Timeline for On the set of non-zero elements in an integral domain whose generating principal ideal is of a special kind
Current License: CC BY-SA 4.0
63 events
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| May 12, 2018 at 21:12 | history | edited | Luc Guyot | CC BY-SA 4.0 |
non-associated (distinct in $R/R^{\times}$) is more precise than "distinct"
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| May 12, 2018 at 20:22 | comment | added | Luc Guyot | @users I have now added a proof of Claim 5. | |
| May 12, 2018 at 20:21 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Added a proof of Claim 5
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| May 12, 2018 at 15:53 | comment | added | user111524 | In your claim 5, is $R$ really any non-field integral domain (no UFD or extra assumption?) ? I don't see how does it follow ... | |
| Apr 8, 2018 at 0:03 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixes Corollary 2's statement and proof
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| Apr 7, 2018 at 22:33 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Erratum for Corollary 2
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| Jan 9, 2018 at 21:21 | comment | added | Luc Guyot | @users Yes if $R$ is a UFD, but I have no idea in general. Considering examples may help. | |
| Jan 7, 2018 at 18:24 | comment | added | user111524 | We know that $S_R$ is always saturated. Let $\hat S_R$ be the multiplicative submonoid generated by $S_R$ and the unit of $R$; is $\hat S_R$ still saturated ? | |
| Dec 6, 2017 at 13:00 | comment | added | Luc Guyot | This is just enough as such prime is maximal. | |
| Dec 6, 2017 at 12:56 | comment | added | user111524 | If $a$ is a product of prime elements then why would $\mathfrak m$ be a principal ideal generated by one of them ? I can only see that $\mathfrak m$ contains a principal ideal generated by one of them ... | |
| Dec 5, 2017 at 20:44 | comment | added | Luc Guyot | Oops, this is a non-zero element of $\mathfrak{m}$, just corrected. | |
| Dec 5, 2017 at 20:43 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixes typo: $a$ belongs to $\mathfrak{m} \setminus \{0\}$.
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| Dec 5, 2017 at 20:22 | comment | added | user111524 | what is your $a$ in the argument ? | |
| Dec 5, 2017 at 20:20 | comment | added | Luc Guyot | Every Abelian group is isomorphic to the ideal class group of some Dedekind domain, so the answer is no in general, although it holds for rings of integers. The only hypothesis is that $R$ is not a PID, so it has a non-principal maximal (= non-zero prime) ideal. | |
| Dec 5, 2017 at 19:49 | comment | added | user111524 | Is every maximal ideal class an element of finite order in the ideal class group of a Dedekind domain ? (you seem to be using so in your corollary about ideal class groups) | |
| Nov 14, 2017 at 22:41 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Simplifies the proof of the "Corollary" and corrects the description of $S_R$ for an atomic domain $R$.
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| Nov 14, 2017 at 22:39 | comment | added | Luc Guyot | @users I simplified the proof of the corollary. Since a Dedekind domain $R$ is atomic, $S_R$ is generated by the units of $R$ and the prime elements $p$ such that $Rp$ is a maximal ideal. | |
| Nov 14, 2017 at 22:35 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Simplifies the proof of the "Corollary"
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| Nov 11, 2017 at 14:12 | comment | added | user111524 | I had some difficulty understanding your corollary about ideal class groups. How do you get that there is a maximal ideal whose ideal class is non-trivial and has finite order ? I can see that from $\mathcal m^n=Ra$ , and $\mathcal m$ not principal, we can conclude that $a$ is not a product of primes, but how does that imply that $a \notin S_R$ ? | |
| Oct 24, 2017 at 16:10 | comment | added | Luc Guyot | @users It is quite the same to me. My 2 cents: relax the atomic assumption until it breaks mulitplicative closedness. You may find then (counter)examples which are GCD or Prüfer. | |
| Oct 16, 2017 at 12:39 | comment | added | user111524 | Okay ... what about Prufer domains ? | |
| Oct 16, 2017 at 9:37 | comment | added | Luc Guyot | @users I am unable to show anything assuming $R$ is a GCD. If I had time, I would try to generalize the claim about multiplicative closedness to some generalizations of atomic rings ("Non-Noetherian commutative ring theory" by S.T. Chapman and S. Glaz looks like a good place to start from) and strive for the search of a non-atomic counter-examples. | |
| Oct 12, 2017 at 19:42 | comment | added | user111524 | Assuming $R$ is a GCD domain i.e. $Ra \cap Rb$ is a principal ideal for every $a,b \in R$ , (en.wikipedia.org/wiki/GCD_domain) , can you show that $S_R$ is multiplicatively closed ? Assuming $R$ is a Prufer domain , (en.wikipedia.org/wiki/Pr%C3%BCfer_domain) , can you show that $S_R$ is multiplicatively closed ? | |
| Sep 28, 2017 at 8:38 | vote | accept | CommunityBot | ||
| Sep 21, 2017 at 6:47 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Removes superfluous right brace
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| Sep 20, 2017 at 21:36 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixed typo in the proof of Claim 2 and added further references to Wiki pages for definitions
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| Sep 20, 2017 at 19:31 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixes typo in the proof of the top Corollary
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| Sep 20, 2017 at 19:27 | comment | added | Luc Guyot | @users The typo is now corrected, thanks. I expanded my answer by giving an example of ring $R$ such that $R^{\times} \subsetneq S_R \subsetneq R \setminus \{0\}$. | |
| Sep 20, 2017 at 19:23 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Gives an explicit example of set $S_R \subsetneq R \setminus \{0\}$
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| Sep 20, 2017 at 16:49 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Gives a non-trivial example of multiplicative set $S_R$
