Prove that this field is a splitting field

Question

I was doing a proof to find the degree of the polynomial $X^4+9$, when I encountered a question. The roots of the polynomial in $\mathbb{C}$ are $$\pm\sqrt{6}\left(\frac{1\pm i}{2}\right)$$.

So $\mathbb{Q}(\sqrt{6},i)$ contains all the roots. But how to I prove that there exists no subfield of $\mathbb{Q}(\sqrt{6},i)$ which contains all the roots? In general how would you do this?

Answer 1

Let $F$ be a sub-field of $\mathbb C$ which contains $\pm\sqrt6(\frac{1\pm i}{2})$. Then $$\sqrt6=\sqrt6(\frac{1+i}2)+\sqrt6(\frac{1-i}2)\in F.$$ Moreover, $$i=\frac{\sqrt6i}{\sqrt6}=\frac{\sqrt6(\frac{1+i}2)-\sqrt6(\frac{1-i}2)}{\sqrt6}\in F.$$ Consequently $\mathbb Q(\sqrt6,i)\subseteq F$.

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Hope this helps.

Answer 2

The splitting field is the field $\Bbb K=\Bbb Q (\pm\sqrt{6}\left(\frac{1\pm i}{2}\right))$. As you said $\Bbb K\subset \Bbb Q (i, \sqrt 6)$. But also you have $\sqrt{6}\left(\frac{1+i}{2}\right)+ \sqrt{6}\left(\frac{1-i}{2}\right)=\sqrt 6$ so $\sqrt 6\in \Bbb K $ and also for difference $i\in\Bbb K$. By double inclusion $\Bbb Q(i, \sqrt 6)= \Bbb K$.