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Which of the following is an irreducible factor of $x^{12}-1$ over $\mathbb Q$?

  1. $x^8+x^4+1$
  2. $x^4+1$
  3. $x^4-x^2+1$
  4. $x^5-x^4+x^3-x^2+x-1$

Is the answer $(1)$ correct? I am not sure which one is.

Silvia Ghinassi
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3 Answers3

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Warning: Probably too advanced for this question, but if you want to jump ahead.

If you know the cyclotomic polynomials, this factors into irreducibles as:

$$x^{12}-1=\Phi_1(x)\Phi_2(x)\Phi_3(x)\Phi_4(x)\Phi_6(x)\Phi_{12}(x)$$

Where $\Phi_d(x)$ is the polynomial having the primitive $d$th roots of unity as roots (the $d$th cyclotomic polynomial.)

This is because $x^{12}-1$ has every primitive $d$th root of unity as a root, for $d=1,2$.

So $$\begin{align} \Phi_1(x)&=x-1\\ \Phi_2(x)&=\frac{x^2-1}{\Phi_1(x)}=x+1\\ \Phi_3(x)&=\frac{x^3-1}{\Phi_1(x)} = x^2+x+1\\ \Phi_4(x)&=\frac{x^4-1}{\Phi_2(x)\Phi_1(x)} = x^2+1\\ \Phi_6(x)&=\frac{x^6-1}{\Phi_3(x)\Phi_2(x)\Phi_1(x)} = x^2-x+1\\ \Phi_{12}(x)&=\frac{x^{12}-1}{\Phi_6(x)\Phi_4(x)\Phi_3(x)\Phi_2(x)\Phi_1(x)} = x^4-x^2+1 \end{align}$$

Proving that Cyclotomic polynomials are irreducible is famously non-trivial.

Thomas Andrews
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  • I have to ask; in order to use this method to factorize $x^{12}-1$ do you need to know the forms of cyclotomic polynomials or are you finding them as you go along? Because I think finding $\Phi_{12}(x)$ and factorizing $x^{12}-1$ required the same effort (looking at your formula) – Not Euler Dec 14 '19 at 13:39
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By repeatedly using the simple factorizations: $$x^{2n}-1=(x^n+1)(x^n-1)$$ and $$x^{3n}-1=(x^{2n}+x^n+1)(x^n-1)$$ and $$x^{3n}+1=(x^{2n}-x^n+1)(x^n+1)$$ We get: $$x^{12}-1=(x^4-x^2+1)(x^2+1)(x^2-x+1)(x+1)(x^2+x+1)(x-1)$$ Ruling out every choice except for number 3: $$x^4-x^2+1$$ Call this $p(x)$. There are many ways to show that $p(x)$ is irreducible. For example, we can note that it has no real roots, and hence it must be a product of two 2nd degree polynomials, if it is not irreducible. But $p(2)=p(-2)=13$ and $p(3)=p(-3)$ and $p(4)=p(-4)=241$, so it is a prime number for at least 6 different integer values of $x$. However, a product of two second degree polynomials with integer coefficients can only be prime for $4$ integer values of $x$, since one of the factors must be either $1$ or $-1$ for $p(x)$ to be prime.

Jesse P Francis
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sbares
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Hint: $$x^{12}-1 = (x^6-1)(x^6+1) = (x^3-1)(x^3+1)(x^6+1)=\cdots$$ Can you continue factorizing? (answer: yes; you can factorize $(x^6+1)$ using complex numbers... but the factors maybe $\not\in\Bbb Q[x]$)

Martín-Blas Pérez Pinilla
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    In particular, then, there is no irreducible factor of degree $8$ just from your first equality... – Thomas Andrews Dec 21 '15 at 18:44
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    You can continue to factorize $x^6+1$, of course, without using complex numbers. – Thomas Andrews Dec 21 '15 at 18:58
  • @ThomasAndrews: what about $(x^4-1)(x^8+x^4+1)$? – Will R Dec 21 '15 at 19:00
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    @WillR But the first line indicates taht $x^8+x^4+1$ is not irreducible. This is because $\mathbb Q[x]$ has unique factorization into irreducibles, so the only irreducible factors of $x^{12}-1$ have to be factors of $x^6-1$ or $x^6+1$. So, in particular, they can't be degree $8$. And if you need proof, multiply out $(x^4+x^2+1)(x^4-x^2+1)$ to see that $x^8+x^4+1$ is not irreducible. – Thomas Andrews Dec 21 '15 at 19:02
  • @ThomasAndrews: Thanks, I understand better now. – Will R Dec 21 '15 at 19:07