# Finding the degree of $\Bbb Q(\zeta_9)$

I know that the degree is $6$. But now I need to prove it without using Galois theory or the theorem about the function $\phi(n)$. May I ask a proof with an explicit minimal polynomial of $\zeta_9$ over $\Bbb Q$ with justification? Thanks!

The minimal polynomial is $$x^6 + x^3 +1.$$ That the polynomial is irreducible over $\mathbb{Q}$, follows from the fact that it is irreducible mod 2.

Following Kaj Hansens's comment: To show that the polynomial is irreducible, you have to check that it has no roots (mod 2) and it cannot be factored by an irreducible polynomial of degree 2 or 3 (mod 2). There are not so many of those. A complete list is the following: $$x^2+x+1, \quad x^3+x^2 +1,\quad x^3+x +1.$$

• May I please ask why it is irreducible mod $2$. I think that since it has degree $6$, we cannot simply look at whether it has roots or not.
– Y.X.
Apr 12, 2017 at 6:50
• @PropositionX: why you think that this is not possible and how this is connected to degree 6? Apr 12, 2017 at 6:53
• @PropositionX, it's a little cumbersome, but there aren't many polynomials of degree $2$ or $3$ over $\mathbb{Z}_2$ at all, and a good fraction of them have roots. If this is reducible, it will splits into a product of two cubics or into a product of a quadratic and a quartic. To check whether this is possible all you'd need to do is check for divisibility by the irreducible quadratics and cubics in $\mathbb{Z}_2$ using polynomial long division. As a side note, I'm curious how one would find $x^6 + x^3 + 1$ as the minimal polynomial a priori. Apr 12, 2017 at 6:55
• I have learnt that only for degree $2$ or $3$, no roots implies irreducible.
– Y.X.
Apr 12, 2017 at 6:55
• @PropositionX: I understand. So you have to work a bit harder (see Kaj Hansen's comment) Apr 12, 2017 at 6:57