Is $\mathbb{F}_{2011^2}[x]/(x^4-6x-12)$ a field? I'm studying for my qualifying exam and this was one of the questions in the question bank under Field and Galois theory section. I'm currently stuck on this question.

Is $\mathbb{F}_{2011^2}[x]/(x^4-6x-12)$ a field?

I'm guessing the answer is no but I don't know how do I prove that $x^4-6x-12$ is an irreducible polynomial over $\mathbb{F}_{2011^2}$. I'm trying to use the following theorem: "let $p$ be a prime. Over $\mathbb{F}_p, x^{p^n}−x$ factors as the product of all monic irreducible polynomials of degrees $d\mid n$."
 A: If it is irreducible over $\mathbb F_{2011}$ then it divides $x^{2011^4}-x.$ But then that means it is a product of distinct quadratic and monic irreducible polynomials in $\mathbb F_{2011^2}.$
This requires the knowledge that:

Over $\mathbb F_q,$ $ x^{q^n}-x$ factors as the product of all monic irreducible polynomials of degrees $d\mid n.$

So if $p(x)$ is an irreducible of even degree over $\mathbb F_q,$ then it factors in $\mathbb F_{q^2}.$

Explicit factorization, taking $\mathbb F_{2011^2}=\mathbb F_{2011}[\sqrt2].$
$$x^4-6x-12=(x^2+15\sqrt{2} x+(225+1810\sqrt{2})(x^2-15\sqrt{2} x+(225-1810\sqrt{2}).$$
A: Extending the comment.

*

*If $f(x)=x^4-6x-12$ is reducible over $\Bbb{F}_{2011}$ then it is also reducible over the extension field. It follows that the quotient ring is not a field.

*But all the extensions of finite fields are normal. So if $f(x)$ is irreducible over $\Bbb{F}_{2011}$ then it splits over the field $K=\Bbb{F}_{2011^4}$. By the basic properties of finite fields $K$ contains a copy of $F=\Bbb{F}_{2011^2}$. So the zeros of $f(x)$ have quadratic minimal polynomials over $F$. Therefore $f(x)$ is reducible over $F$, and we can conclude that this quotient ring is not a field irrespective of whether $f(x)$ is irreducible over the prime field or not.


This was a trick question in the sense that the same argument works for any quartic in place of $f(x)$. All because $\gcd(4,2)>1$.


A general related result is that a degree $m$ polynomial $g(x)$, irreducible in $\Bbb{F}_q[x]$, remains irreducible over $K=\Bbb{F}_{q^n}$ if and only if $\gcd(m,n)=1$. The proof is similar. The roots of $g(x)$ reside in $\Bbb{F}_{q^m}$ which is a subfield of $L=\Bbb{F}_{q^\ell}$, where $\ell=\operatorname{lcm}(m,n)$. Therefore the minimal polynomials of those roots over $K$ have degree $\ell/n=m/\gcd(m,n)$. Those minimal polynomials are the factors of $g(x)$ in $K[x]$.
A: Explicitly, let $\mathbb F_{2011^2}=\mathbb F_{2011}[u]/(u^2-7u+3)$.
Then, $x^4-6x-12=(x^2+(688+378u)x-548-641u)(x^2-(688+378u)x+998+641u)$ in $\mathbb F_{2011^2}$.
(Used Macaulay 2)
