The polynomial $x^6+x^3+1$ is irreducible over $\mathbb{Q}[x]$

Question

To solve Dummit-Foote's exercise I want to check that whether the polynomial $p(x)= x^6+x^3+1\in \mathbb{Z}[x]$ is reducible over $\mathbb{Q}$ or not.

I've seen that this polynomial has no zero in $\mathbb{Q}$. But the degree of this polynomial is other than 2 and 3. Therefore, I can not say directly that this polynomial is irreducible over $\mathbb{Q}$.

How do I show that the polynomial $x^6+x^3+1\in \mathbb{Z}[x]$ is irreducible over $\mathbb{Q}$? Can anyone give me some hints?

Thank You in advance.

Answer 1

I would do the following:

1. Note that $p(x)$ is irreducible if and only if $p(x+1)$ is irreducible.

2. Compute $p(x+1)$. We have \begin{align} p(x+1)&=(x+1)^{6}+(x+1)^{3}+1\\ &=\sum_{k=0}^{6}\binom{6}{k}x^{k}+\sum_{k=0}^{3}\binom{3}{k}x^{k}+1\\ &=(x^{6}+6x^{5}+15x^{4}+20x^{3}+15x^{2}+6x+1)+(x^{3}+3x^{2}+3x+1)+1\\ &=x^{6}+6x^{5}+15x^{4}+21x^{3}+18x^{2}+9x+3 \end{align} Can you see that this polynomial is irreducible over $\mathbb{Q}$?

Answer 2

studiosus answer has a much better argument, but in my experience my students always had trouble finding good substitutitions. So here is a harder, less elegant, but more staightforward way.

Note that it is an integer polynomial, so it is enough to decide irreduciblity in $\mathbb{Z}[X]$ by one of the many results which claim the name 'Gauss Lemma'.

If $p$ were reducible over $\mathbb{Z}$, then it would also be reducible after 'going mod 2', i.e. in $\mathbb{F}_2[X]$. However in $\mathbb{F}_2[X]$ it is easy to go through the different cases. Clearly $\overline{p}$ has no roots, i.e. no factors of degree 1. The only irreducible polynomial of degree $2$ is $X^2+X+1$ and one can simply check that $p=(X^2+X+1)(X^4+X^3)+1$. Also there are just two irreducible polynomials of degree 3, namely $X^3+X+1$ and $X^3+X^2+1$. All possible products of those two come out to be different from $p$.

So $p$ does not have factors of degree 1, 2, or 3. So it must be irreducible in $\mathbb{F}_2[X]$, hence in $\mathbb{Z}[X]$, hence in $\mathbb{Q}[X]$.