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

Question

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 $d\mid n$."

Answer 1

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 $\mathbb F_{q^2}.$

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Explicit factorization, taking $\mathbb F_{2011^2}=\mathbb F_{2011}[\sqrt2].$

$$x^4-6x-12=(x^2+15\sqrt{2} x+(225+1810\sqrt{2})(x^2-15\sqrt{2} x+(225-1810\sqrt{2}).$$

Answer 2

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)

Answer 3

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 $\gcd(4,2)>1$.

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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 $K[x]$.