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| Sep 2, 2017 at 16:31 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixes typo spotted by 'users'
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| Sep 2, 2017 at 13:01 | comment | added | user111524 | Your proof of claim 2 still contains a typo .. $1$ can be written as a linear combination of $p$ and $x$ , not $a$ and $x$ ... as far the motivation of the question regrads , I was just tryinv to see various properties related to domains having some good factorial or arithmetical prooerties and then trying to isolate the elements with special properties ... I am thinking about vatious othet related properties , I will post about it here if I need help seeing as good a response as yours ... | |
| Sep 1, 2017 at 17:11 | comment | added | Luc Guyot | @users I corrected and simplified the proof of Claim 2. | |
| Sep 1, 2017 at 17:08 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Corrected and simplified the proof of Claim 2
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| Aug 31, 2017 at 16:15 | comment | added | Luc Guyot | @users I just added Claim 5 which also relates to your new question. | |
| Aug 31, 2017 at 16:14 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Added Claim 5 about multi-variate polynomials and an instance of $S_R \neq N_R$
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| Aug 30, 2017 at 23:17 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixes claim about regular local ring
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| Aug 30, 2017 at 23:12 | history | edited | Luc Guyot | CC BY-SA 3.0 |
added 83 characters in body
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| Aug 30, 2017 at 21:51 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixed typo ($x'$ instead of $e$), removed useless mention to $R$, removed obsolete sentence
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| Aug 30, 2017 at 21:46 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixed typo ($x'$ instead of $e$), removed useless mention to $R$
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| Aug 30, 2017 at 21:26 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Widen the scope of Corollary 2
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| Aug 30, 2017 at 21:23 | comment | added | Luc Guyot | @users I just modified my answer to settle the case of atomic domains. It answers your new question for domains which satisfy the ascending condition on principal ideals. | |
| Aug 30, 2017 at 21:13 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Added a wiki link for atomic domains
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| Aug 30, 2017 at 21:00 | history | edited | Luc Guyot | CC BY-SA 3.0 |
widen the scope of the answer to atomic domains
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| Aug 30, 2017 at 20:40 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Rewording of a sentence in the proof of Claim 3
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| Aug 30, 2017 at 20:13 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Positive answer in the Noetherian case
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| Aug 30, 2017 at 20:07 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Positive answer in the Noetherian case
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| Aug 30, 2017 at 19:58 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Positive answer in the Noetherian case
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| Aug 30, 2017 at 17:11 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Added Corollary 2 about Noetherian local domains
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| Aug 29, 2017 at 20:22 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Add remarks about the case of a local domain
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| Aug 29, 2017 at 10:15 | comment | added | user111524 | Suppose for an integral domain $R$ , the set in the question say $S_R$ is multiplicatively closed ; then is $S_{R[X]}$ also multiplicatively closed in $R[X]$ ? (We already know it is saturated ... ) | |
| Aug 29, 2017 at 10:05 | comment | added | user111524 | Here is the proof that the set you mention (call it $S$ say) is saturated : Let $ab \in S$ and let $t\in R$ ; then either $ab | tb $ or $tb | ab$ ; now , since $b \neq 0$ , so if $ab | tb$ then $a|t$ and if $tb |ab$ then $t|a$ ; so either $a|t$ or $t|a$ and $t \in R$ was arbitrary , so $a \in S$ | |
| Aug 29, 2017 at 0:41 | history | edited | Luc Guyot | CC BY-SA 3.0 |
deleted 7 characters in body
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| Aug 29, 2017 at 0:36 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Corrects assumption of Claim 2; edited body
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| Aug 28, 2017 at 21:37 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixed Claim 2
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| Aug 28, 2017 at 21:27 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Existence of constant $u$-polynomial is trivial.
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| Aug 28, 2017 at 21:16 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Precise reference to Weintraub's example
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| Aug 28, 2017 at 21:09 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Added Claim 2 as an application of Claim 1 to polynomial rings over domains
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| Aug 28, 2017 at 18:00 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixes "saturation" and typo
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| Aug 28, 2017 at 15:35 | comment | added | user111524 | I also don't know of any examples ... but that shouldn't stop us from trying to prove or disprove the claim ... your proof of saturated ness is nice and simple , though I know this trick , I totally missed it here .. | |
| Aug 27, 2017 at 21:25 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Removes signs of uncertainty
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| Aug 27, 2017 at 20:54 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Right multiplication is more consistent
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| Aug 27, 2017 at 19:56 | history | answered | Luc Guyot | CC BY-SA 3.0 |