Example: Krull dimension 1 but not a PID

Question

It's easy to prove that if $A$ is a PID which is not a field then $\dim A= 1$. What is a counterexample to the converse? Thanks for any insight.

Answer 1

I'm not sure how much algebraic geometry you know, but one way to think about failure of unique factorization is that there is some singularity in the associated variety.

So an example would be $A=k[t^2,t^3]$. This has Krull dimension 1 (since its fraction field is $k(t)$), the ideal $(t^2,t^3)$ cannot be generated by less than two elements.

Answer 2

$\mathbb Z[\sqrt{-n}]$, for $n\ge 3$ and square-free, is one-dimensional (why?) and it's not a UFD.