# 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 roadmap: 1) What is the answer if $x^4-6x+12$ happens to be reducible over the prime field $\Bbb{F}_{2011}$. 2) Assuming that it is irreducible over the prime field, what can you say about the extension degree of its splitting field? 3) In view of item 2 you should be able to answer the question also when it is irreducible over the prime field. Jul 11 at 3:22
• @JyrkiLahtonen How do you actually check that this is irreducible modulo such a large prime, with pencil and paper?
– user147556
Jul 11 at 3:27
• @MichaelBarz I don't want to do that with paper and pencil :-) There are algorithms though. Anyway, at the risk of spoiling things, the answer to the question is independent of whether it is irreducible over the prime field or not. Jul 11 at 3:31
• Fractorization: $$x^4-6x-12=(x^2+15\sqrt{2} x+(225+1810\sqrt{2})(x^2-15\sqrt{2} x+(225-1810\sqrt{2})$$ where we repressent $\mathbb F_{2011^2}$ as $\mathbb F_{2011}[\sqrt{2}]$ Jul 11 at 5:46
• Did your exam exist 10 years ago? Jul 11 at 23:35

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 $$\mathbb F_{2011^2}=\mathbb F_{2011}[\sqrt2].$$

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

• in the statement that you quoted, is it necessary for $q$ to be prime? Jul 11 at 3:40
• $q$ can be any prime power. (A common variable choice when talking finite fields.) @FreePawn The whole point here is that it apples when $q=2021$ and $q=2021^2.$ Jul 11 at 3:43
• Thanks for your answer. Actually, I knew your quoted result in the case of $q$ being a prime, but I didn't know that it also holds for power of prime and it didn't come to my mind to prove the general case (for power of prime). Jul 11 at 3:55
• The same reasoning apply to any $q.$ If $f$ is irreducible of degree $n,$ then the quotient field is of size $q^n$ and thus any $y^{q^n}=y$ in that field, so $f$ must divide $x^{q^n}-x.$ And since $x^{q^{nm}}-x$ is divisible by $x^{q^n}-x,$ you get the other divisors. Then you need to show these are the only ones. Jul 11 at 4:03
• That last step require you to know that $$\gcd(q^n-1,q^m-1)=q^{\gcd(m,n)}-1$$ and a polynomial equivalent. Jul 11 at 4:07

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 $$K[x]$$.

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)

• Easier to do it with a simple square extension, like $\mathbb F[\sqrt{2}].$ Then the factorization has to be of the form $(x^2+(a+b\sqrt2)x+(c+d\sqrt 2))(x^2+(a-b\sqrt2)x+(c-d\sqrt{2}))$ and you quickly get $a=0,c=b^2$ and $b^4-2d^2=-12, 2bd=3.$ This was mostly by hand, but then I used a computer program to solve for $b,d.$ Jul 11 at 6:01