# How to find a root field of $x^4-2$

Find the root field of $x^4-2$ first over $\mathbb Q$ then $\mathbb R$.

I am struggling to understand root fields and how to form them. I thought we took the root of a polynomial if it is reducible, and then if the root is irreducible among the field then we adjoin the two (field extension). However, I am not understanding the formation part.

For example, the root field of $a(x)=(x^2-3)(x^3-1)$ is $\mathbb Q(\sqrt{3},i)$ because $\pm \sqrt{3}, 1, \frac{1}{2}(-1\pm\sqrt{3}i)$ are the roots. How did they take the roots and find that field?

• Is "root field" the same as "splitting field"? – lhf Dec 5 '17 at 15:00

Hint: The roots of $x^4-2$ in $\mathbb C$ are $\omega\sqrt{2}$, where $\omega$ is a complex number such $\omega^4=1$.
The splitting field of $x^4-2$ is $\mathbb Q(\pm \sqrt{2},\pm i \sqrt{2})=\mathbb Q(\sqrt{2},i)$.
• So a root field is $\mathbb Q(\sqrt{2})$? Is that all I need? – K Math Dec 5 '17 at 15:11
• @KMath, no because $\mathbb Q(\sqrt{2}) \subseteq \mathbb R$ and there are complex roots. – lhf Dec 5 '17 at 15:27
• So maybe a better question I should ask is what makes an extension a root field over $\mathbb Q$? In this instance I thought it was that it included all rational numbers and then element of the form $a+b\sqrt{2}+c\sqrt{2^2}$ where $a,b,c,\in \mathbb Q$. – K Math Dec 5 '17 at 17:08