I choose my title "choosing" since I found two splitting field (I know splitting fields are unique!) Here is my solution to the question:
First, observe that $$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$ and I found that the set of solution is: $$\lbrace{\dfrac{-1-i\sqrt{3}}{2},\;\dfrac{-1+i\sqrt{3}}{2},\;\dfrac{1-i\sqrt{3}}{2},\;\dfrac{1+i\sqrt{3}}{2}\rbrace}$$
So I conclude that the splitting field over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt3,i)$.
But then I saw a solution saying splitting field over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{-3})$. At first I thought that they are equal but they aren't since degree of their minimal polynomial are 4 and 2, respectively. So, which one is true? Thanks in advance!
Best Answer
Here’s a way of understanding your polynomial $f(x)=x^4+x^2+1$. Its roots are clearly the square roots of the roots of $g(x)=x^2+x+1$. That is, $f(x)=g(x^2)$.
Now I leave it to you to recognize that the roots of $g$ are the primitive cube roots of unity, $\frac{-1\pm\sqrt{-3}}2$. Their square roots all are sixth roots of unity, though not all of these are primitive. Indeed, as you have recognized, the roots of $f$ fall into two classes: roots of $x^2+x+1$ (primitive cube roots of unity) and roots of $x^2-x+1$ (primitive sixth roots of unity).
I think that you see that by adjoining the roots of $x^2-x+1$, you automatically get the roots of the other factor as a bonus. So your splitting field is quadratic, $\Bbb Q(\sqrt{-3\,}$ .