The irreducibility of $X^3-X^2+1$ over $\mathbb{F}_9[X]$ 
Let $K$ be the field with $9$ elements. I am asked to study the irreducibility of the polynomial $f:=X^3-X^2+1$ over $K[X]$.

I proceeded as follows. $\mathbb{Z}[i]/3\mathbb{Z}[i]$ is a field with $9$ elements, so $K\cong \mathbb{Z}[i]/3\mathbb{Z}[i]$. If $\phi$ is an isomorphism from $K$ to $\mathbb{Z}[i]/3\mathbb{Z}[i]$, then we may construct a ring homomorphism $\overline{\phi}: K[X] \to (\mathbb{Z}[i]/3\mathbb{Z}[i])[X]$ such that $\overline{\phi}(X)=X$ and $\overline{\phi}(a)=\phi(a)$ for any $a \in K$ via the universal property of the polynomial ring . It is pretty straightforward to show that $\overline{\phi}$ is an isomorphism.
Since $\overline{\phi}(f)=X^3-X^2+\hat{1}$ and $X^3-X^2+\hat{1}$ is irreducible over $(\mathbb{Z}[i]/3\mathbb{Z}[i])[X]$ (it has no roots in $\mathbb{Z}[i]/3\mathbb{Z}[i]$), it follows that $f$ is irreducible over $K[X]$: if it were reducible, we would have $f=gh$ for some nonconstant polynomials $g, h \in K[X]$. This would imply that $\overline{\phi}(f)=\overline{\phi}(g)\overline{\phi}(h)$, so since our $\overline{\phi}$ obviously preserves degree, we would get that $X^3-X^2+\hat{1}$ is reducible over $(\mathbb{Z}[i]/3\mathbb{Z}[i])[X]$, which would be a contradiction.
I would like to know if my solution is correct or if I am missing something.
 A: Suppose it is reducibile over $\mathbb{F}_9[X]$. Then exists such $a\in \mathbb{F}_9$  that: $a^3=a^2-1;\;(*)$. Clearly $a^9=a$ and $a\ne 0$. So we have \begin{align}
    (a^2-1)^3 =a&\implies a^6-3a^4+3a^2-a=1\\ 
&\implies a^4-3a^3+a^2+2a=0\\
&\implies 2a^2=a+1\\
&\implies a^2 = a+2\\
&\implies a=-3\\
(*)&\implies 0=-1\\
\end{align}
A contradiction.
A: Your solution is ok. If a polynomial factors over some field, its homomorphic image surely factors over the isomorphic image of the field.

The taxing part in your solution was to check that the polynomial has no zeros in that field of nine elements. I mean, it was taxing, if you plugged in all the nine elements. A trick to get around that would be to observe that an element $a$ of $K$ is a root of either a linear polynomial or a quadratic polynomial $g(x)$ with coefficients of $f(x)$ in $\Bbb{Z}/3\Bbb{Z}$. All depending on whether $a$ is in the prime field or not (ask, if you don't know how to prove this).
Consequently $g(x)$ must be a factor of $f(x)$. Hence $f(x)=g(x)h(x)$, but this time with both $g(x)$ and $h(x)$ with coefficients in the prime field. One of the factors, $g$ or $h$, is linear. Hence has a zero in the prime field. Hence so does $f$. A contradiction.
