Finding the splitting field of a polynomial. I am confused on how to find the splitting field of a polynomial. For example, consider the polynomial $$p(x)=x^2+2.$$ I know that the splitting field is the smallest field that contains the roots of the polynomial, here $\pm \sqrt{2} i$. I know that the splitting field here is $\mathbb{Q}(\sqrt{2}, i)$ but shouldn't it be $\mathbb{Q} (\sqrt{2} i) $ ?
 A: I don't know how you "know that the splitting field is $\mathbf Q(\sqrt 2, i)$" but it isn't. You may be confusing what the splitting field is in the general case:
The splitting field of $x^n - a$ is $$\mathbf Q(\sqrt[n]{a}, \omega)$$
where $\omega = e^{2\pi i/n}$ is a primitive $n$-th root of unity. This is because the roots of $x^n - a$ are $\sqrt[n]{a}, \omega \sqrt[n]{a}, \dots ,\omega^{n-1} \sqrt[n]{a}$ and so if we have all the roots of $x^n - a$ then we also have
$$ \omega = \frac{\omega \sqrt[n]{a}}{\sqrt[n]{a}}. $$
And conversely, if we have $\sqrt[n]{a}$ and $\omega$ then we have $\sqrt[n]{a}, \omega \sqrt[n]{a}, \dots ,\omega^{n-1} \sqrt[n]{a}$.
When $n = 2$ and $a = - 2$, then $\omega = e^{\pi i} = -1$ (not $\omega = i$).  Therefore the splitting field of $x^2 + 2$ is just $\mathbf Q(\sqrt{-2})$ but you can also verify that this is correct without looking at the general case.
A: A splitting field of a polynomial $f(x) \in F[x]$ is defined as follows: Suppose that the polynomial splits over the a field extension $K/F$ (equivalently, $K$ contains the roots of $f(x)$). Then the splitting field of $f(x)$ is the minimal subfield of $K$ containing $F$ such that $f(x)$ splits.
If we have such a field extension $K$, and $\alpha_1,\cdots,\alpha_n$ are the roots of $f(x)$, then a splitting field will be $F(\alpha_1,\cdots,\alpha_n)$. 
One can show that for any field and polynomial, there is a splitting field and all splitting fields are isomorphic (hence, we refer to "the" splitting field of a polynomial). 
So in your problem you have $x^2+2 \in \Bbb{Q}[x]$. Your polynomial splits over $\Bbb{C}$ and has roots $\pm i\sqrt{2}$. So the splitting field will be $\Bbb{Q}(i\sqrt2,-i\sqrt2)=\Bbb{Q}(i\sqrt2)$
