$R/\mathfrak p$ not always a UFD

Question

I am looking for a nice counterexample that for a UFD $R$ and $\mathfrak p\subset R$ a prime ideal, $R/\mathfrak p$ is not always a UFD as well.

Answer 1

Pick any finitely generated domain which is not a UDF and write it as a quotient of $\mathbb Z[X_1,\dots,X_n]$. The ideal will be prime.

Answer 2

A good geometric example is $R=k[x,y]$ and $\mathfrak p=(y^2-x^3)$.