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.
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1Actually, you might also have dimension $0$ for a PID, but this is not really important for the question (and it is easy to find examples of dimension $0$ nonPIDs). – Tobias Kildetoft Jun 22 '15 at 08:43

Take the ring of integers of a number field where unique factorization fails. – David Benjamin Lim Jun 22 '15 at 09:17

@Mr.Chip: remember that a Dedkind domain (as the ring of integers of a number field) is a PID iff it is a UFD — this could help. – Watson Aug 23 '16 at 11:47
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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.
Fredrik Meyer
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$\mathbb Z[\sqrt{n}]$, for $n\ge 3$ and squarefree, is onedimensional (why?) and it's not a UFD.