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Is there any counterexample that will disprove that every unique factorization domain (UFD) is also a principal ideal domain (PID)? I mean, any PID is a UFD, does the converse hold?

Thanks in advance!

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    $\begingroup$ $\mathbb C[x, y]$ is a UFD but not a PID. $\endgroup$ – Dustan Levenstein Mar 3 '14 at 4:48
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    $\begingroup$ $\mathbb{Z}[x]$ is another common example of a UFD that is not a PID. $\endgroup$ – BW. Mar 3 '14 at 4:51
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    $\begingroup$ PID = UFD of dimension 1 (i.e. non-zero prime ideals are maximal). $\endgroup$ – Cantlog Mar 3 '14 at 7:53
  • $\begingroup$ See this math.stackexchange.com/questions/16754/… $\endgroup$ – FedeB Mar 3 '14 at 10:42
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    $\begingroup$ This Q is surely a duplicate. Perhaps someone can find a link (including the standard example $K[x,y]$). $\endgroup$ – Martin Brandenburg Mar 3 '14 at 11:55
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A PID is always of dim $1$. So you can find a lot of UFD's that are not PID; for example:
Every polynomial ring in more than one variable with coeffcients in a field.
Every regular local ring of dim greater than $1$ (i.e the maximal ideal can not generated with one element).

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