I am working on a problem from A Book Of Abstract Algebra by Pinter Chapter 31.A.4 which is
Explain: $\mathbb{Q}(i,\sqrt{2})$ is the root field of $x^4 - 2x^2 + 9$ over $\mathbb{Q}$ and is the root field of $x^2 - 2\sqrt{2}x + 3$ over $\mathbb{Q}(\sqrt{2})$.
The root field of a polynomial over a field is the same thing as I have seen elsewhere referred to as a splitting field of a polynomial over a field. I believe they are equivalent definitions but maybe there is a subtle difference?
I can find the roots of $x^4 - 2x^2 + 9$ by factoring into $(x^2 -1)^2 + 8$ so that the roots are $\pm \sqrt{1 \pm 2i\sqrt{2}}$ so that the root field is $\mathbb{Q}(\sqrt{1 \pm 2i\sqrt{2}})$
I am not sure how to argue that this root field is equal to $\mathbb{Q}(i,\sqrt{2})$ I assume it is using the closure properties of the field but can't work the details. I think the square roots are throwing me off?
We have $$X^4-2X^2+9=(X^2-2\sqrt2 X+3)(X^2+2\sqrt 2X+3).$$ The roots of these quadratic polynomials are $\sqrt{2}\pm i$ and $-\sqrt{2}\pm i$. It is then clear that the splitting field of $X^4-2X^2+9$ over $\mathbf{Q}$ (or "root field" as Pinter calls it) as well as the splitting field of $X^2-2\sqrt{2}X+3$ over $\mathbf{Q}(\sqrt{2})$ is indeed $\mathbf{Q}(\sqrt{2},i)$.
The thing you didn't see was that in fact $$(\sqrt{2}\pm i)^2=2\pm 2\sqrt{2}i-1=1\pm 2\sqrt{2}i,$$ so one can simplify the nested radical $\sqrt{1+2\sqrt{2}i}=\sqrt{2}+i$.