Quotient ring is a field

I am asked whether $\mathbb{F}_4[x]/(x^2+x+1)$ is a field, where $\mathbb{F}_4$ is the field of $4$ elements.

I'm not too sure where to go with this. Since $\mathbb{F}_4$ is a field, it is a PID, and that polynomial is an ideal. However I'm not able to determine whether it is maximal or not, as I'm not sure how $\mathbb{F}_4$ looks like since it's not $\mathbb{Z}_4$ obviously. If there's some type of eisenstein's criteron for irreducibility in this field or anything I'd greatly appreciate some help.

• "That polynomial is an ideal" - if $a$ is an element of a ring $R$, the notation $(a)$ means the ideal generated by $a$: $\{ra\mid r\in R\}$. You don't need the fact that $\mathbb{F}_4[x]$ is a PID to get that $(x^2+x+1)$ is an ideal - it's an ideal by definition! Rather, the fact that this ring is a PID means that every ideal has the form $(p(x))$ for some polynomial $p(x)$. – Alex Kruckman Apr 23 '18 at 0:20
• If $x^2+x+1$ is reducible, it factors into two polynomials of degree $1$. $\mathbb{F}_4$ only has four elements, so there are only $12$ non-constant polynomials of degree $1$. So if you can write down a multiplication table for $\mathbb{F}_4$, then you can just check by hand whether the polynomial is irreducible. – Alex Kruckman Apr 23 '18 at 0:22
• @AlexKruckman yes you are correct. I realize I may have been throwing out extraneous information to get me on some track as to whether that ideal is maximal. also why $12$ and not $16$? If linear polynomials of form $ax + b$ where both $a,b \in \mathbb{F}_4$ – blanchey Apr 23 '18 at 0:25
• If $a=0$, the polynomial doesn't have degree $1$. – Alex Kruckman Apr 23 '18 at 0:26

$x^2+x+1$ is reducible with roots $\alpha$ and $\beta$ where $\alpha$ and $\beta$ are the elements of $\mathbb{F}_4$ not in $\mathbb{F}_2$.
We can also note every polynomial of degree 2 with coefficients in $\mathbb{F}_2$ has a root in a degree 2 extension of $\mathbb{F}_2$, therefore has a root in $\mathbb{F}_4$ which is the only degree 2 extension of $\mathbb{F}_2$.
Therefore $\mathbb{F}_4[x]/(x^2+x+1)$ is not a field.
If $\Bbb F_4=\{0,1,\omega,\omega+1\}$, where $\omega$ and $\omega+1$ are (primitive) cube roots of unity, both of them being roots of $X^2+X+1$, then we get, over $\Bbb F_4$, $$X^2+X+1=(X+\omega)(X+\omega+1)\,,$$ and it turns out that $\Bbb F_4[X]/(X^2+X+1)\cong\Bbb F_4\oplus\Bbb F_4$, where the isomorphism is induced by $X\mapsto(\omega,\omega+1)\in\Bbb F_4\oplus\Bbb F_4$.