In the polynomial ring $\mathbb Z_3[x]$, the ideal generated by $x^6+1$ is a prime ideal.
Theorem: Let $F$ be a field and $f(x)\in F[x]$. The ideal is $\langle f(x) \rangle$ is maximal iff $f(x)$ is irreducible over $F$.
Here $Z_3$ is a field, and the ideal generated by $x^6+1$ is maximal since $x^6+1$ is irreducible over $\mathbb Z_3$. So, $\langle x^6+1 \rangle$ is maximal and hence prime as every maximal ideal is prime ideal. Hence, the ideal generated by $x^6+1$ is a prime ideal.
Theorem:$F[x]$ is a PID iff $F$ is a field.
Theorem: Let $R$ be a PID. For an non-zero ideal $I$ such that $I \neq R$, then $I$ is prime ideal iff $I$ is maximal ideal. Using the above proof again, $I=\langle x^6+1 \rangle$ is maximal and hence prime.
But my answer key says this is false, am I doing something wrong here? Please try to explicitly point out where I am wrong and also what would be the correct approach. Thanks !