Timeline for Countable Maximal Ideals
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 3, 2013 at 13:24 | comment | added | Eric Wofsey | Incidentally, by this answer, my example does not require the axiom of choice (such a set $S$ can be exhibited explicitly). @dimo: Any element of $\ker(\varphi)$ has finite support, and every multiple of it must have smaller support. | |
| Sep 3, 2013 at 6:31 | vote | accept | CommunityBot | ||
| Sep 3, 2013 at 6:31 | history | bounty ended | CommunityBot | ||
| Sep 2, 2013 at 10:48 | comment | added | dimo | Nice, and how do you show that $ker(\phi)$ is not finitely generated ? | |
| Aug 29, 2013 at 18:13 | comment | added | YCor | The current edited post has nothing to do with the original answer, you should have written another post. | |
| Aug 29, 2013 at 17:18 | history | edited | Eric Wofsey | CC BY-SA 3.0 |
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| Aug 29, 2013 at 17:10 | history | edited | Eric Wofsey | CC BY-SA 3.0 |
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| Aug 29, 2013 at 17:05 | history | edited | Eric Wofsey | CC BY-SA 3.0 |
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| Aug 25, 2013 at 3:04 | history | edited | Eric Wofsey | CC BY-SA 3.0 |
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| Aug 25, 2013 at 3:03 | comment | added | Eric Wofsey | Oh, yes, I was implicitly assuming everything was commutative, perhaps because you tagged the question "commutative rings". | |
| Aug 24, 2013 at 19:00 | comment | added | user38138 | I think you need to use commutativity assumption to conclude that $i$ is an idempotent generator of $I$. | |
| Aug 24, 2013 at 14:34 | history | answered | Eric Wofsey | CC BY-SA 3.0 |