The degree of the splitting field of $X^6+X^3+1$

Question

Suppose $L \subset \mathbb{C}$ is a splitting field for the polynomial $X^6+X^3+1$ over $\mathbb{Q}$. Determine $[L:\mathbb{Q}]$

I have several solutions for the problem. However I'm having trouble understanding one of them. It goes as follows:

We have shown that $z=\exp(2\pi i/9)$ is a root and all the roots are of the form $z^k$ where $k=1,2,4,5,7,8$. Suppose $f(X)$ is the minimum degree polynomial for $z$. Then there is a standardized polynomial $g(X)$ over $\mathbb{Q}$ s.t.

$$X^6+X^3+1=f(X)g(X)$$

(Now comes the part I have trouble understanding, this solution was provided by teacher): If $z^4$ where a root of $g(X)$ then $g(z^2)=0$ (why?). In mod 2 calculus we would then get that $g(z)^2=0$ and so $g(z)=0$ (mod 2), so $z$ is a root in both f and g mod 2 which implies that the polynomial has root of degree 2 or more contradicting that they are all distinguish.

For the first I don't really understand why $z^4$ of $g$ implies that so is $z^2$. Also does it hold that if $g(z^p)=0$ where p is a prime, then $g(z)^p=0$ mod p (I guess so). Finally, I don't really understand why it is sufficient just to assume that $g$ has the root $z^4$ or is my teacher just skipping the other cases, because the are almost identical to what he provided?

Answer 1

The polynomial you show is of the form $(x^9 - 1)/(x^3-1)$, so it is clear that any 9th root of unity which is not a cubic root (i.e. of order EXACTLY 9) is a root of your polynomial, which contains precisely these roots and no other.

Your polynomial is what is called a cyclotomic polynomial.

Cyclotomic polynomials are irreducible, as one can see in Lang's Algebra or in Ireland and Rosen's Modern Introduction to Classical Number Theory. Clearly, the degree of the cyclotomic polynomial of degree $n$ is $\varphi(n)$ (Euler phi function). I will assume that you know that these polynomials have integer coefficients, Gauss's Lemma and so on (any irreducible factor can be assumed to have integer coefficients, too).

The argument by Dedekind is as follows. Assume that $\phi_n(x)$, your cyclotomic polynomial of $n$-th degree, has an irreducible factor $f(x)$, and choose a root $\zeta$ of $f(x)$. One need prove that, for every $p$ not dividing $n$, $\zeta^p$ is also a root of $f(x)$, and this will imply that $\phi_n(x)= f(x).$ Clearly, $\phi_n(x)$ is a separable polynomial, and so is its reduction (mod $p$) for every $p$ not dividing $n$ (simply because it divides $x^n-1$ which has no multiple roots (mod $p$) for such $p$'s).

Now, if $\phi_n(x)= f(x) h(x)$ where $h(\zeta^p)=0$ (i.e. reductio ad absurdum), one has on $\overline{F}_p[x]$ that

$$f(\zeta^p)\equiv f(\zeta)^p\equiv 0,$$

and so the corresponding root $\zeta^p$ (mod $p$) is a root both of the reductions of $f, h$, hence the polynomial $\phi_n(x)$ has a multiple root (mod $p$), which leads to a contradiction: i.e, if $f(\zeta)=0$ and $p$ is a prime non-factor of $n$, then the root $\zeta^p$ is really a root of $f(x)$, and so $f(x)$ has all roots of order precisely $n$, i.e. $f(x)=\phi_n(x)$.

Hope this and the above references help, else you have a nice argument on page 6 of the following link, which is fully devoted to this matter for all $n$.

http://www.lehigh.edu/~shw2/c-poly/several_proofs.pdf

[I took a little licence here, reducing $\zeta$ (mod $p$), which isn't really such, but the above references will prove fully satisfactory to you :) That is, either you take the ring of integers of the number field in question, or go to the algebraic closure of $F_p$ and take one such root, and so on. This is made clear in the references, Ireland - Rosen if I remember rightly.]