Tensor product of finite type UFD algebras over an algebraically closed field is again UFD? Let $K$ be an algebraically closed field, $A$ and $B$ two finite type $K$-algebras which are assumed to be UFD. Is $A \otimes_K B$ again a UFD?
This question has been already asked here and here, but no (convincing) answer was given. I would like to add a few examples to show that the answer might be slightly less standard than it appears.
$\bullet$ If we drop the finite type assumption, the result is wrong. Indeed, take $\require{enclose}
     \enclose{horizontalstrike}{A = K[[T]]}$ and $\require{enclose}
     \enclose{horizontalstrike}{B = K[X,Y,Z]\Big/ \left(X^5 + Y^7-Z^2 \right)}$. Then Samuel proves in his lecture notes Lecture on unique factorization domains that $A$ and $B$ are UFD (see Theorem 8.1 in chapter 1 and Theorem 2.1 in chapter 2). On the other hand, he also proves that $\require{enclose}
     \enclose{horizontalstrike}{A \otimes_K B = B[[T]]}$ is not a UFD (see Theorem 9.1 in chapter 1).
Edit : as mentionned by Friedrich Knop, $A \otimes_K B$ is only a subalgebra of $B[[T]]$ (my stupid mistake) and so it's not clear wether it is a UFD or not.
$\bullet$ Even with the finite type assumption, the result doesn't seem to be known (at least at the time when Samuel wrote his lecture notes). Indeed, he states his Theorem 8.1 as:
Let $A$ be a UFD and $A[X_1, \ldots, X_n]$ be graded by assigning weights $\omega_i$ to $X_i$ ($\omega_i >0)$. Let $F(X_1, \ldots, X_n)$ be an irreducible homogeneous (for the grading) polynomial, $c$ be a positive integer, prime to the degree of $F$, and:
$$B = A[X_1,\ldots,X_n,Z] \Big/ \left(Z^c-F \right).$$
Then $B$ is a UFD if one of the following conditions is satisfied:

*

*$c \equiv 1 \ [\omega]$


*Every finitely generated projective $A$-module is free.
If Samuel knew that the tensor product of two finite type $K$-algebras (with $K$ algebraically closed) which are UFD is again a UFD, he would certainly have mentionned that:
$$B = A[X_1,\ldots,X_n,Z] \Big/ \left(Z^c-F \right) = A \otimes_K K[X_1, \ldots, X_n,Z]\Big/(Z^c-F)$$
is always a UFD when $A$ is a UFD which is a finite type algebra over $K$ algebraically closed (and he doesn't). At the end of the lecture notes, he gives various examples of such rings which are not UFD, but I can't find one where $K$ is an algebraically closed field and the second condition is the one that is not satisfied.
All in all, I would be very grateful to have a definitive answer on the initial question (even with a highly technological argument or counter-example).
 A: My previous answer gives a partial result under some additional hypotheses, but these are not needed.
Theorem (Boissière–Gabber–Serman). If $X$ and $Y$ are locally factorial varieties over an algebraically closed field $k$, then so is $X \times Y$. If $\operatorname{Cl}(X) = \operatorname{Cl}(Y) = 0$, then $\operatorname{Cl}(X \times Y)=0$.
See [BGS, Thm. 5.1 and Cor. 5.4]. $\square$
Taking $X = \operatorname{Spec} A$ and $Y = \operatorname{Spec} B$ for unique factorisation domains $A$ and $B$ shows that $A \otimes_k B$ is a unique factorisation domain.

