Timeline for (Krull) dimension of any associated graded ring of a ring R equals the dimension of R
Current License: CC BY-SA 2.5
9 events
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Jul 12, 2014 at 1:02 | comment | added | ABIM | Thanks Mariano :) You've been of immense help as always! | |
Jul 11, 2014 at 22:27 | comment | added | Mariano Suárez-Álvarez | @CSA, you can find en C&E or in Weibel or in many other places a construction of the Chevalley-Eilenberg resolution of the trivial left module of an enveloping algebra. There is an analogue of that for bimodules. | |
Jul 11, 2014 at 20:39 | comment | added | ABIM | Cool, thanks Mariano!.. but just so I can understand the nooks and crannies of that argument would you have a reference (that particular algebra is extremely important for a counterexample Im working on at you just confirmed my intuition that it indeed does work) | |
Jul 11, 2014 at 18:53 | comment | added | Mariano Suárez-Álvarez | @CSA, that algebra is the enveloping algebra of the unique non-abelian Lie algebra of dimension 2. It's global dimension is equal to is bimodule projective dimension is equal to 2. | |
Jul 11, 2014 at 15:53 | comment | added | ABIM | hm... then would you know of a technique for calculating the hochschild c. dimension of $k<x,y>/(xy-yx-x)$? Since the result in your paper (from C&E's Homological algebra) does not apply and neither does this (^false hypothesised) result... | |
Jul 11, 2014 at 3:44 | comment | added | Mariano Suárez-Álvarez | @CSA, it is not true, so no :-). For example, suppose $2\neq0$ in the field $k$; then the algebra $k[x]/(x^2-1)$ has zero dimension as a bimodule over itself, yet there is an increasing filtration on it such that the associated graded algebra is $k[x]/(x^2)$, which has infinite H dimension. In any case, there are several different things you can mean by «analogous»... | |
Jul 11, 2014 at 3:00 | comment | added | ABIM | Is there a reference which deals with the analogous question but for Hochschild cohomological dimension? | |
Nov 17, 2010 at 5:02 | vote | accept | Timothy Wagner | ||
Nov 17, 2010 at 5:02 | |||||
Nov 17, 2010 at 0:39 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 2.5 |