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Is $x^{14}+x^7+1$ irreducible over $\mathbb{Q}[x]$?

I think it is, but I'm not able to justify it using any existing criterion (e.g. Eisenstein). Any help?

Indeed, this is a question I encounter on a linear algebra one. The original question gives a $8 \times 8$ real matrix satisfying $A^{21}=I$ and asks to prove that $\mathbb{R}^8$ can be decomposed into the direct sum of 4 2-dimensional vector subspace invariant w.r.t. $A$.

My attempt was to find the minimal polynomial then discuss several cases, which need the factor of $x^{21}-1$. Any hint on the original question is also appreciated.

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    $\begingroup$ $x^2+x+1$ is a factor $\endgroup$ Jan 20, 2020 at 4:28
  • $\begingroup$ @J.W.Tanner yeah you are right.. $\endgroup$
    – Lamda8
    Jan 20, 2020 at 4:37
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    $\begingroup$ The roots of $x^2 + x + 1$ are the primitive third roots of unity. They are roots of the above equation. $\endgroup$
    – green frog
    Jan 20, 2020 at 4:37
  • $\begingroup$ Note: $(x^{14}+x^7+1)(x^7-1)=x^{21}-1$; $(x^2+x+1)(x-1)=x^3-1$ divides $x^{21}-1$ $\endgroup$ Jan 20, 2020 at 4:37

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No, it is not irreducible. Indeed, as you observe, $X^{21}-1 = (X^{7}-1)(X^{14}+X^{7}+1)$. On the other hand, $X^{21}-1 = \prod_{d \mid 21} \Phi_{d}(X)$, where $\Phi_{d}(X)$ denotes the $d$th cyclotomic polynomial.

In this case, the divisors of $21$ are very nice: they are $1, 3, 7$ and $21$. It is easy to compute $\Phi_{d}$ for $d$ prime by a familiar Eisenstein computation; we find:

$$\Phi_{d}(X) = \sum_{i=0}^{d-1} X^{i}$$

for $d$ prime.

This gives $\Phi_{3}(X) = X^{2}+X+1, \Phi_{7}(X) = X^{6}+X^{5}+\cdots+1$. Since $\Phi_{1}(X) = X-1$, we can compute $\Phi_{21}(X)$ by long division, for example; it is a polynomial of degree $12$. By unique factorization, we must therefore have that $X^{14} + X^{7}+1$ is irreducible. By degree considerations (or just factoring $X^{7}-1$ as indicated above), one sees that $X^{14}+X^{7}+1 = (X^{2}+X+1)\Phi_{21}(X)$, which confirms the result in the comments by J.W. Tanner.

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  • $\begingroup$ (J.W. Tanner's comments explain the factorization theorem above, and the slick way to arrive at the answer to this question: the main point is that for any integer $n$, $X^{d}-1$ divides $X^{n}-1$ if and only if $d$ divides $n$. But I wanted to explain that in this case, you can explicitly write down the factorization if you really want to!) $\endgroup$ Jan 20, 2020 at 4:55
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You can also apply the technique used in this answer by Will Jagy to a similar question. In our case, we note that $14 \equiv 2 \pmod 3$ and $7 \equiv 1 \pmod 3$ so $\zeta_3^{14} = \zeta_3^2$ and $\zeta_3^7 = \zeta_3$.

Then $\zeta_3^{14} + \zeta_3^7 + 1 = \zeta_3^2 + \zeta_3 + 1 = 0$, so $\zeta_3$ is a root of $X^{14} + X^7 + 1$.

Therefore this polynomial must be divisible by the minimal polynomial of $\zeta_3$ which is $X^2 + X + 1$.

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Noticing that $0,7,14$ is an AP we have

$$x^{14}+x^7+1=\frac{x^{21}-1}{x^7-1}=\frac{\Phi_{21}(x)\Phi_{7}(x)\Phi_3(x)(x-1)}{\Phi_7(x)(x-1)}=\Phi_{21}(x)\Phi_3(x)$$ so the LHS factors as the product of two irreducible (cyclotomic) polynomials with degrees $12$ and $2$.
$$ x^n-1=\prod_{d\mid n}\Phi_d(x) $$ has been used.

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$x^{14}+x^7+1=(x^7-\omega)(x^7-\omega^2)$ $$=(x^7-\omega^7)(x^7-\omega^{14}) ----(1)$$ Factorizing each factors further we get $$(x^7-\omega^7)=(x-\omega)(x^6+x^5\omega +x^4\omega^2 +x^3 +x^2\omega +x \omega^2+1)$$ $$=((1+x^3+x^6)+(x^2+x^5)\omega +(x+x^4)\omega^2 ---(2)$$ Similarly $(x^7-\omega^{14})=(x-\omega^2)((1+x^3+x^6)+(x^2+x^5)\omega^2+(x+x^4)\omega) -----(3)$ $$let\,p(x)=(1+x^3+x^6)\,,\,q(X)=(x^2+x^5)\,,\,r(x)=(x+x^4)$$ Using (1) , (2) , and (3) we get $$x^{14}+x^7+1=(x-\omega)(x-\omega^2)(p(x)+q(x)\omega +r(x)\omega^2)(p(x)+q(x)\omega^2 +r(x)\omega)$$ $$=(1+x+x^2)((p(x))^2+(q(x))^2+(r(x))^2-p(x)q(x)-q(x)r(x)-r(x)p(x))$$

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