How to prove an element belongs to a splitting field?

Question

Let $K$ be the splitting field of $x^2 + 2$ over $\mathbb{Q}$.

Prove or disprove that $ i = \sqrt{-1}$ is an element of $K$.

Q. How can I prove that? And, in general, how can I prove an element belongs to a splitting field?

Answer 1

If $i\in \mathbb Q(\sqrt{-2})$, there are $a,b\in\mathbb Q$ s.t. $$a+b\sqrt{-2}=i\implies a^2+2\sqrt{-2}ab-2b^2=-1\implies \frac{2b^2-1-a^2}{2ab}=\sqrt{-2}$$$$\implies -2=\left(\frac{2b^2-1-a^2}{2ab}\right)^2,$$ which is a contradiction.

Answer 2

Since $x^2+2$ has degree $2$, its splitting field $K$ is obtained by just adding a root of it, so $K=\mathbb{Q}(\sqrt{-2})=\mathbb{Q}(i \sqrt{2})$.

If $i$ were in $K$ the same would be true for $\sqrt{2}$, so $\mathbb{Q}(\sqrt{2}, \, i) \subseteq K$ which is absurd since the first extension has degree $4$ over $\mathbb{Q}$. In fact, we have a tower of extensions $$\mathbb{Q} \subset \mathbb{Q}(i) \subset \mathbb{Q}(i, \, \sqrt{2}),$$ the second strict inclusion coming from the fact that $$\sqrt{2} \notin \mathbb{Q}(i)= \{z \in \mathbb{C}\, | \, \mathrm{Re}(z), \, \mathrm{Im}(z) \in \mathbb{Q} \}.$$