Timeline for For which monic irreducible $f \in \mathbb{C}[x,y][T]$, $\mathbb{C}[x,y][T]/(f)$ is a UFD?
Current License: CC BY-SA 3.0
20 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
|
|
Jun 12, 2017 at 8:17 | comment | added | user237522 | Thanks for your comment (unfortunately I have not fully understood it due to my lack of knowledge of algebraic geometry). Please, what do you think about Ben Webster's answer in mathoverflow.net/questions/5591/… | |
Jun 12, 2017 at 5:46 | comment | added | მამუკა ჯიბლაძე | For some more pessimistic considerations, see also answers to mathoverflow.net/q/82830/41291 (basically that on singular varieties no useful geometric interpretation of $\operatorname{Cl}=0$ seems to be possible) | |
Jun 12, 2017 at 2:19 | history | edited | user237522 | CC BY-SA 3.0 |
deleted 340 characters in body
|
Jun 9, 2017 at 0:27 | comment | added | user237522 | Thanks for your comment. Yes, I am familiar with Serre's criterion (which can be found, for example, in Matsumura's CRT, Theorem 23.8). Unfortunately, I do not know how to apply it. Please, have you noticed my late "important remark"? Is it easier to apply Serre's criterion in the special case of my "important remark"? | |
Jun 9, 2017 at 0:17 | comment | added | R. van Dobben de Bruyn | For normality, you have to check that $R$ is regular in codimension one (by Serre's criterion, noting that $S_2$ is always satisfied for hypersurfaces). In this case that means that the singular locus is only a finite set of points in $\mathbb A^3$. You can use the Jacobian criterion to compute the singular locus for any given polynomial, but it's not so obvious when it has codimension at least $2$. There is no direct parametrisation of polynomials satisfying this. | |
Jun 8, 2017 at 23:04 | history | edited | user237522 | CC BY-SA 3.0 |
added 35 characters in body
|
Jun 8, 2017 at 22:55 | history | edited | user237522 | CC BY-SA 3.0 |
added 60 characters in body
|
Jun 8, 2017 at 22:48 | history | edited | user237522 | CC BY-SA 3.0 |
added 398 characters in body
|
Jun 8, 2017 at 22:26 | history | edited | user237522 | CC BY-SA 3.0 |
edited title
|
Jun 8, 2017 at 22:11 | comment | added | user237522 | @R.vanDobbendeBruyn, thanks for your comment. Please, do you think that you can answer my above question with 'normal' (= integrally closed in its field of fractions) instead of 'UFD'? (I guess no? since you have written: "Neither of these can be easily read of from $f$"). | |
Jun 8, 2017 at 22:04 | history | edited | user237522 | CC BY-SA 3.0 |
added 12 characters in body
|
Jun 8, 2017 at 21:59 | comment | added | user237522 | I am not sure if and how the following paper may help solve my question: ams.org/journals/proc/1997-125-01/S0002-9939-97-03663-0/… | |
Jun 8, 2017 at 21:56 | comment | added | user237522 | Perhaps I should first ask my question with one less variable. Even in this case, I am not sure if it is possible to describe all irreducible polynomials in two variables, only bring several (nice) sufficient conditions, which are brought in mathoverflow.net/questions/14076/… | |
Jun 8, 2017 at 21:52 | comment | added | user237522 | @JohannesHahn, thanks. Yes, I had in mind an integral domain $B$, so I will add to my question the assumption that $f$ is a prime (=irreducible) element of $\mathbb{C}[x,y,T]$. I guess it is quite complicated to describe all the irreducibles of the polynomial ring in three variables. (In order to be a UFD there should be some additional restrictions). | |
Jun 8, 2017 at 21:07 | comment | added | R. van Dobben de Bruyn | A Noetherian domain $R$ is a UFD if and only if it is normal with $\operatorname{Cl}(R) = 0$. Neither of these can be easily read of from $f$. For example, there are polynomials of arbitrary degree for which $R$ is and is not a UFD: $T^n - x$ always gives a UFD (namely $\mathbb C[y,T]$), and $T^n - xy$ never does if $n > 1$ (as $T^n = xy$). | |
Jun 8, 2017 at 19:44 | comment | added | Johannes Hahn | I assume that you want $f$ to be a prime in order for $B$ to be an integral domain, right? Otherwise, would you please clarify how "UFD" is defined in the presence of zero divisors? | |
Jun 8, 2017 at 19:11 | history | edited | user237522 | CC BY-SA 3.0 |
added 184 characters in body
|
Jun 8, 2017 at 18:27 | history | edited | user237522 | CC BY-SA 3.0 |
added 68 characters in body
|
Jun 8, 2017 at 17:19 | history | asked | user237522 | CC BY-SA 3.0 |