I need to determine which, if any, of the following ideals in $\mathbb{Z}_{3}[x]$are maximal:
$$\mathbf{(2)},\, \mathbf{(x+1)},\, \text{and}\, \mathbf{(x^{2}+x+1)}$$
(i.e., the principal ideals generated by the polynomials $2$, $x+1$, and $x^{2}+x+1$, respectively).
This is my attempt:
- For $(x+1)$:
$\mathbb{Z}_{3}$ is a field, since $3$ is prime. Therefore, $\mathbb{Z}_{3}[x]$ is an integral domain.
Moreover, since $\mathbb{Z}_{3}$ is a field, it is vacuously a unique factorization domain. Therefore, $\mathbb{Z}_{3}[x]$ is also a unique factorization domain.
Now, for $f = x+1 = x - (-1) = x-2$ in $\mathbb{Z}_{3}[x]$, the content is $C(f) = 1$, which is a unit of $\mathbb{Z}_{3}$. Therefore, $x+1$ is primitive in $\mathbb{Z}_{3}[x]$, which implies that $x+1$ is irreducible in $\mathbb{Z}_{3}[x]$. Since $\mathbb{Z}_{3}[x]$ is an integral domain, $(x+1)$ is a maximal ideal of $\mathbb{Z}_{3}$
- For $(x^{2}+x+1)$:
In $\mathbb{Z}_{3}[x]$, $g = x^{2}+x+1=(x+2)(x+2)$, so $g$ is not irreducible there. Therefore, since $\mathbb{Z}_{3}[x]$ is an integral domain, $(x^{2}+x+1)$ is not maximal there.
- For $(2)$:
For $h = 2$, the content $C(h) = 2$, which is a unit of $\mathbb{Z}_{3}$. Hence, $h$ is primitive in $\mathbb{Z}_{3}[x]$
Any possible factorization of $2$ in $\mathbb{Z}_{3}$ will include $2$, so $2$ is an irreducible element of $\mathbb{Z}_{3}[x]$, which implies that $(2)$ is maximal, since $\mathbb{Z}_{3}[x]$ is an integral domain.
Is this correct? If not, what can I do to fix them? I'm worried I might be missing some subtlety here as to why a nonzero element $k \in \mathbb{Z}_{3}[x]$ irreducible implies $(k)$ maximal. I have a theorem that says that such "$k$ in an integral domain $R$ is irreducible if and only if $(k)$ is a maximal ideal in the set of all principal ideals of $R$", but does this necessarily mean that it is a maximal ideal of $R$ period?