Is a smooth ring extension of a UFD a UFD?

Question

Let $A \subseteq B$ be noetherian integral domains, $A$ a UFD, and $B$ a smooth $A$-algebra (=the definition of a smooth algebra can be found in https://mathoverflow.net/questions/207595/when-a-smooth-algebra-is-regular).

Is $B$ a UFD too?

Actually, I have a specific example in mind where $B$ is normal (=integrally closed), but I wish to know if $B$ is a UFD.

Also, in my specific example, the fraction field of $A$ is NOT assumed to equal the fraction field of $B$, so I prefer an answer not assuming this.

Answer 1

The answer is "no".
Indeed $A=\mathbb R[X]$ is a PID, and hence a UFD. However if you take $$B=\mathbb R[X,Y,Z]/\langle Y^2+Z^2-1\rangle$$ you obtain a smooth $A$-algebra which is not a UFD and this example satisfies all your requirements: $A$ and $B$ are noetherian normal domains with different fraction fields.

Edit
At the OP's request here is an example with $\mathbb C$ as the base field:
Take $A=\mathbb C[X]$ and $B=\mathbb C[X,Y,Z,T]/\langle Y^2+Z^2+T^2-1\rangle$.
Again $A,B$ are noetherian, normal integral domains and $B$ is a smooth $A$-algebra, and yet $B$ is not a UFD.

Note carefully that the obvious candidate $B'=\mathbb C[X,Y,Z]/\langle Y^2+Z^2-1\rangle$ for answering the OP's new request for an example over $\mathbb C$ doesn't work because $B'$ is factorial.

Answer 2

The answer is "no" even if $B$ is étale over $A$.

Let $X$ be a smooth projective curve of positive genus over $\mathbb{C}$. Let $f \colon X \to \mathbb{P}^1$ be a nonconstant morphism. Fix a nonempty open subvariety $U \subsetneq \mathbb{P}^1$ such that $f^{-1} U \to U$ is étale. Write $U = \operatorname{Spec} A$ and $f^{-1} U = \operatorname{Spec} B$. Then $A$ and $B$ are Dedekind domains, and the class group $\operatorname{Pic}(A)$ is a quotient of $\operatorname{Pic} (\mathbb{C}[x]) = 0$, but $\operatorname{Pic}(B)$ is a quotient of the uncountable group $\operatorname{Pic}(X)$ by a finitely generated subgroup, hence uncountable.