Isn't $ x^2+1$ irreducible in $\mathbb Z$, then why is $\langle x^2+1 \rangle$ not a maximal ideal in $\mathbb Z[x]$?

$ x^2+1$ cannot be broken down further non trivially in $\mathbb Z[x]$. hence, it's irreducible in $\mathbb Z[x]$. Hence, shouldn't $\mathbb Z[x]/\langle x^2+1 \rangle$ be a field and hence, $ x^2+1$ a maximal ideal in $\mathbb Z[x]$

Thanks for your help.

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    $\begingroup$ Your «hence» is just not true: it is true that if $K$ is a field and $p\in K[x]$ is irreducible, then $K[x]/(p)$ is a field. Yet $\mathbb Z$ is not a field. $\endgroup$ – Mariano Suárez-Álvarez Jul 27 '14 at 2:33
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    $\begingroup$ For the same reason that $(x)$ isn’t a maximal ideal. $\endgroup$ – Lubin Jul 27 '14 at 2:37
  • $\begingroup$ Got it. I completely forgot the fact that $Z$ is not a field. Thank you for your comments $\endgroup$ – MathMan Jul 27 '14 at 2:38

Because the quotient is not a field, as you can easily check!

For example, the class of $2$ is neither zero nor invertible in $\mathbb Z[x]/(x^2+1)$.

  • $\begingroup$ Yes, I was able to confirm that the quotient is not a field. But, we have a theorem that : Let $F$ be a field and let $p(x) \in F[x]$. then $\langle p(x) \rangle$ is a maximal ideal in $F[x]$ if and only if $p(x)$ is irreducible over $F$ $\endgroup$ – MathMan Jul 27 '14 at 2:34
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    $\begingroup$ Well, $\mathbb Z$ is not a field... $\endgroup$ – Mariano Suárez-Álvarez Jul 27 '14 at 2:34
  • $\begingroup$ Ohhh :( .. I just lost it . Thank you for the answer $\endgroup$ – MathMan Jul 27 '14 at 2:35

More directly, the ideal $\langle x^2\!+\!1 \rangle$ is properly contained in the ideal $\langle x^2\! +\! 1, \ 3 \rangle$. By definition, it cannot be maximal.

In particular, it is a theorem that no principal ideal in $\mathbb{Z}[x]$ is maximal. In fact, all maximal ideals of $\mathbb{Z}[x]$ are of the form $\langle p, f(x) \rangle$ where $p$ is prime and $f(x)$ is irreducible over $\mathbb{F}_p$.

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    $\begingroup$ Glad I could help! $\endgroup$ – Kaj Hansen Jul 27 '14 at 7:18

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