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Show that $R = \mathbb{Z}[\sqrt{-17}]$ is not a Euclidean ring. To do this I tried showing that the ring is not a principal ideal domain. I wonder if this is enough and how to actually show that it is not a PID. Thanks!

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Reminder. A euclidean ring is factorial.

Hint. Notice that: $$18=2\times 3\times 3=(1+i\sqrt{17})(1-i\sqrt{17}).$$ If you can prove that $2,3,1-i\sqrt{17}$ and $1+i\sqrt{17}$ are irreducible, you are done!

Remark. If you can show that $\mathbb{Z}[i\sqrt{17}]$ is not a PID, you are also done. Try to consider the ideal generated by $2$ and $1+i\sqrt{17}$ and proceed by contradiction. What if there exists $z=a+ib\sqrt{17}$ such that: $$(z)=(2,1+i\sqrt{17}).$$ What will happen if you take the norm of $z$ which is $a^2+17b^2$?

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  • $\begingroup$ Thanks for the answer, I was indeed specifically wondering about the PID way of showing the desired result! $\endgroup$ – Slugger Jun 15 '16 at 15:15
  • $\begingroup$ In this case you say "A euclidean ring". Another expection is with "u" when pronounced like "you" as in "a uniform" for example. $\endgroup$ – Gregory Grant Jun 15 '16 at 15:15
  • $\begingroup$ @GregoryGrant Thank you for this correction! $\endgroup$ – C. Falcon Jun 15 '16 at 15:17

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