I am trying to do the problem: is $\mathbb{F}_{2011^2}[x] /(x^4 -6x -12)$ a field?
I know that this is a field if and only if $(x^4 - 6x - 12)$ is a maximal ideal, if and only if $x^4 - 6x - 12$ is irreducible over $\mathbb{F}_{2011^2}$. I also tried to use Rabin's algorithm:
Let $p_1, \ldots, p_k$ be all the prime divisors of $n$, and denote $n / p_i = n_i$, for $1 \leq i \leq k$. A polynomial $f \in \mathbb{F}_q [x]$ of degree $n$ is irreducible in $\mathbb{F}_q [x]$ if and only if gcd $(f, x^{q^ni} - x \mod f )=1$, for $1 \leq i \leq k$, and $f$ divides $x^{q^n}-x$.
In this case that would mean checking if $x^4 - 6x - 12$ divides $x^{2011^8}-x$ which doesn't sound like the most efficient thing to do.
Any help is much appreciated!
It can't be irreducible. The main obstacle is that it has coefficients in the very bottom field $\mathbb F_{2011}$, as well as the fact that a finite field has one and only one extension of each degree (inside a fixed algebraic closure).
First, if it is reducible over $\mathbb F_{2011}$ then it is a fortiori reducible over $\mathbb F_{2011^2}$. But if it is irreducible over $\mathbb F_{2011}$ then a root of it generates a degree $4$ extension of $\mathbb F_{2011}$. This degree $4$ extension must contain the unique degree $2$ extension, $\mathbb F_{2011^2}$, so it is a degree 2 extension of that field. Thus $f$'s roots have degree at most $2$ over the latter field, and so $f$ cannot be irreducible, since irreducibility would tell you that its roots degree $4$.