Determing all maximal ideals in $\mathbb Z / 4\mathbb Z[x]/(x^4 + 2x + 1)$ Say I have a quotient ring $\mathbb Z/4\mathbb{Z}[x]/(x^4 + 2x + 1)$.
What is the process for finding the maximal ideas of this ring?
 A: I assume you mean $\mathbb{Z}_4[x]$, the ring of polynomials over the integers modulo $4$. Let $p(x)= x^4 + 2x+1$. We seek the maximal ideals of $R=\mathbb{Z}_4[x]/(p(x))$.
First, notice that $p(3)=0$, so $p(x)$ is divisible by $x+1$. We have 
$p(x)=(x+1)(x^3 + 3 x^2 + x + 1)$. The second factor, $q(x) = x^3 + 3 x^2 + x + 1$, is 
irreducible mod 4 because it is of degree $3$ but has no roots mod 4.
We know that a polynomial ring modulo a polynomial is a field if the modulus is prime and the polynomial being modded out is irreducible. Using the isomorphism theorems, you can show that 
$R/(\overline{q(x)}, 2) \cong \mathbb{Z}_2[x] / (\overline{q(x)})$ and 
$R/(\overline{x+1}, 2) \cong \mathbb{Z}_2[x] /( x+1)$. Now,
 $\mathbb{Z}_2[x] / (\overline{q(x)})$ is not a field because $q(1)\equiv 0 \pmod{2}$ implies $q(x)$ is not irreducible mod 2 (in fact it is divisible by $x-1$). However, $\mathbb{Z}_2[x] /( x+1)$ is clearly a field. Therefore of these two ideals, only $(\overline{x+1}, 2)$ is a maximal ideal of $R$. On the other hand, by the isomorphism theorems, the ideals of $R$ correspond to the ideals of $\mathbb{Z}_4[x]$ containing $(p(x))$, which are obtained by the factorization of $p(x)$ and of the modulus $4$, so those were the only two candidates for maximal ideals, and so we are done.
A: If you know about the prime ideals of $\Bbb Z[x]$, and you can recognize this ring as $\Bbb Z[x]/(4,x^4+2x+1)$, then the maximal ideals (among the prime ideals) will be $(p, f(x))$ where $p$ is a prime divisor of $4$ and $f(x)$ mod $p$ is an irreducible polynomial over $\Bbb F_p$ dividing $x^4+2x+1$.
So we've got just $2$ as a candidate for $p$, and then we are looking to see how $x^4+1\pmod{2}$ factors. Luckily, this is easy: $x^4+1=(x+1)^4$. So, you've got one maximal ideal: $(2,x+1)$
A: A slight variation of the answer by rschwieb. Since the element $2$ of  $R=\mathbb Z_4[x]/(x^4 + 2x + 1)$ is nilpotent, it is contained in every prime ideal. Therefore asking for the prime ideals or $R$ is equivalent to asking for the prime ideals of $R/2R\cong \mathbb Z_2[x]/(x^4 + 1)$. Now $\mathbb Z_2[x]$ is a PID, so the non-zero prime ideals are generated by irreducible polynomials. Since $x^4+1=(x+1)^4$ has only one irreducible factor, the only prmie ideal containing it is generated by that factor$~(x+1)$.
Concluding, $R$ has just one prime ideal, generated by $2$ and $x+1$; it is of course also maximal.
