Finding all $c\in \mathbb{Z}_5$ for which $\mathbb{Z}_5[x]/\langle x^3+2x+c\rangle$ is a field. Finding all $c\in \mathbb{Z}_5$ for which $\mathbb{Z}_5[x]/\langle x^3+2x+c\rangle$ is a field.  
I have worked out $0$ is not because it factors to $x(x^2+2)$.
I believe that $c=1,2,3,4$ are all fields because it is irreducible, but I am not confident in this answer.  Can anyone help?
 A: Note that $\mathbb Z_5[x]$ is a PID, so all prime ideals are maximal, i.e. $\mathbb Z_5[x]/\langle x^3+2x+c\rangle$ is a field iff $x^3+2x+c$ is prime (and since PIDs are UFDs, it suffices to show that $x^3+2x+c$ is irreducible). Since $x^3+2x+c$ has degree $3$, if it is not irreduible it must factor into a linear and a quadratic polynomial, thus it has a root. So it suffices to determine for which $c$ it has a root in $\mathbb Z_5$. Since $\mathbb Z_5$ is so small, the easiest way is probably to plug in each element of $\mathbb Z_5$ and see for what $c$ they are roots. This gives us:
$$\begin{align}
x=0 &: c=0\\
x=1 &: 3+c=0\\
x=2 &: 2+c=0\\
x=3 &: 3+c=0\\
x=4 &: 2+c=0\\
\end{align}$$
thus $x^3+2x+c$ is irreducible for $c=1,4$.
A: $\Bbb Z_5[x] / \langle x^3 + 2x + c \rangle$ is a field iff $f(x) = x^3 + 2x + c$ is irreducible in $\Bbb Z_5$. This is because $\Bbb Z_5$ is a field, so $\Bbb Z_5[x]$ is a Euclidean domain. In particular, every prime ideal in $\Bbb Z_5[x]$ is maximal, and an ideal is prime iff it's irreducible.
Since $\operatorname{deg} f(x) = 3$, $f(x)$ is irreducible iff it doesn't have any roots in $\Bbb Z_5$.
For each $k \in \Bbb Z_5$, solve $f(k) = 0$ for $c$ and dismiss this value of $c$. The values that are left in the end give a field. (Hint: Your current answer is wrong.)
