This question is from a 2002 Harvard qualifier:
Let $R=\mathbb{Z}[x]/(f)$ where $f(x)=x^4 - x^3 + x^2 - 2x + 4$. Let $I = 3R$ be the principal ideal of $R$ generated by $3$. Find all prime ideals $\mathcal{P}$ of $R$ that contain $I$. (Give generators for each $\mathcal{P}$.)
My first thought was using the correspondence Theorem. Consider $\pi:\mathbb{Z}[x]\rightarrow \mathbb{Z}[x]/(f)$ the canonical projection. Then, ideals of $\mathbb{Z}[x]$ containing $(f)$ are in a one to one correspondence with ideals of $R$. Furthermore, this correspondence preserves prime ideals because if $(f)\subset I$ is such that $I$ is prime, then $\mathbb{Z}[x]/(f)$ is a domain and so is:
$$(\mathbb{Z}[x]/(f))/(I/(f))\approx \mathbb{Z}[x]/I$$
So $I/(f)$ is prime in $R$. We have $\pi^{-1}(\langle 3+(f)\rangle)=\langle 3\rangle+\langle f\rangle$. So it suffices to study prime ideals $\mathcal{P}$ in $\mathbb{Z}[x]$ that contain $3$ and $f$ and then we may map these to $R$ via $\pi$.
I tried writing $f(x)=(x^2+Ax+B)(x^2+Zx+D)$ in $\mathbb{Z}[x]$ but this is not possible so $f$ is irreducible (it clearly also has no roots in $\mathbb{Z}$). I am a bit lost on how to find prime ideals with $\langle 3\rangle+\langle f\rangle\subset \mathcal{P}$.
EDIT 1: With Eric's hint we are almost done:
Factoring $f$ in $\mathbb{Z}_3$ yields $(x-1)(x+1)(x^2-x+2)$. So in $\mathbb{Z}[x]$ $f=(x-1)(x+1)(x^2-x+2)+3q\in \mathcal{P}$ and $(x-1)(x+1)(x^2-x+2)\in \mathcal{P}$. Therefore, we have three cases to consider:
Case I: $(x-1)\in \mathcal{P}$. Because $3\in \mathcal{P}$, we need $\langle 3,x-1\rangle\subset \mathcal{P}$, but because of maximality (not sure why it is maximal though), $\langle 3,x-1\rangle= \mathcal{P}$.
Case II: $(x+1)\in \mathcal{P}$. Similarly, $\mathcal{P}=\langle 3,x+1\rangle$.
Case III: $(x^2-x+2)\in \mathcal{P}$. In this case, $\mathcal{P}=\langle 3,x^2-x+2\rangle$.
Maping this via $\pi$ means we are done by the correspondence theorem.
EDIT 2: $\langle p,f\rangle$ is indeed maximal if $f$ is irreducible mod p.
$\phi: \mathbb{Z}[x]\rightarrow \mathbb{Z}_p[x]/(f_p)$ given by the obvious map $\phi(g)=g_p+(f_p)$ where $g_p$ is $g$ modulo $p$ allow us to conclude:
$$\mathbb{Z}[x]/\langle p,f\rangle\approx \mathbb{Z}_p[x]/(f_p)$$
Because $f_p$ is irreducible, then $(f_p)$ is maximal and we have a field! So $\langle p,f\rangle$ is maximal