Timeline for Height of maximal homogeneous ideals
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| S Jul 15, 2015 at 20:44 | history | suggested | user 1 | CC BY-SA 3.0 |
tags addition
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| Jul 15, 2015 at 19:42 | review | Suggested edits | |||
| S Jul 15, 2015 at 20:44 | |||||
| Dec 16, 2011 at 15:25 | history | edited | Ralph | CC BY-SA 3.0 |
Editorial adjustment
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| Dec 16, 2011 at 11:56 | comment | added | Neil Epstein | A related problem that comes up in these situations is equidimensionality. One might naively think that in this situation, all maximal chains of homogeneous primes ending in the homog. maximal ideal have the same length (such a ring would be called equidimensional). But this is false; consider the ring $R=k[x,y_1, ..., y_n]/ (xy_1, xy_2, ..., xy_n)$, where $k$ is any field and $n\geq 2$ an integer. The minimal primes are $(x)$ and $(y_1, ..., y_n)$, and if $M$ is the homogenous max ideal, we have height $M/(x) = n$ but height $M/(y_1, ..., y_n) = 1$ | |
| Dec 16, 2011 at 9:51 | comment | added | Ralph | In fact, the answer is yes, if $R_0$ is in addition indecomposable. For, since an indecomposable Artinian ring is local, $R_0$ has a unique max. ideal and hence $R$ has a unique homogeneous max. ideal. | |
| Dec 16, 2011 at 9:35 | comment | added | Ralph | Nice counterexample, thanks. Maybe a positive result can be obtained, if one requires $R_0$ to be indecomposabel. But the counterexample iis sufficient for me. Won't you post it as answer, I'll accept it. | |
| Dec 16, 2011 at 5:45 | comment | added | user2035 | You might want to add some condition to exclude trivial counterexamples like $k[X,Y]\times k[Z]$ where $X,Y,Z$ have degree $1$. | |
| Dec 15, 2011 at 22:41 | history | edited | Ralph | CC BY-SA 3.0 |
Added that Krull dimension is finite
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| Dec 15, 2011 at 22:24 | history | edited | Ralph | CC BY-SA 3.0 |
Added polynomial ring example
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| Dec 15, 2011 at 22:02 | history | asked | Ralph | CC BY-SA 3.0 |