Factor $x^{6}+2$ as a product of irreducible polynomials over C and R

Question

As Title.

It's really a two part question.

Part one is to show if R is an integral domain and $\omega$ is a primitive n th root unity in R, then for any $a\in R$, Show $x^{n}-a^{n}= \prod_{k=0}^{n-1}(x-a\omega^{k})$

Which I've already done.

I don't see the connection between the two questions, could someone please provide some hints?

*Edit: typo and formatting.

Answer 1

For $n=6$ and $\omega=\zeta_6$ the complex roots of $X^6+2=0$ are $\xi_k=\omega^k\sqrt[6]{-2}$ for $k=1,\ldots ,6$. Hence over $\Bbb C$, we have $$ X^6+2=(X-\xi_1)(X-\xi_2)\cdots (X-\xi_6). $$

Answer 2

let $x^3=t$, you have a quadratic equation in $t$, so: $$t^2=-2$$ The solutions to this are: $$t_1=\sqrt{2}i \lor t_2=-\sqrt{2}i$$ Now, you can substitute $z^3$, and you have $z^3=\sqrt{2}i$: $$z^3=r^3(\cos(3\theta)+i\sin(3\theta)=\sqrt{2}(\cos(\pi/2)+i\sin(\pi/2)$$ From here, you have: $$\theta_k=\frac{\pi}{6}+\frac{2k}{3}\pi, k=0,1,2 \: \; \land \; \; r=\sqrt[6]{2}$$

In the second, substitute $z^3=t$ and obtain: $$z^3=r^3(\cos(3\theta)+i\sin(3\theta)=\sqrt{2}(\cos(3\pi/2)+i\sin(3\pi/2)$$ From here, you have: $$\theta_k=\frac{\pi}{2}+\frac{2k}{3}\pi, k=0,1,2 \: \; \land \; \; r=\sqrt[6]{2}$$

In conclusion, you can say that: $$z^6+2=\left(z-\sqrt[6]{2}\left(\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)\right)\right)\cdot\left(z-\sqrt[6]{2}\left(\cos\left(\frac{5\pi}{6}\right)+i\sin\left(\frac{5\pi}{6}\right)\right)\right)\cdot\left(z-\sqrt[6]{2}\left(\cos\left(\frac{3\pi}{2}\right)+i\sin\left(\frac{3\pi}{2}\right)\right)\right)\cdot\left(z-\sqrt[6]{2}\left(\cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)\right)\right)\cdot\left(z-\sqrt[6]{2}\left(\cos\left(\frac{7\pi}{6}\right)+i\sin\left(\frac{7\pi}{6}\right)\right)\right)\cdot\left(z-\sqrt[6]{2}\left(\cos\left(\frac{11\pi}{6}\right)+i\sin\left(\frac{11\pi}{6}\right)\right)\right)$$

Answer 3

Over $\Bbb C$ it splits, by the fundamental theorem of algebra. Use a primitive sixth root of unity and $(-2)^{1/6}$, to get the six roots.

Over $\Bbb R$, we get quadratic factors. The roots come in conjugate pairs, three times.

We get $(x^2+\sqrt[3]2)(x^2-\sqrt[6]2\sqrt3x+\sqrt[3]2)(x^2+\sqrt[6]2\sqrt3x+\sqrt[3]2)$.