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

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?

• Please add context to your question. Here are some tips on how to ask a good question. – André 3000 Jan 31 at 16:29
• Hey dude, that's three Problem-Statement-Questions asking for "show irreducible" in two hours. Before you carpet bomb us with any more, please go back and add context to the questions, or at least learn from the earlier responses before you pepper us with more. Thanks – rschwieb Jan 31 at 18:06
• I'm really very sorry for my unexpected manner of questioning. I think I've edited my question properly now. Thanks every one for commenting here. – BijanDatta Feb 1 at 8:17

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 $$\mathbb{Q}$$?

• Eisenstein criterion when p=3, right? – pendermath Feb 1 at 13:43
• @pendermath Exactly – studiosus Feb 1 at 13:44
• However, how did you come up with the substitution p(x+1)? There are many of these to prove (eg, 1, -1, 2, -2, etc.), and success is not even guaranteed. – pendermath Feb 1 at 13:45

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 $$\mathbb{Q}[X]$$.