Does $x^6 + x^2 + 1$ factor over an extension of $\mathbb{Z}$ by roots of unity?

Question

For example, $x^2 - 1$ factors over $\mathbb{Z}[\sqrt{-1}]$.

I would like to ask if the polynomial $x^6 + x^2 + 1$ can be factored over an extension of $\mathbb{Z}$ by roots of unity.

EDIT: The community has requested I elaborate on what I've already tried.

I have tried many things, such as analyzing how the polynomial factors over $\mathbb{F}_p$, but nothing seems like it leads anywhere. The following is the motivation for the question:

The polynomial $x^6 + x^2 + 1$, evaluated at x = 2, returns $69 = 3 \times 23$. Performing this in the inverse for $3$ leads to $x + 1$, and for $23$ leads to $x^4 + x^2 + x + 1$, for coefficients in $\mathbb{Z}/2\mathbb{Z}$. Looking at the polynomial representation for binary representations of semiprimes, most of them seem to have some multiplicative-ish structure going on. For example, $21 = 3 \times 7$ leads to $(x^2 - x + 1)(x^2+x+1)$, which evaluate to 3 and 7 respectively at x = 2.

By asking the question, I wish to come across more intuition as to whether this multiplicative structure holds for extensions of $\mathbb{Z}$.

The given polynomial is among the first corresponding to an odd semiprime that does not factor in $\mathbb{Z}$:

While some of these factors include negative coefficients, to ensure uniqueness corresponding to the binary representation of $pq$, the full expansion of the polynomial enforced coefficients of $0$ or $1$.

Answer 1

No.

We have $x^6+x^2+1=y^3+y+1$ where $y=x^2$. Now $y^3+y+1=0$ has Galois group $S_3$ (guaranteed by one real root and a pair of complex conjugates), which is not abelian, and from here that means its zeroes cannot be expressed in terms of roots of unity. Thus $y^3+y+1$ where $y=x^2$ cannot be factored in terms of such roots, leaving no hope for $x^6+x^2+1$.

Your polynomial for 55 ($x^5+x^4+x^2+x+1$) won't work either. This quintic factors modulo $5$ to give an irreducible cubic factor ($(x+3)(x+4)(\color{blue}{x^3+4x^2+3})$), so the Galois group must include a $3$-cycle. With an irreducible quintic that means it cannot be solved by radicals, whereas a factorization into lower-degree polynomials with roots of unity would imply solvability by radicals.

Answer 2

$u=x^2$

We need a $u$ so that: $u^3+u+1=0$. By Descartes' Sign Rule, there are one negative real root and two complex roots.

$(u+a)(u-b-ci)(u-b+ci)=0$

$(u+a)(u^2-bu+icu-bu+b^2-icb-icu+ibc+c^2 )=(u+a)(u^2+b^2+c^2-2bu)=0$

$u^3+ub^2+uc^2-2bu^2+au^2+ab^2+ac^2-2abu=0$

$u^3+(a-2b)u2+u(b^2+c^2-2ab)+a(b^2+c^2)=0$

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$a=2b$

$b^2+c^2-2ab=1$

$a(b^2+c^2)=1$

$c^2-3b^2=1$ :condition for complex number b+ic to solve $u^3+u+1=0$

$c^2+b^2=1$ : necessary condition for complex solution to be a root of unity.

Simultaneous solutions only occur for $b^2=0$ and $c^2=1 \implies u=\pm i$.

By inspection no values of $u$ having modulus 1 solve $u^3+u+1=0$. Taking a square root preserves the modulus and halves the argument, so no modulus=1 solutions for one imply no such solutions for the other.

Finally, the reals 1 and -1 do not solve $x^6+x^2+1=0$.

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Alternatively:

There are only 8 possible candidates of modulus 1 and they don't work.

$x^6+x^2+1=0$

$z^6+z^2+1=0$

$\bar{z}^6+\bar{z}^2+1=0$

$(z^6-\bar{z}^6)+(z^2-\bar{z}^2)=0$

$(z^3-\bar{z}^3)(z^3+\bar{z}^3)+(z+\bar{z})(z-\bar{z})$

$(z^2-\bar{z}^2)=0 \implies $ z is real or totally imaginary. Only candidates of modulus one are $1,-1,i, -i$.

$(z-\bar{z})(z^2+z\bar{z}+\bar{z}^2)(z+\bar{z})(z^2-z\bar{z}+\bar{z}^2)+(z+\bar{z})(z-\bar{z})=0$

$(z^2+z\bar{z}+\bar{z}^2)(z^2-z\bar{z}+\bar{z}^2)+1=0$

$(z^2+r^2+\bar{z}^2)(z^2-r^2+\bar{z}^2)+1=0$

$z^4-z^2r^2+z^2\bar{z}^2+r^2z^2-r^4+r^2\bar{z}^2+\bar{z}^2z^2-r^2\bar{z}^2+\bar{z}^4+1=0$

$z^4+r^4+\bar{z}^4+1=0$

$r^4e^{4i\theta}+r^4+r^4e^{-4i\theta}+1=0$

Let $ r=1$

$2\cos 4\theta+2=0$

$\cos 4\theta=-1$

$\theta = \pi/4 + 2\pi k/4, k\in \{0,1,2,3\}$

These produce complex numbers of the form $\pm\frac{\sqrt{2}}{2}\pm\frac{\sqrt{2}}{2}$ which aren't extension of the integers without introducing radicals.