Is $\mathbb{Q}(\sqrt[3]{3},\sqrt{3},i)$ a splitting field over $\mathbb{Q}$?
I am thinking that if i prove $\mathbb{Q}(\sqrt{3})$ is a splitting field over $\mathbb{Q}$ and $\mathbb{Q}(\sqrt[3]{3},i)$ is a splitting field over $\mathbb{Q}$ too, then $\mathbb{Q}(\sqrt{3},\sqrt[3]{3},i)$ is a splitting field over $\mathbb{Q}$. Proving for $\mathbb{Q}(\sqrt{3})$ was easy ,but I'm having trouble for $\mathbb{Q}(\sqrt[3]{3},i)$. If it is,doesnt that mean that $\mathbb{Q}(\sqrt[3]{3},ω\sqrt{3},ω^2\sqrt{3})=\mathbb{Q}(\sqrt{3},\sqrt[3]{3},i)$ ,$ω=\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}$?
Yes, it's the splitting field of the product of minimal polynomials of the generators:
$f(x)=(x^3-3)(x^2-3)(x^2+1)$
The extension is generated by roots of $f$, so we just have to prove that $f$ splits in the extension field. And indeed, if we let $L=\mathbb{Q}(\sqrt[3]{3},\sqrt{3}, i)$ then clearly $x^2+1$ and $x^2-3$ split in $L[x]$. For $x^3-3$ it is a bit less trivial, but it splits too because over $\mathbb{C}$:
$x^3-3=(x-\sqrt[3]{3})(x-\omega\sqrt[3]{3})(x-\omega^2\sqrt[3]{3})$
Where $\omega=e^{\frac{2\pi i}{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$. Note that $\omega,\sqrt[3]{3}\in L$, and so $x^3-3$ indeed splits in $L[x]$.