When a quotient of a UFD is also a UFD?

Question

Let $R$ be a UFD and let $a\in R$ be nonzero element. Under what conditions will $R/aR$ be a UFD?

A more specific question:

Suppose $R$ is a regular local ring and let $I$ be a height two ideal which is radical. Can we find an element $a\in I$ such that $R/aR$ is a UFD?

Answer 1

This is indeed a complicated question, that has also been much studied. Let me just more or less quote directly from Eisenbud's Commutative Algebra book (all found in Exercise 20.17):

The Noether-Lefschetz theorem: if $R = \mathbb{C}[x_1, \ldots x_4]$ is the polynomial ring in $4$ variables over $\mathbb{C}$, then for almost every homogeneous form $f$ of degree $\ge 4$, $R/(f)$ is factorial.

On the other hand, in dimension $3$, there is a theorem of Andreotti-Salmon:

Let $(R,P)$ be a $3$-dimensional regular local ring, and $0 \ne f \in P$. Then $R/(f)$ is factorial iff $f$ cannot be written as the determinant of an $n \times n$ matrix with entries in $P$, for $n > 1$.