Salmon
answered Apr 20, 2010 at 4:10
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$\begingroup$ Thanks, it was very helpful. This reasoning applies to modules as well, by the way. $\endgroup$ Commented Apr 20, 2010 at 14:29
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$\begingroup$ What is an example of $I$ such that $I \not\subset {\mathrm{rad}}\,R$? $\endgroup$ Commented Jun 27, 2016 at 1:52