# simplifying root fields / splitting field on Pinter 31.A.4 problem

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?

• Did you mean $\mathbb{Q}(i,\sqrt{2})$ where you typed $\mathbb{Q}(1,\sqrt{2})$? – J. W. Tanner Apr 20 '20 at 14:16
• Yes, thanks. I'll fix it. – topoquestion Apr 20 '20 at 14:17
• Try. thinking of succesive adjunctions – Chris Leary Apr 20 '20 at 14:31

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 $$\sqrt{1+2\sqrt{2}i}=\sqrt{2}+i$$.
• That's. I didn't see that simplification. I'm not clear fully on how to compare root fields but I think theorem 2 of this chapter shows $\mathbb{Q}(i + \sqrt{2}) = \mathbb{Q}(i,\sqrt{2})$ – topoquestion Apr 20 '20 at 17:08
• That is true, but the reason here is the following. The splitting field is the field extension of $\mathbf{Q}$ obtained by adjoining all roots: $\mathbf{Q}(i+\sqrt{2},-i+\sqrt{2},i-\sqrt{2},-i-\sqrt{2})=\mathbf{Q}(i+\sqrt{2},i-\sqrt{2})=\mathbf{Q}(i,\sqrt{2})$. The last equality is true because $\boxed{\subset}$ is obvious and $\boxed{\supset}$ is true because $\frac12((i+\sqrt{2})+(i-\sqrt{2}))=i$ and $\frac12((i+\sqrt{2})-(i-\sqrt{2}))=\sqrt{2}$. Is is clearer now? :) – rae306 Apr 20 '20 at 20:30
• That makes sense, thanks. I missed the $i-\sqrt(2)$ part and couldn't get $i$ and $\sqrt{2}$ separated without it. – topoquestion Apr 20 '20 at 20:41