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Will $Z[X] / (x^2 +1)$ be a PID?

My Try : It is not a PID. If $R$ = $Z[X] / (x^2 +1)$ were a PID then $2$ will be an irreducible element in $R$ and $R/(2)$ is a field. But $x$ will not have an inverse element in $R/(2)$. contradiction

Have I gone wrong anywhere? Please correct me if I have.

Thank You in Advance.

INDIAN
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    The Gaussian integers $\Bbb Z[i]$ are well-known to form a Euclidean domain. – Angina Seng May 07 '18 at 12:33
  • I could not connect $Z[i]$ to my question?@LordSharktheUnknown – INDIAN May 07 '18 at 12:36
  • @INDIAN $\Bbb Z[X]/(X^2 + 1) \cong \Bbb Z[i]$ and you can use various methods to show that this is a principal ideal domain; e.g. use the Minkowski bound to show that the class group of $\Bbb Z[i]$ is trivial, or the proof that $\Bbb Z[i]$ is a Euclidean domain and then use that every Euclidean domain is a principal ideal domain. – Edward Evans May 07 '18 at 12:39
  • I got you. Actually I want to learn. Can you please point out where did I wrong?@ÍgjøgnumMeg – INDIAN May 07 '18 at 12:42
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    In the Gaussian integers, $2$ is not irreducible. It is the product $(1+i)(1-i)$ – G Tony Jacobs May 07 '18 at 12:46

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