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Let $f=f(T) \in \mathbb{C}[x,y][T]$ be a monic irreducible polynomial: $f=T^n+a_{n-1}T^{n-1}+\cdots+a_1T+a_0$, $a_j \in \mathbb{C}[x,y]$, $0 \leq j \leq n-1$.

Denote $B=\mathbb{C}[x,y,T]/(f)=\mathbb{C}[x,y][w]$, where $w^n+a_{n-1}w^{n-1}+\cdots+a_1w+a_0=0$.

Of course, $\mathbb{C}[x,y]$ is a UFD; however, it seems that $B$ does not have to be a UFD (for example, if $f=T^2-xy$, then $B$ is not a UFD, since $w^2=xy$).

(1) Is it possible to find a general form of $f$, for which $B$ is a UFD?

(2) What happens if we do not assume that $f$ is monic? (I guess being monic may yield a nicer answer?).

(3) It seems (at least to me) that the answer of Ben Webster to this question is relevant.

Perhaps relevant questions are the following: 1 and 2 (which asks when a quotient of a UFD is a UFD).

Thank you very much!

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  • $\begingroup$ 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? $\endgroup$ – Johannes Hahn Jun 8 '17 at 19:44
  • $\begingroup$ 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 $T^n = xy$). $\endgroup$ – R. van Dobben de Bruyn Jun 8 '17 at 21:07
  • $\begingroup$ @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). $\endgroup$ – user237522 Jun 8 '17 at 21:52
  • $\begingroup$ 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/… $\endgroup$ – user237522 Jun 8 '17 at 21:56
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    $\begingroup$ 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. $\endgroup$ – R. van Dobben de Bruyn Jun 9 '17 at 0:17

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