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My proof goes as follows:

Suppose for contradiction that $\mathbb{Z}[x]$ is a PID. Then the ideal generated by any irreducible element is maximal. We know that $x^2+1$ is irreducible in $\mathbb{Z}[x]$ so $(x^2+1)$ should be maximal; $(x^2+1) \ne \mathbb{Z}[x]$ since $1 \notin (x^2+1)$. However, $(x^2+1) \subset (x^2+1, x)$, and the containment is strict since $x \notin (x^2+1)$. Thus $(x^2+1)$ is not maximal, and $\mathbb{Z}[x]$ is not a PID, and therefore also not an Euclidean domain.

I feel like I made a mistake somewhere because $x^2+1$ is also irreducible over $\mathbb{Q}$ so the same argument should work for $\mathbb{Q}[x]$, but I know that $\mathbb{Q}[x]$ is an Euclidean domain (and thus a PID) since $\mathbb{Q}$ is a field. Where is my mistake? Am I wrong to say $(x^2+1) \subset (x^2+1, x)$ when considered as ideals over $\mathbb{Q}[x]$?

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  • $\begingroup$ The ideal generated by $x^2+1$ and $x$ is the full ring $\mathbb{Z}[x]$. $\endgroup$ – André Nicolas Nov 10 '15 at 5:40
  • $\begingroup$ Ah, because $\gcd(x^2+1, x) = 1$? $\endgroup$ – Kevin Sheng Nov 10 '15 at 5:42
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    $\begingroup$ Or more simply because $x^2+1-(x)(x)=1$. Same idea. $\endgroup$ – André Nicolas Nov 10 '15 at 5:43
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    $\begingroup$ Try the ideal generated by $x$ and $2$. $\endgroup$ – André Nicolas Nov 10 '15 at 5:48
  • $\begingroup$ Yes, that is the proof my book gives but I was just trying to come up with an alternative approach. Thanks for the help. $\endgroup$ – Kevin Sheng Nov 10 '15 at 5:49
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The example does not work, because the ideal generated by $x^2+1$ and $x$ is all of the ring.

Try the ideal generated by $2$ and $x$. It should not be hard to show that this ideal is not principal.

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Well, "the ideal generated by any irreducible element is maximal" is not true.
Instead, "the ideal generated by any irreducible element is prime" is always true.

In fact, irreducibility means "$fg∈I \Rightarrow f∈I$ or $g∈I$", which is the property of prime ideal.

EDIT I was also wrong: A counterexample is $3 \in Z[\sqrt{-5}]$ (referred to in Wikipedia).
The truth is "the ideal generated by any prime element is prime".

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  • $\begingroup$ But if you're in a UFD, then irreducible = prime! $\endgroup$ – oxeimon Nov 15 '15 at 2:37
  • $\begingroup$ Yes, so I had to say Z[X] is UFD, from the fact that if R is UFD then R[X] is also UFD. $\endgroup$ – aerile Nov 16 '15 at 13:40

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