Timeline for Maximal subideal of an ideal
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 7, 2019 at 4:17 | comment | added | Pavel Čoupek | Maybe it's worth pointing out that there is a popular class of rings with the above property for all (nonzero) iprincipal deals: Namely, local rings. | |
| May 1, 2018 at 6:09 | vote | accept | Antonyoo | ||
| May 1, 2018 at 6:09 | |||||
| Feb 14, 2018 at 14:16 | comment | added | მამუკა ჯიბლაძე | Well in the first comment I also was not precise enough - I wanted to say there that in your second claim one should restrict to nonzero principal ideals. After that comment, I realized that one could do better by changing from "there are maximal proper subideals" to "every proper subideal is contained in a maximal one" (instead of restricting to nonzero principal ideals only). | |
| Feb 13, 2018 at 22:05 | comment | added | Sándor Kovács | @მამუკაჯიბლაძე: OK, I finally understood your point and updated the claim. Just for the record, your first comment is talking about "excluding" which seems totally irrelevant (at the end you need to exclude the generator to get the upper bound). | |
| Feb 13, 2018 at 22:03 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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| Feb 13, 2018 at 20:07 | comment | added | მამუკა ჯიბლაძე | Sorry for being unclear. Given a proper subideal in a principal ideal $P$ I have to find a maximal proper subideal containing it. If there is a proper subideal in $P$ then $P=(t)$ for some nonzero $t$. Then the argument in your second claim shows that chains of proper subideals have upper bounds. Zorn's lemma then gives that every proper subideal is contained in a maximal proper subideal. | |
| Feb 13, 2018 at 19:54 | comment | added | Sándor Kovács | @მამუკაჯიბლაძე: I don't understand your argument. Are you saying that any subideal of a principal ideal is also principal? How about $R=k[x,y,z]$ and $(xy,xz)\subseteq (x)$? | |
| Feb 13, 2018 at 18:04 | comment | added | მამუკა ჯიბლაძე | Well I admit it is sort of cheating but in fact it is not :D If you can give me a proper subideal of a principal ideal to try me, then this proper subideal can be generated by a nonzero $t$ and then I will argue as you did | |
| Feb 13, 2018 at 17:37 | comment | added | Sándor Kovács | @მამუკაჯიბლაძე: How do you get the upper-bound without the exclusion? | |
| Feb 13, 2018 at 2:08 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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| Feb 12, 2018 at 13:45 | comment | added | მამუკა ჯიბლაძე | Actually the standard conclusion of Zorn's lemma gives a correct statement without excluding anything: any proper subideal (of a principal ideal) is contained in a maximal proper subideal | |
| Feb 12, 2018 at 11:15 | comment | added | YCor | @PeterHeinig the second statement "Claim" of the OP "Let $T$ be a principal ideal... then $T$ contains [...] proper subideals" is false for $T=0$. To say that $\{0\}$ is maximal among proper subideals of $\{0\}$ is not vacuously true, it is false, because it says in particular that $\{0\}$ is a proper subset of $\{0\}$, which means the existence of an element in the empty set. | |
| Feb 12, 2018 at 9:45 | comment | added | მამუკა ჯიბლაძე | Well to be entirely accurate one has to exclude the zero ideal from this... | |
| Feb 12, 2018 at 9:12 | history | edited | Peter Heinig | CC BY-SA 3.0 |
Small grammatical corrections. In particular, 'differ by a unit multiple' is a much more common way to put it. Added usual technical term 'associate' .Content and style respected.
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| Feb 12, 2018 at 7:25 | history | answered | Sándor Kovács | CC BY-SA 3.0 |