# Is the ideal $(2,x^4+x^2+1)<\mathbb{Z}[x]$ maximal?, principal?

I'm trying to solve the following problem:

Let I=$(2,x^4+x^2+1)<\mathbb{Z}[x]$ be an ideal.

• Is $I$ maximal?
• Is $I$ principal?

Any help would be appreciated.

• You might be able to use that $x^6-1=(x^2-1)(x^4+x^2+1)=(x^3+1)(x^3-1)$ Aug 6, 2014 at 8:38
• Maximal ideals of $\mathbb{Z}[x]$ are of the form $(p,f(x))$, where $p$ is a prime number and $f(x)$ is irreducible modulo $p$. Can you show why? (for starters, take an arbitrary maximal ideal $M$, show that $M\cap\mathbb{Z}\ne 0$, and pick a suitable quotient of $\mathbb{Z}$ that injects into $\mathbb{Z}[x]/M$). Aug 6, 2014 at 8:46

Extending the comment of Warren Moore, we see that for $p=2$ we have $x^4+x^2+1\equiv (x^2+x+1)^2 \mod p$. Hence the polynomial is reducible over $\mathbb{F}_2$; and the quotient $\mathbb{Z}[x]/I$ is not a field, so that $I$ is not maximal. Regarding principal ideals, it could help reviewing the classical example $I=(2,x)$.

• Note that $(x^4+x^2+1)=(x^2+x+1)(x^2-x+1)$ over $\mathbb Z$, so will be reducible for any choice of $p$. Aug 6, 2014 at 9:14

With your help, here is my solution:

Let $p(x) := x^4+x^2+1$ a polynomial in $(\mathbb{Z}/2\mathbb{Z})[x]$.

$\mathbb{Z}[x]/I\cong(\mathbb{Z}/2\mathbb{Z})[x]/(p(x))$

$p(x)=(x^2+x+1)(x^2+x+1) \Rightarrow \mathbb{Z}[x]/I$ is not integral $\Rightarrow \mathbb{Z}[x]/I$ is not a field.

$\Rightarrow I$ is not maximal.

By contradiction, if $I$ is principal there is a polynomial $q(x)\in \mathbb{Z}[x]$ such that $(q)=I$.

We must have $2\in(q)\Rightarrow \exists p_1\in \mathbb{Z}[x]$ such that $2=qp_1\Rightarrow$ deg(q)=0

$q$ cannot be $0$ so $q=1\Rightarrow (q)=\mathbb{Z}[X]\Rightarrow \mathbb{Z}[X]/I=\{0\}$ which is a contradiction.

$\Rightarrow$ $I$ is not principal.

Please, tell me if this is correct or if there is mistakes.

• Your decomposition of $p(x)$ only holds modulo $2$. Aug 6, 2014 at 11:22
• Yes, it's true, $p(x)\in(\mathbb{Z}/2\mathbb{Z})[x]$ and $\mathbb{Z}[x]/I\equiv(\mathbb{Z}/2\mathbb{Z})[x]/(p(x))$. Is there a mistake? Aug 6, 2014 at 12:10
• No, there is no mistake, I think ! Aug 6, 2014 at 12:25