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

Question

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.

Answer 1

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 $\Bbb Z[x]$.

Answer 2

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 $\mathbb Z/13 \mathbb Z$.

Answer 3

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.