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

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 $$b = 2$$.

• Same comment as above, plus 2 remarks: 1) since you agreed with the correction, you should edit your post to register it, for future readers 2) when asking for a solution verification, you are supposed to specify which step you are doubting and why: math.stackexchange.com/tags/solution-verification/info "For posts looking for feedback or verification of a proposed solution. "Is my proof correct?" is off topic (too broad, missing context)..." Commented Jan 3 at 7:37
2. Plugging in $$x=2$$ should yield the congruence $$2b+13\equiv0\pmod3$$.