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

Question

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.

Answer 1

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 $I=(2,x)$.

Answer 2

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.