Determining the reducibility of $x^4+2x-2$ in $\mathbb{F}_9[x]$ I am trying to solve Problem 6.10.10 from Berkeley Problems in Mathematics.
Here is the problem statement：

Let $M$ be one of the following fields: $\mathbb{R}, \mathbb{C}, \mathbb{Q}$, and $\mathbb{F}_9$ (the field with nine elements). Let $I \subset M[x]$ be the ideal generated by $x^4+2x-2$. For which choices of $M$ is the ring $M[x]/I$ a field?

My questions:

*

*I have solved the case for $M = \mathbb{R}, \mathbb{C}$ and $\mathbb{Q}$, could you check my solution?

*Could you provide some hints or references for me to solve the case when $M = \mathbb{F}_9$?

Here is my partial solution:
Because $M$ is a field, $M[x]$ is a principal ideal domain, so it suffices to determine for which $M$ the polynomial $p(x):= x^4+2x-2$ is irreducible. For $\mathbb{Q}$, by Eisenstein's criterion we see that $p(x)$ is irreducible in $\mathbb{Z}[x]$ and therefore irreducible in $\mathbb{Q}[x]$. By the fundamental theorem of algebra, all polynomials of degree greater than 2 are reducible in $\mathbb{R}[x]$ and all polynomials of degree greater than 1 are reducible in $\mathbb{C}[x]$, so $p(x)$ is reducible in both polynomial rings.
Now, for $M = \mathbb{F}_9$, I am not familiar with finite fields and after a bit of research I found that we can construct it as $\mathbb{F}_9 = \mathbb{Z}_3[y]/(y^2+1)$. The elements of this field are $0, 1, 2, y, 2y, y+1, y+2, 2y+1, 2y+2$. I verified that none of these elements are roots of $p(x)$, so $p(x)$ does not have a linear factor. How do I proceed from here? Do I write $p(x) = (a_1x^2+b_1x+c_1)(a_2x^2+b_2x+c_2)$ and try to solve it/arrive at a contradiction? Or is there a more straightforward way to do this? In general, how do we determine the reducibility of a polynomial in $\mathbb{F}_n[x]$ where $n$ is not a prime?
Update: I made a computational mistake. We do actually have $p(2) \equiv 0 \bmod 3$ so $p(x)$ has a linear factor and hence is reducible. But I am still interested in the general case regarding how to determine the reducibility of a polynomial over a finite field that doesn't have a linear factor.
 A: Let $f(x)=x^4+2x-2$.

Since $\mathbb{F}_9$ has characteristic $3$, and $f(-1)\equiv 0\;(\text{mod}\;3)$, it follows that $(x+1){\,\mid\,}f$ in $\mathbb{F}_9[x]$, hence $f$ is reducible in $\mathbb{F}_9[x]$.

That was easy, but as we show below, every monic polynomial $f(x)$ of degree $4$ with integer coefficients is reducible in $\mathbb{F}_9[x]$. 

More generally, let $p$ be prime, let $q=p^m$ where $m > 1$, and let $f(x)$ be a non-constant monic polynomial with integer coefficients of degree $n$ where $m{\,\mid\,}n$.

Claim:$\;f$ is reducible in $\mathbb{F}_q[x]$.

Proof:

Assuming the given hypothesis, suppose$\;f$ is irreducible in $\mathbb{F}_q[x]$.

Our goal is to derive a contradiction.

Let $K$ be an algebraic closure of $\mathbb{F}_p$, and let $\theta\in K$ be a root of $f$.

Since $f$ is irreducible when regarded as an element of $\mathbb{F}_q[x]$,$\;f$ must also be irreducible when regarded as an element of $\mathbb{F}_p[x]$.

Thus we get both $[\mathbb{F}_q(\theta):\mathbb{F}_q]=n$ and $[\mathbb{F}_p(\theta):\mathbb{F}_p]=n$.

From $[\mathbb{F}_p(\theta):\mathbb{F}_p]=n$, we get $\mathbb{F}_p(\theta)\cong \mathbb{F}_r$, where $r=p^n$.

But then since $m{\,\mid\,}n$, it follows that $\mathbb{F}_q$ is a subfield of $\mathbb{F}_r$, so $\mathbb{F}_p(\theta)=\mathbb{F}_q(\theta)$, hence
$$
n
=
[\mathbb{F}_p(\theta):\mathbb{F}_p]
=
[\mathbb{F}_q(\theta):\mathbb{F}_p]
=
[\mathbb{F}_q(\theta):\mathbb{F}_q]
[\mathbb{F}_q:\mathbb{F}_p]
=
nm
 >
n
$$
contradiction.
