# Finding the splitting field of $x^6-8$ over $\mathbb{Q}$

Finding the splitting field of $$x^6-8$$ over $$\mathbb{Q}$$.

So, this polynomial factors as $$(x^2-2)(x^4+2x^2+4)$$ and so all the roots will be $$\pm \sqrt{2}$$ and $$\pm \sqrt{-1 \pm i\sqrt{3}}$$. So a splitting field would be $$\mathbb{Q}(\sqrt{-1 \pm i\sqrt{3}},\sqrt{2})$$.

How do I tell if these added roots are linearly independent? Or if adding $$\sqrt{-1 + i\sqrt{3}}$$ includes $$\sqrt{-1 - i\sqrt{3}}$$? And then once I get the splitting field written in it's nicest form can somebody help me calculate the degree of the extension? I'd really appreciate it thanks!

## 1 Answer

Let $$a=\sqrt{2}$$, and let $$b=\sqrt{-3}$$.

Let $$\omega=\exp\left(i\left({\large{\frac{\pi}{3}}}\right)\right)$$.

Noting that

• $$\omega$$ is a primitive $$6$$-th root of $$1$$.$$\\[4pt]$$
• $$\sqrt{8}=\sqrt{2}=a$$.

it follows that the roots of $$x^6-8$$ in $$\mathbb{C}$$ are $$a,a\omega,a\omega^2,a\omega^3,a\omega^4,a\omega^5$$ hence $$K=\mathbb{Q}(a,\omega)$$ is a splitting field of $$f$$.

Then since $$\omega = \cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right) = \frac{1}{2}+\frac{1}{2}i\sqrt{3} = \frac{1}{2}+\frac{1}{2}b$$ it follows that we can also write $$K=\mathbb{Q}(a,b)$$, or explicitly, $$K=\mathbb{Q}\left(\sqrt{2},\sqrt{-3}\right)$$.

Note that $$[\mathbb{Q}(a):\mathbb{Q}]=2$$ and $$[\mathbb{Q}(b):\mathbb{Q}]=2$$.

Since $$a\in\mathbb{R}$$, and $$b\not\in\mathbb{R}$$, it follows that $$[\mathbb{Q}(a,b):\mathbb{Q}(a)] > 1$$.

From $$[\mathbb{Q}(b):\mathbb{Q}]=2$$, we get $$[\mathbb{Q}(a,b):\mathbb{Q}(a)]\le 2$$, hence $$[\mathbb{Q}(a,b):\mathbb{Q}(a)]=2$$

Hence, we get $$[K:\mathbb{Q}] = [\mathbb{Q}(a,b):\mathbb{Q}] = [\mathbb{Q}(a,b):\mathbb{Q(a)}] {\,\cdot\,} [\mathbb{Q}(a):\mathbb{Q}] = 2{\,\cdot\,}2 = 4$$