# Something wrong my proof that $\mathbb{Z}[x]$ is not a PID nor an Euclidean domain?

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 $\mathbb{Q}[x]$?

• The ideal generated by $x^2+1$ and $x$ is the full ring $\mathbb{Z}[x]$. – André Nicolas Nov 10 '15 at 5:40
• Ah, because $\gcd(x^2+1, x) = 1$? – Kevin Sheng Nov 10 '15 at 5:42
• Or more simply because $x^2+1-(x)(x)=1$. Same idea. – André Nicolas Nov 10 '15 at 5:43
• Try the ideal generated by $x$ and $2$. – André Nicolas Nov 10 '15 at 5:48
• Yes, that is the proof my book gives but I was just trying to come up with an alternative approach. Thanks for the help. – Kevin Sheng Nov 10 '15 at 5:49

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.

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".

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