Timeline for A $0$-dimensional ring that is not noetherian
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 9, 2015 at 1:44 | comment | added | rschwieb | This is a very simple answer. To prove the prime ideals are maximal, just note that this ring is Von Neumann regular. The quotient by a prime ideal is a Von Neumann regular domain, end it is easy to show this is a field. So prime ideals are all maximal. | |
| Apr 6, 2012 at 9:54 | comment | added | Martin Brandenburg | Sequences with finite support. | |
| Apr 6, 2012 at 9:48 | comment | added | Filippo Alberto Edoardo | Wow, really? I confess I have always thought it was true. But where does the argument saying that for every element in the ideal, all its components also lie in the ideal because you can multiply the element by $e_i$ (at least in the countable-index set-case)? Do you have an example of an ideal in $\prod_\mathbb{N}k$ for $k$ a field which is not of the form I wrote? Thanks! | |
| Apr 6, 2012 at 9:27 | comment | added | Martin Brandenburg | No, the ideals are more complicated. They correspond to filters on the index set. | |
| Apr 6, 2012 at 9:07 | comment | added | Filippo Alberto Edoardo | I might be wrong, but I argue as follows. It is not noetherian because the chain of ideals $I_n=<e_i>_{1\leq i \leq n}$ is ascending and never stabilizes ($e_i$ is the $\infty$-ple having $1$ at the i-th place and $0$ elsewhere). It is of dimension $0$ because every ideal is product of ideals (general in any product of rings) and a quotient of my ring is a domain iff is a field, so every prime is maximal. | |
| Apr 6, 2012 at 8:49 | comment | added | Martin Brandenburg | You can find a proof in my answer. | |
| Apr 6, 2012 at 8:49 | comment | added | jmc | I thought I had a prove that, that will not work. Maybe I made a mistake. Can you prove yours? | |
| Apr 6, 2012 at 8:36 | history | answered | Filippo Alberto Edoardo | CC BY-SA 3.0 |