Let $\alpha = 2\cos(\frac{2\pi}{7})$. What is the minimal polynomial of $\alpha$ over $\mathbb{Q}$?

Question

Let $\alpha = 2\cos(\frac{2\pi}{7})$. What is the minimal polynomial of $\alpha$ over $\mathbb{Q}$?

After a lot of calculation, I found a polynomial such that $\alpha $ is a root: $f(x) = x^6 + 2x^5 -3x^4 - 6x^3 + 2x^2 + 4x + 1$

Now the problem is, I can't tell if this polynomial is reducible or not. I feel like it's irreducible but all my attempts to prove it have either failed, or become so tedious (I believe there must be a better way). If it's irreducible, is there any way I can show that without resorting to a huge system of equations?

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Edit: So according to the comment, $f(x) = (x^3 + x^2 - 2x - 1)^2$. This makes me wonder, is there a way to directly show that $\alpha^3 + \alpha^2 - 2\alpha - 1 = 0$?

The way I originally found the polynomial was by using Chebyshev's method on $\cos(\frac{2\pi}{7})$ and then removing the extra factor $(x-1)$ and then dividing the coefficients by powers of 2 so that $2\cos(\frac{2\pi}{7})$ is a root. However looking at the new polynomial, it looks so simple so I think there must have been a better way, but I don't know what it is.

Answer 1

Yes, you can directly show this. This is a classic trick which you can generalise to many similar examples. Let $\zeta=\exp\left(2\pi i\cdot\frac{1}{7}\right)$.

So $\alpha=\zeta+\zeta^{-1}$. Consider that $\zeta^6=\zeta^{-1},\,\zeta^5=\zeta^{-2},\,\zeta^4=\zeta^{-3}$. The symmetry overlaps at $3$, so I am motivated to consider the polynomial: $$(x-(\zeta+\zeta^{-1}))(x-(\zeta^2+\zeta^{-2}))(x-(\zeta^3+\zeta^{-3}))$$Let's use Vieta's formulae. The sum of roots here is: $$\zeta+\zeta^2+\zeta^3+[\zeta^{-3}=\zeta^4]+\zeta^5+\zeta^6=-1$$(consider the geometric series, e.g., it is known that the sum of the roots of the unity is always zero). The product of the first two roots is: $$(\zeta+\zeta^{-1})(\zeta^2+\zeta^{-2})=\zeta^3+\zeta^{-3}+\zeta+\zeta^{-1}=\zeta+\zeta^3+\zeta^4+\zeta^6$$Similarly, the other three products are: $$\zeta^2+\zeta^3+\zeta^4+\zeta^5,\,\zeta+\zeta^2+\zeta^5+\zeta^6$$If we sum all of these we get two copies of each power of $\zeta$, so the sum is $2\cdot(-1)=-2$. You can calculate the product of all three roots as $1$. Vieta then tells us: $$x^3-(-1)x^2+(-2)x-(1)=x^3+x^2-2x-1$$Is the desired polynomial. It is also clear that if you remove any of the roots, we get non-real coefficients. So this cubic is the minimal polynomial that you want!

Answer 2

Use $\cos 4\theta = \cos 3\theta$. This is satisfied by $4\theta = 2n \pi \pm \cos 3\theta$, so by $\cos\big(\frac{2n \pi}{7} \big)$, but only four of these are distinct (say for $n$ = $0$, $1$, $2$ and $3$). But the equation can also be written $8 \cos^4 \theta -8\cos^2 \theta + 1 = 4 \cos^3 \theta - 3 \cos \theta$, which, on taking out the factor corresponding to the root 1, and dividing the coefficients by powers of 2 as you describe, gives the polynomial you have.