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.

  • 1
    $\begingroup$ Actually, 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 $0$ non-PIDs). $\endgroup$ Jun 22, 2015 at 8:43
  • $\begingroup$ Take the ring of integers of a number field where unique factorization fails. $\endgroup$ Jun 22, 2015 at 9:17
  • $\begingroup$ @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. $\endgroup$
    – Watson
    Aug 23, 2016 at 11:47

2 Answers 2


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.

  • 4
    $\begingroup$ Can you explain why $Frac(A)=k(t)$ implies that $dim(A)=1$? $\endgroup$
    – Ninja
    Nov 21, 2017 at 1:24

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .