Splitting field justification 
Is $\mathbb Q(\sqrt[3]{2})$ (where $\sqrt[3]{2}$ is the real $\sqrt[3]{2}$) a root field over $\mathbb Q$? Additionally, is $\mathbb Q(i+\sqrt{3})$ a root field over $\mathbb Q$?

I don't think $\mathbb Q(\sqrt[3]{2})$ is a root field because the polynomial $x^3-$, where $\sqrt[3]{2})$ comes from has complex roots that will not be in the extension. 
And $i+\sqrt{3}$ comes from the polynomial $x^4 - 4 x^2 + 16$ which I believe also has complex roots but I think they are included. So I think it may be a root field.
Am I on the right track
 A: Now, $\;\Bbb Q(\alpha)\;$ is always a field whenever $\;\alpha\;$ is a root in, say $\;\Bbb C\;$ ( or any other algebraically closed field containing $\;\Bbb Q\;$ ) of some non-zero rational polynomial.
In our case, the minimal (and thus irreducible) polynomial of $\;\sqrt[3]2\;$ is $\;x^3-2\;$ . If $\;\Bbb Q(\sqrt[3]2\;)\;$ were a splitting field over $\;\Bbb Q\;$ , then it'd contain all the roots of any irreducible rational polynomial having one root in it. Yet, the given field is real, meaning $\;\Bbb Q(\sqrt[3]2)\subset\Bbb R\;$ , whereas the other two (non-real) roots of $\;x^3-2\;$ are complex non-real: $\;\sqrt[3]2\,\omega\,,\,\,\sqrt[3]2\,\omega^2\;$ , with $\;\omega=e^{2\pi i/3}=\;$ a (primitive) root of unity of order three.
Thus, $\;\Bbb Q(\sqrt[3]2)\;$ is a field but not a splitting field over $\;\Bbb Q\;$ as it doesn't contain all the roots of the irreducible $\;x^3-2\in\Bbb Q[x]\;$ , but containing one of them.
A: For the second field $F=\mathbf Q(i+\sqrt 3)$, the minimal polynomial of the generator is $x^4-4x^2+16$. Its roots  are 
$$ \bigl\{i+\sqrt 3,\, -i-\sqrt 3,\, i-\sqrt 3,\,-i+\sqrt 3 \bigr\}.$$
Let's set $\omega=i+\sqrt 3$. The second root lies in $F$, and the third (and therefore the fourth) root too. 
Indeed  try to write
$$i-\sqrt 3=x+y\,\omega+z\,\omega^2+t\,\omega^3=x+y(i+\sqrt 3)+2z(1-i\sqrt3)+8ti. $$
This leads to the system of equations
$$\begin{cases}
x+\sqrt 3\,y+2z=-\sqrt3 ,\\
y-2z\sqrt3+8t=1,
\end{cases}\qquad(x,y,z,t\in\mathbf Q).$$
Since $\sqrt 3$ is irrational, we deduce
$$\begin{cases}
x+2z=0,& y=-1 ,\\
y+8t=1,& z=0.
\end{cases}\quad\text{whence}\quad x=z=0,\;y=-1,\;t=\frac14.$$
As a conclusion, $\mathbf Q(i+\sqrt 3)$ is the splitting field of $x^4-4x^2+16$.
