Show the ideal $\langle x^2+1, 5\rangle$ isn't maximal

Question

Show the ideal generated by $x^2+1$ and $5$ isn't maximal in $\mathbb{Z}[x]$.

I thought of the following:

An element of $\mathbb{Z}[x] /I$ is $f(x) + \langle x^2 +1 , 5 \rangle = ax+b + I$ where $a , b \in \mathbb{Z}_5$.

$((x+2) + I)((x+3) + I) = x^2 + 1 + I = I$, so the factor ring has zero divisors. It's not a field, and so I isn't maximal.

Is this correct? Is there any other ideas?

Answer 1

Observe that $$ \Bbb{Z}[x]/\langle x^2 + 1, 5 \rangle \simeq \Bbb{F}_5[x] / \langle x^2 + 1 \rangle $$ where $\Bbb{F}_5 = \Bbb{Z}/5\Bbb{Z}$ is the field with $5$ elements. Now note that $x^2 + 1 = (x - 2)(x + 2)$ over $\Bbb{F}_5$, because $(\pm 2)^2 = 4 \equiv -1 \pmod{5}$. Hence $$ \Bbb{F}_5[x] / \langle x^2 + 1 \rangle \simeq \Bbb{F}_5[x] / \langle x + 2 \rangle \oplus \Bbb{F}_5[x] / \langle x - 2 \rangle $$ and this has non-zero zero-divisors, for example $(1,0)(0,1) = (0,0)$.

Answer 2

We have the inclusion of ideals $$ \langle x^2+1,5\rangle \subset \langle x+2,5\rangle\subset \mathbb Z[x]. $$ Indeed, this follows from $x^2+1=(x+2)(x-2)+5$.

Next, observe that $x+2\not\in \langle x^2+1,5\rangle$. Indeed, the only degree $1$ polynomials in the latter ideal are of the form $5(ax+b)$. Hence $\langle x^2+1,5\rangle$ is not maximal.