Show $(29,x^2+1)\subset\mathbb{Z}[x]$ is not a maximal ideal

Question

I know the maximal ideals of $\mathbb{Z}[x]$ are of the form $(p,f(x))$ where $f(x)$ is irreducible mod $p$. Here, $17$ is a root of $x^2+1$ mod $29$, so $f(x)$ is not irreducible mod $29$. Thus $(29,x^2+1)$ cannot be a maximal ideal. However, I was hoping to find some "more direct" proofs of this, without leveraging as much machinery. Perhaps we can find some ideal properly contained in $\mathbb{Z}$ that also properly contains $(29,x^2+1)$? I was thinking something like $(29,x^2+1,x)$, but haven't been able to show it properly contains $(29,x^2+1)$. Maybe another way to do this is to assume $(29,x^2+1)$ is indeed maximal, and then derive a contradiction from the fact that $\mathbb{Z}[x]/(29,x^2+1)$ is a field? I'd just like to see a way to do this without using the full classification of the maximal ideals of $\mathbb{Z}[x]$.

Answer 1

We have $x^{2}+1 = (x-17)(x+17) + 290$, hence $x^{2}+1 \in (29,x-17)$; hence $(29,x^{2}+1) \subseteq (29,x-17)$. To show that this inclusion is proper, we show that $x-17 \not\in (29,x^{2}+1)$. For the sake of contradiction, suppose $x-17 = 29 \cdot f + (x^{2}+1) \cdot g$ for some $f,g \in \mathbb{Z}[x]$; plug in $x = -17$ to get $-34 = 29 \cdot f(-17) + 290 \cdot g(-17)$ which means $29$ divides $34$, which is a contradiction.