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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!

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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 $1$.$\\[4pt]$
  • $\sqrt[6]{8}=\sqrt{2}=a$.

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

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