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Jun 15, 2020 at 7:27 history edited CommunityBot
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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
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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
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Jun 8, 2017 at 22:55 history edited user237522 CC BY-SA 3.0
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Jun 8, 2017 at 22:48 history edited user237522 CC BY-SA 3.0
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Jun 8, 2017 at 22:26 history edited user237522 CC BY-SA 3.0
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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
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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
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Jun 8, 2017 at 18:27 history edited user237522 CC BY-SA 3.0
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Jun 8, 2017 at 17:19 history asked user237522 CC BY-SA 3.0