Timeline for When does prime elements remain prime in certain integral extension
Current License: CC BY-SA 3.0
21 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Feb 25, 2020 at 4:55 | comment | added | user237522 | @R.vanDobbendeBruyn, please, could you help me with my following question mathoverflow.net/questions/353057/…, which is a variation/generalization of the above question. Thank you. | |
| May 20, 2018 at 18:02 | comment | added | user111492 | I just wanted to comment that recently I have found out that the assertion of the question I was making is correct if $R$ is a Noetherian domain. This is Lemma 4.7 in ; On finite generation of $R$-subalgebras of $R[X]$ , Amartya K. Dutta; Nobuharu Onoda; Journal of Algebra 320 (2008) 57- 80. google.co.in/url?sa=t&source=web&rct=j&url=http://… | |
| Feb 8, 2018 at 21:32 | comment | added | Georges Elencwajg | @Neil: With pleasure! | |
| Feb 8, 2018 at 18:57 | comment | added | Neil Epstein | @GeorgesElencwajg I have an answer to your question about height. Could you post it as a regular question? | |
| Feb 3, 2018 at 17:11 | comment | added | Georges Elencwajg | Thanks a lot for your comment, Remi, and sorry for bothering you so much. The subject seems to be unexpectedly and frustratingly hard. | |
| Feb 3, 2018 at 15:38 | comment | added | R. van Dobben de Bruyn | @GeorgesElencwajg: I have thought about this, but so far this has been unsuccessful. Most non-equidimensional domains I can think of are a localisation of some finite type domain, so height behaves the same as in the finite type case. | |
| Feb 3, 2018 at 7:11 | comment | added | Georges Elencwajg | Yes, your new answer convinces me [but I can't upvote you again :-)]. Do you know an example of a Noetherian domain $R$ and a prime $\mathfrak P \subset \tilde R$ whose height in $\tilde R $ is not the same as the height of $\mathfrak P\cap R$ in $R$? Or even an example in the situation $R\subset S$ where $S$ is just supposed to be a domain integral over $R$, but not necessarily its integral closure? | |
| Feb 3, 2018 at 4:04 | comment | added | R. van Dobben de Bruyn | @GeorgesElencwajg: I'm fairly certain my current argument should fix it (at the expense of somewhat strengthening the hypotheses). | |
| Feb 3, 2018 at 3:50 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Fixed a gap pointed out by Georges Elencwajg in the comments.
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| Feb 2, 2018 at 19:46 | comment | added | R. van Dobben de Bruyn | @misao: there are many examples if $\mathfrak p$ is not principal, even if it still has height $1$. For example, consider the resolution of the nodal or cuspidal cubic. (See also Raymond's comment above.) | |
| Feb 2, 2018 at 19:45 | comment | added | R. van Dobben de Bruyn | @GeorgesElencwajg: I wanted to do the following: if $\operatorname{ht}(\mathfrak p') > 1$, then there is a chain $0 \subsetneq \mathfrak p'' \subsetneq \mathfrak p'$. Applying going up gives a chain of length at least $2$ in $(\tilde R)_{\mathfrak p'}$. But I just realised that this is not necessarily contained in $\mathfrak r$, so the argument indeed seems to have a gap. I'm not sure how to fix it at this moment; I hope it doesn't break the argument. | |
| Feb 2, 2018 at 19:42 | comment | added | user111492 | For the kinds of rings you described, do you think for any prime ideal $P$ in $R$, $P\tilde R$ is prime in $\tilde R$ also ? | |
| Feb 2, 2018 at 19:35 | comment | added | Georges Elencwajg | Dear Remi, I'm sorry but in line 16 I don't understand to what inclusion of prime ideals in $R_{\mathfrak p'}$ you apply the going-up theorem for the morphism $R_{\mathfrak p'} \to (\tilde R)_{\mathfrak p'}$ | |
| Jan 28, 2018 at 23:50 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Simplified the argument.
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| Jan 28, 2018 at 23:32 | comment | added | R. van Dobben de Bruyn | @GerhardPaseman: it means that for every finite extension $K = \operatorname{Frac} R \to L$, the integral closure of $R$ in $L$ is finite over $R$. We only need the case $K = L$. | |
| Jan 28, 2018 at 23:22 | comment | added | Gerhard Paseman | OK. What does it mean for a ring to be Japanese? Gerhard "Enquiring Mind Seeking Knowledge, Arigato" Paseman, 2018.01.28. | |
| Jan 28, 2018 at 23:18 | history | undeleted | R. van Dobben de Bruyn | ||
| Jan 28, 2018 at 23:18 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Fixed gap in proof.
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| Jan 28, 2018 at 21:48 | history | deleted | R. van Dobben de Bruyn | via Vote | |
| Jan 28, 2018 at 21:29 | history | answered | R. van Dobben de Bruyn | CC BY-SA 3.0 |