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

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 $\mathbb Z[x]$

• 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. – Mariano Suárez-Álvarez Jul 27 '14 at 2:33
• For the same reason that $(x)$ isn’t a maximal ideal. – Lubin Jul 27 '14 at 2:37
• Got it. I completely forgot the fact that $Z$ is not a field. Thank you for your comments – 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 $\mathbb Z[x]/(x^2+1)$.

• 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 $F$ – MathMan Jul 27 '14 at 2:34
• Well, $\mathbb Z$ is not a field... – Mariano Suárez-Álvarez Jul 27 '14 at 2:34
• Ohhh :( .. I just lost it . Thank you for the answer – 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 $\mathbb{F}_p$.

• Glad I could help! – Kaj Hansen Jul 27 '14 at 7:18