Show that the ideal generated by $13$ and $x^2 − x + 1$ in $\mathbb Z[x]$ is not maximal. 
I need to show that the ideal generated by $x^2 − x + 1$ and 13 in $\mathbb{Z}[x]$ is not maximal.

I think that I need to check first that the polynomial is irreducible, but not sure how to follow from that.
 A: We have $\mathbb{Z}[x]/(13) \cong \mathbb{Z}_{13}[x]$. In $\mathbb{Z}_{13}[x]$, 
$$x^2 - x + 1 = (x + 3)(x + 9). $$
Thus, the quotient $\mathbb{Z}_{13}[x]/(x^2 - x + 1)$ has proper zero divisors $\bar{x} + 3$ and $\bar{x} + 9$, where $\bar{x}$ is the image of $x$ in the quotient. Hence, it is not even an integral domain.
A: The maximal ideals of $\Bbb Z[x]$ are well-known to be of the form $(p, f(x))$, where $p \in \Bbb P$ is a prime and $f(x)$ is irreducible $\mod p$; see this essay for a proof.  Applying this criterion to $(13, x^2 - x + 1)$, we see that $x^2 - x + 1$ is reducible $\mod 13$, since $4 \in \Bbb Z_{13}$ is a root:
$4^2 - 4 + 1 = 16 - 4 + 1 = 13 \equiv 0 \mod 13; \tag 1$
thus
$x - 4 \mid x^2 - x + 1 \; \text{in} \; \Bbb Z_{13}; \tag 2$
indeed, in $\Bbb Z_{13}$,
$(x - 4)(x - 10) = x^2 - (4 + 10)x + 40$
$= x^2 - 14x + 40 \equiv x^2 - x + 1 \mod 13. \tag 3$
Since $x^2 - x + 1$ is reducible $\mod 13$, by the cited result $(13, x^2 - x + 1)$ is not maximal in $\Bbb Z[x]$.
A: Hint Consider the composition of the two functions
$$f: \mathbb Z[X] \to (\mathbb Z/13 \mathbb Z) [X]\\
f(P)= P \pmod{13} \\
\pi: (\mathbb Z/13 \mathbb Z) [X] \to \frac{(\mathbb Z/13 \mathbb Z) [X]}{\langle X^2-X+1 \rangle}  \\
\pi(Q)= Q +\langle X^2-X+1\rangle$$
Show that $\pi \circ f$ is an onto Ring homomorphism. What is its Kernel.
Hint 2: Show that $X^2-X+1$ has roots in $\mathbb Z/13 \mathbb Z$.
