My question is stated below:
Let $K=\mathbb Z_3 [x]$ and $p(x) ∈ \mathbb Z_3 [x]$ be defined by $p(x) = x^4 + x + 2$. Consider the field extension $\mathbb Z_3 [x]/(p(x))$. Define $q(x) ∈ \mathbb Z_3[x]$ by $q(x) = x^4 + 2x^3 + 2$. Find all the roots of the polynomial $q$ in the field extension $\mathbb Z_3 [x]/(p(x))$, if there is any at all. Justify your answer.
I attempted to prove that there is no roots of the polynomial $q$ in the field extension $\mathbb Z_3 [x]/(p(x))$.
If there is a root in the field extension, then there would exists a isomorphism $\phi:\mathbb Z_3 [x]/(p(x)) \to \mathbb Z_3 [x]/(q(x))$ such that $\phi (a)=a$ for all $a \in \mathbb Z_3$ and $\phi (\alpha) = \beta$, where $\alpha$ and $\beta$ are roots of $p(x)$ and $(q(x)$ respectively. But then it implies that $0=\phi (0)=\phi (x^4 + x + 2) = \phi(x)^4 + \phi (x) +2$. Hence $\phi (x)$ and $\beta$ are in the same residue class, and $\beta$ satisfies $p(x)$ and $q(x)$ simultaneously. Contradiction.
Is my "proof" correct?
Very good work and very good comments by Captain Lama, as he already solved the problem in a very nice way.
You can check as follows: let $\;w\;$ be a root of $\;p(x)\;$ in $\;L:=\Bbb Z_3[x]/\langle\; p(x)\;\rangle\;$ (say, $\;w= x+\langle\;p(x)\;\rangle\;$ ) , so that $\;(*)\;\;w^4=-w-2=2w+1\pmod3\;$ , and from here that all the elements of $\;L\;$ are polynomials in $\;\Bbb Z_3[x]\;$ of degree$\,\le3\;$ in $\;w\;$, with usual addition and multiplication modulus $\;3\;$ with the relation $\;(*)\;$.
Thus for example
$$(w+1)^4+2(w+1)^3+2=w^4+\color{red}{w^3}+w+1+\color{red}{2w^3}+2+2\stackrel{(*)}=$$
$$=2w+1+w+1+2+2=0$$
and thus $\;w+1\;$ is a root of $\;g(x)\;$ ... You can play around with the $\;16\;$ elements in $\;L\;$ and discover the other roots both of $\;p(x)\;$ or $\;g(x)\;$ (or any other polynomial you can come up with if degree four or two)