Quotient Ring of $\mathbb{Z}_3\lbrack x \rbrack$

Question

Find all values of $b$ in $\mathbb{Z}_3$ such that the quotient ring,

\begin{align*} R:= \frac{\mathbb{Z}_3\lbrack x \rbrack}{(x^3+x^2+bx+1)}, \end{align*}is a field.

I would appreciate a double check on my thinking.

Since $\mathbb{Z}_3$ is commutative, $R$ will be a field if and only if the ideal $(x^3+x^2+bx+1)$ is maximal. Since $\mathbb{Z}_3$ is a field, $\mathbb{Z}_3\lbrack x\rbrack$ is a PID, hence any irreducible element in $\mathbb{Z}_3\lbrack x\rbrack$ will generate a maximal ideal. Thus, it suffices to find all $b$ such that $x^3 + x^2 + bx + 1$ is irreducible. Since this polynomial is degree 3, it is sufficient to show that it has no roots in $\mathbb{Z}_3$. Evaluating the polynomial for $x = 0$ will never have a root. For the $x = 1$ and $x = 2$, we see: \begin{align*} b+3 &\equiv 0 \textrm{ mod } 3 \implies b \equiv -3 \textrm{ mod } 3 \equiv 0 \textrm{ mod } 3\\ b + 13 &\equiv 0 \textrm{ mod } 3 \implies b \equiv -13 \textrm{ mod } 3 \equiv 1\textrm{ mod } 3. \ \end{align*}

Since $(x^3 + x^2 + bx + 1)$ is reducible if $b \equiv 0 \textrm{ mod } 3$ or $1 \textrm{ mod }$, the only value making $R$ a field is if $b = 2$.

Answer 1

Your approach is good, just two details are off:

1. You state that any irreducible element will generate a maximal ideal. But is every maximal ideal generated by an irreducible element?
2. Plugging in $x=2$ should yield the congruence $2b+13\equiv0\pmod3$.