# Splitting field of $x^{4 }-3$ over $\mathbb{Q}$

I'm having trouble in understanding "the form" of a splitting field.

The problem is: Construct the splitting field over $\mathbb{Q}$ of the following polynomials. One of the polynomials is $x^{4}-3$

So, the roots of this polynomial are $\pm \sqrt[4]{3}$ and $\pm \sqrt[4]{3}i$. Therefore, the splitting field over $\mathbb{Q}$ of this polynomial is $\mathbb{Q}[\sqrt[4]{3},i]$, right? But what is the form of these elements?

Like, we have that $\mathbb{Q}[\sqrt{2}]=\{a_1+a_2\sqrt{2};a_1,a_2 \in \mathbb{Q}\}$. So, $\mathbb{Q}[\sqrt[4]{3},i]$ = ?

Remember that $\;Q(\sqrt[4]3\,,\,i)\;$ is a vector space over $\;Q\;$ of dimension $\;4\cdot 2=8\;$ , and it has a very nice basis (putting $\;w:=\sqrt[4]3\;$ for simplicity, we get):
$\;\{1\,,\,w\,,\,w^2\,,\,w^3\,,\,i\,,\,wi\,,\,w^2i\,,\,w^3i\}\;$ , so you can conveniently write any element in the above field in the form
$$a+bw+cw^2+dw^3+ewi+fw^2i+gw^3i\;,\;\;a,b,c,d,e,f,g\in\Bbb Q$$
• @Carol: $$\Bbb Q\le\Bbb Q(w:=\sqrt[4]3)\le \Bbb Q(w,i)$$ so by a simple lemma in fields extensions, the products of a basis of the second over the first times a basis of the third over the second is a basis for the third over the first one: $$\Bbb Q(w)/\Bbb Q=\text{Span}\,\{1,w,w^2,w^3\}\;,\;\;\Bbb Q(w,i)/\Bbb Q(w)=\text{Span}\,\{1,i\}\implies$$ $$\Bbb Q(w,i)/\Bbb Q=\text{Span}\,\{1,w,w^2,w^3,i,wi,w^2i,w^3i\}$$ – DonAntonio Oct 28 '13 at 19:53