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

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.

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

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.

• (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!) Jan 20, 2020 at 4:55

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 $$X^2 + X + 1$$.

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.

$$x^{14}+x^7+1=(x^7-\omega)(x^7-\omega^2)$$ $$=(x^7-\omega^7)(x^7-\omega^{14}) ----(1)$$ 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 $$x^{14}+x^7+1=(x-\omega)(x-\omega^2)(p(x)+q(x)\omega +r(x)\omega^2)(p(x)+q(x)\omega^2 +r(x)\omega)$$ $$=(1+x+x^2)((p(x))^2+(q(x))^2+(r(x))^2-p(x)q(x)-q(x)r(x)-r(x)p(x))$$