References.
[BGS] S. Boissière, O. Gabber, O. Serman, Sur le produit de variétés factorielles ou $\mathbf Q$-factorielles. Preprint (2011). arXiv:1104.1861.
A: Recall that $A$ is a UFD if and only if $A$ is a normal domain with $\operatorname{Cl}(A) = 0$. The tensor product of normal domains over an algebraically closed field is a normal domain [Tags 06DF and 09P9], so the question is about the class group. I will write $U = \operatorname{Spec} A$ and $V = \operatorname{Spec} B$ and treat the more general case of normal varieties with $\operatorname{Cl}(U) = \operatorname{Cl}(V) = 0$.
When $U$ is locally factorial, we get $\operatorname{Cl}(U) \cong \operatorname{Pic}(U)$, and the question becomes substantially easier. Here is a positive result when $U$ and $V$ are locally factorial and admit compactifications $X$ and $Y$ that are locally factorial. (It seems that Mohan was already aware of this, at least in the smooth case, in the cited Math SE post, but no details were provided.)
Lemma. If $X$ and $Y$ are proper and locally factorial varieties over an algebraically closed field $k$ and $U \subseteq X$ and $V \subseteq Y$ are dense open subschemes with $\operatorname{Pic}(U) = \operatorname{Pic}(V) = 0$, then $\operatorname{Pic}(U \times V) = 0$.
Proof. The Picard schemes $\mathbf{Pic}_X$ and $\mathbf{Pic}_Y$ are representable, and there is an isomorphism
$$\mathbf{Pic}_{X\times Y}\cong\mathbf{Pic}_X\times\mathbf{Pic}_Y\times\mathbf{Hom}(\mathbf{Alb}_X,\mathbf{Pic}^0_Y).$$
(See for instance §4.4 of my dissertation [vDdB] for details and further references.) Denote the complements of $U$ and $V$ by $D$ and $E$ respectively (they may or may not be divisors). Writing $d = \dim X$ and $e = \dim Y$, we get exact sequences [Fulton, Prop. 1.8] (see also [Hartshorne, Prop. II.6.5])
\begin{align*}
\operatorname{CH}_{d-1}(D) &\to \operatorname{Cl}(X) \to \operatorname{Cl}(U) \to 0,\\
\operatorname{CH}_{e-1}(E) &\to \operatorname{Cl}(Y) \to \operatorname{Cl}(V) \to 0.
\end{align*}
Since $\operatorname{Cl}(U) = \operatorname{Cl}(V) = 0$ and $\operatorname{CH}_{d-1}(D)$ and $\operatorname{CH}_{e-1}(E)$ are finitely generated (by the cycle classes of the codimension $1$ components), we conclude that $\operatorname{Cl}(X)$ and $\operatorname{Cl}(Y)$ are finitely generated. In particular, $\operatorname{Pic}^0(X)$ and $\operatorname{Pic}^0(Y)$ are finitely generated, so they have to vanish since a positive-dimensional commutative group scheme over an algebraically closed field $k$ has infinitely many torsion elements (and infinite rank if $k \not\cong \bar{\mathbf F}_p$). Thus we see $\operatorname{Pic}(X \times Y) \cong \operatorname{Pic}(X) \times \operatorname{Pic}(Y)$. Again the sequence [Fulton, Prop. 1.8] gives
$$\operatorname{CH}_{d+e-1}\big((X \times Y) \setminus (U \times V)\Big) \to \operatorname{Pic}(X \times Y) \to \operatorname{Pic}(U \times V) \to 0.$$
But if the divisorial components of $D$ and $E$ are $D_1,\ldots,D_m$ and $E_1,\ldots,E_n$ respectively, then the codimension $1$ components of $X \times Y \setminus U \times V$ are $D_1 \times Y, \ldots, D_m \times Y$ and $X \times E_1, \ldots, X \times E_n$. These generate $\operatorname{Pic}(X \times Y)$, so we conclude that $\operatorname{Pic}(U \times V) = 0$. $\square$

References.
[Fulton]  W. Fulton, Intersection theory. Ergebnisse der Mathematik und ihrer Grenzgebiete 2. Springer-Verlag, 1984.
[Hartshorne]  R. Hartshorne, Algebraic geometry. Graduate Texts in Mathematics 52. Springer-Verlag, 1983.
[vDdB]  R. van Dobben de Bruyn, Dominating varieties by liftable ones. PhD Thesis, Columbia University, 2018.
