Irreducible Polynomial Field Extensions with Root $\cos \frac{2\pi}{7}$

Question

Show that $\theta = \frac{2k\pi}{7}$ satisfies the equation $\cos 4\theta − \cos 3\theta =0$ for each integer $k$. Hence find an irreducible polynomial over $\Bbb Q$ with $\cos \frac{2\pi}{7}$ as a root.

I'm not sure how to start this one... The book I am using doesn't go over how to deal with transcendental numbers in this way.

Thank you in advance!

Answer 1

Using Chebyshev polynomials, we can write $\cos 4\theta − \cos 3\theta =0$ as $T_4(\cos \theta) - T_3(\cos \theta)=0$.

Now, $$ T_4(x)- T_3(x) = (8x^4-8x^2+1) - (4x^3-3x) = (x - 1) (8 x^3 + 4 x^2 - 4 x - 1) $$ Finally, $8 x^3 + 4 x^2 - 4 x - 1$ is irreducible over $\mathbb Q$ because it has no rational root. (Check!) Alternatively, it has no root mod $3$.

Answer 2

From the comments below it appears that OP is also interested in the first half of the question (which I had originally overlooked as just a substitution).

Hint For the first part of the question, we are essentially asked to show that $$\cos \left(4 \cdot \frac{2 \pi k}{7}\right) - \cos \left(3 \cdot \frac{2 \pi k}{7}\right) = 0$$ for all integers $k$. Here are two options:

1. Substitute in the sum-to-product formula $\cos \alpha - \cos \beta = -2 \sin (\alpha + \beta) \sin (\alpha - \beta)$. (This method shows, by the way, how we might extend the problem when we replace $7$ with a general nonzero integer.)

2. Alternatively, since $\cos$ has period $2 \pi$, it suffices to check just for $0 \leq k < 7$. Then, using that $\cos$ is even, and hence $\cos (2 \pi - \theta) = \cos \theta$, it suffices to check just for $k = 0, 1, 2, 3$, and checking these cases manually is straightforward and only requires using the identities already discussed.

For the second part, lhf has already given a (good) canonical answer. For a reader who isn't already familiar with Chebyshev polynomials, here's a method that uses only elementary trigonometric identities (this essentially amounts to computing the relevant Chebyshev polynomials by hand).

If we can find a polynomial $f$ such that $$f(\cos \theta) = \cos 4 \theta - \cos 3 \theta, $$ then since $\theta = \frac{2 \pi}{7}$ is a root of the right-hand side, $\cos \frac{2 \pi}{7}$ is a root of $f$.

Expanding $\cos 4 \theta$ using the double angle formula gives $$\cos 4 \theta = 2 \cos^2 2 \theta - 1 = 2 (2 \cos^2 \theta - 1)^2 - 1 = 8 \cos^4 \theta - 4 \cos^2 \theta + 1.$$

Proceeding similarly gives $\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta.$ Putting this together gives that $\frac{2 \pi}{7}$ is a root of $$8 \cos^4 \theta - 4 \cos^3 \theta - 8 \cos^2 \theta + 3 \cos \theta + 1$$ and thus $\cos \frac{2 \pi}{7}$ is a root of $$f(x) := 8 x^4 - 4 x^3 - 8 x^2 + 3 x + 1 .$$ Now, we already know that $f$ is not irreducible: Since $\theta = 0$ is a root of $\theta \mapsto \cos 4 \theta - \cos 3 \theta$, $\cos 0 = 1$ is a root of $f$ and hence $x - 1$ is a factor of $f$. Performing polynomial long division gives $$(x - 1) (\color{#df0000}{\underbrace{8 x^3 + 4 x^2 - 4 x - 1}_{g(x)}}) .$$ Since $\frac{2 \pi}{7}$ is not an integer multiple of $2 \pi$, $\cos \frac{2 \pi}{7} \neq 1$ and hence $\cos \frac{2 \pi}{7}$ is a root of the cubic polynomial $\color{#df0000}{g(x)}$. Applying the Rational Root Theorem shows that $g$ is irreducible (alternatively, use that $g(x + 1)$ is Eisenstein at $7$), so it satisfies the required condition.

Answer 3

Another way is to use the fact that $\zeta_7 = \cos \frac{2\pi}{7} + i \sin \frac{2\pi}{7}$ is a $7$-th root of unity. Now $\cos \frac{2\pi}{7} = \frac{\zeta_7 + \zeta_7^{-1}}{2}$. Now the $7$-th cyclotomic polynomial is:

$$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$$

This is satisfied by $\zeta_7$. Divide by $\zeta_7^3$ to get:

$$(\zeta_7^3 + \zeta_7^{-3}) + (\zeta_7^2 + \zeta_7^{-2}) + (\zeta_7 + \zeta_7^{-1}) + 1 = 0$$

$$(\zeta_7+\zeta_7^{-1})^3 + (\zeta_7+\zeta_7^{-1})^2 + (\zeta_7+\zeta_7^{-1}) + 1 - (3\zeta_7 + 3\zeta_7^{-1} + 2) = 0$$

$$(\zeta_7+\zeta_7^{-1})^3 + (\zeta_7+\zeta_7^{-1})^2 - 2(\zeta_7+\zeta_7^{-1}) - 1 =0$$

Now divide by $8$ to get that:

$$\left(\frac{\zeta_7 + \zeta_7^{-1}}{2}\right)^3 + \frac 12 \left(\frac{\zeta_7 + \zeta_7^{-1}}{2}\right)^2 - \frac 12 \left(\frac{\zeta_7 + \zeta_7^{-1}}{2}\right)- \frac 18 = 0$$

Hence $\cos \frac{2\pi}{7}$ satisfy $x^3 + \frac 12 x^2 - \frac 12 x - \frac 18 = 0$ and is not hard to prove that it's irreducible.

Remark: It's not hard to see that the other roots are $\cos \frac{4\pi}{7} = \frac{\zeta_7^2 + \zeta_7^{-2}}{2}$ and $\cos \frac{6\pi}{7} = \frac{\zeta_7^3 + \zeta_7^{-3}}{2}$

Answer 4

The identity $\cos4\theta-\cos3\theta=0$ for $\theta=2k\pi/7$ is not needed (but is not difficult to prove).

The numbers $z=e^{2ik\pi/7}$ for $0<k<7$ satisfy $$ z^6+z^5+z^4+z^3+z^2+z+1=0 $$ hence also $$ z^3+\frac{1}{z^3}+z^2+\frac{1}{z^2}+z+\frac{1}{z}+1=0 $$ Recall that $$ z^2+\frac{1}{z^2}=\left(z+\frac{1}{z}\right)^2-2 \qquad z^3+\frac{1}{z^3}=\left(z+\frac{1}{z}\right)^3-3\left(z+\frac{1}{z}\right) $$ and notice that $$ z+\frac{1}{z}=z+\bar{z}=2\cos(2k\pi/7) $$ Substituting, you get $$ 8\cos^3(2k\pi/7)-6\cos(2k\pi/7)+4\cos^2(2k\pi/7)-2+2\cos(2k\pi/7)+1=0 $$ and finally $$ 8\cos^3(2k\pi/7)+4\cos^2(2k\pi/7)-4\cos(2k\pi/7)+1=0 $$ The polynomial $8x^3+4x^2-4x+1=0$ is easily seen to be irreducible: set $y=2x$, so you get $y^3+y^2-2y+1=0$ and the rational root test applies easily.