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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 $d\mid n$."

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    $\begingroup$ 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. $\endgroup$ Jul 11, 2021 at 3:22
  • $\begingroup$ @JyrkiLahtonen How do you actually check that this is irreducible modulo such a large prime, with pencil and paper? $\endgroup$
    – user147556
    Jul 11, 2021 at 3:27
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    $\begingroup$ @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. $\endgroup$ Jul 11, 2021 at 3:31
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    $\begingroup$ 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 $\mathbb F_{2011}[\sqrt{2}]$ $\endgroup$ Jul 11, 2021 at 5:46
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    $\begingroup$ Did your exam exist 10 years ago? $\endgroup$ Jul 11, 2021 at 23:35

3 Answers 3

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

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

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  • $\begingroup$ in the statement that you quoted, is it necessary for $q$ to be prime? $\endgroup$
    – It'sMe
    Jul 11, 2021 at 3:40
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    $\begingroup$ $q$ 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 $q=2021^2.$ $\endgroup$ Jul 11, 2021 at 3:43
  • $\begingroup$ 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). $\endgroup$
    – It'sMe
    Jul 11, 2021 at 3:55
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    $\begingroup$ 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. $\endgroup$ Jul 11, 2021 at 4:03
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    $\begingroup$ That last step require you to know that $$\gcd(q^n-1,q^m-1)=q^{\gcd(m,n)}-1$$ and a polynomial equivalent. $\endgroup$ Jul 11, 2021 at 4:07
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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 $K[x]$.

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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)

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    $\begingroup$ 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 $b,d.$ $\endgroup$ Jul 11, 2021 at 6:01

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