# For which monic irreducible $f \in \mathbb{C}[x,y][T]$, $\mathbb{C}[x,y][T]/(f)$ is a UFD?

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!

• 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 19:44
• 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 21:07
• @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 21:52
• 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 21:56
• 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 9, 2017 at 0:17