Show that $\mathbb{Z}_3\left [ x \right ]/\left \langle x^{2}+x+1 \right \rangle$ is not a field Show that $\mathbb{Z}_3\left [ x \right ]/\left \langle x^{2}+x+1 \right \rangle$ is not a field.
It suffice to show that the principal ideal $\left \langle x^{2}+x+1 \right \rangle$ is not a maximal ideal. 
However, I am unable to get the ball rolling. Any hint is appreciated.
 A: Observe that in $\Bbb Z_3 [x]$ we have
$(x - 1)^2 = x^2 -2 x + 1 = x^2 + x + 1, \tag{1}$
since $-2x = x$ in $\Bbb Z_3 [x]$ .  (1) not only shows that $1 \in \Bbb Z_3$ is a zero of $x^2 + x + 1 \in \Bbb Z_3[x]$, but also that
$(x -  1)^2 = 0 \in \Bbb Z_3[x]/\langle x^2 + x + 1 \rangle; \tag{2}$
however,
$x - 1\ne 0 \in  \Bbb Z_3[x]/\langle x^2 + x + 1 \rangle, \tag{3}$
so we see that $ \Bbb Z_3[x]/\langle x^2 + x + 1 \rangle$ contains non-zero nilpotent elements; not only is  $\Bbb Z_3[x]/\langle x^2 + x + 1 \rangle$ not a field, it is not even an integral domain!
As a final note, if $x - 1$ were invertible in $\Bbb Z_3[x]/\langle x^2 + x + 1 \rangle$, i.e. if we had $y \in \Bbb Z_3[x]/\langle x^2 + x + 1 \rangle$ with
$(x - 1)y = 1, \tag{4}$
then
$x - 1 = (x - 1)(1) = (x - 1)((x - 1)y) = (x - 1)^2y = 0, \tag{5}$
clearly not the case for $x - 1 \in \Bbb Z_3[x]/\langle x^2 + x + 1 \rangle$.
A: Since $1$ is a root of $x^2+x+1$, this polynomial is not irreducible in $\mathbb{Z}_3[x]$ and therefore the ideal $\langle x^2+x+1\rangle$ is not maximal.
A: Expanding on Lord Shark the Unknown's comment (what a wonderful name btw), a field has the property that every nonzero element of the field has a multiplicative inverse.
i.e. $\forall p(x) \in \mathbb{Z}_3\left [ x \right ]/\left \langle x^{2}+x+1 \right \rangle \backslash \{0\}, \exists p^{-1}(x): p(x)p^{-1}(x) = p^{-1}(x)p(x) = 1$
If you take the element $x-1$, you will recognize that there is no element that multiplies against $x-1$ to yield $1$. This can be exhaustively checked, or if you're clever, you can use a fact about zero divisors, to show that since $(x-1)^2 = 0$, there can't be a multiplicative inverse. 
