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. $\endgroup$ Mar 3, 2014 at 4:48
  • 10
    $\begingroup$ $\mathbb{Z}[x]$ is another common example of a UFD that is not a PID. $\endgroup$
    – Ben West
    Mar 3, 2014 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, 2014 at 7:53
  • $\begingroup$ See this math.stackexchange.com/questions/16754/… $\endgroup$
    – FedeB
    Mar 3, 2014 at 10:42
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    $\begingroup$ This Q is surely a duplicate. Perhaps someone can find a link (including the standard example $K[x,y]$). $\endgroup$ Mar 3, 2014 at 11:55

1 Answer 1


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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