Splitting field of $x^3-11$ in $\Bbb Q$

Question

Could somebody please show me the answers and how to get to this answer? Find the splitting field of $f(x)= x^3-11$ in $\mathbb{Q}$ And also how do I find the vector space bases and primitive element?

Thanks!

Answer 1

The roots of $x^3-11$ are $\sqrt[3]{11}, \zeta\sqrt[3]{11}$ and $\zeta^2\sqrt[3]{11}$ where $\zeta$ is the primitive $3$'th root of unit, i.e. $\zeta = e^{2\pi/3}=\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})$.

The splitting field is then $\mathbb{Q}(\sqrt[3]{11},\zeta)$.

Now, note that since $\sqrt[3]{11}$ is algebraic over $\mathbb{Q}$ with mynimal polynomial $x^3-11$, then the elements $1,\sqrt[3]{11}$ and $\sqrt[3]{11}^2$ form a basis for $\mathbb{Q}(\sqrt[3]{11})$ over $\mathbb{Q}$.

Also, $\zeta$ is algebraic over $\mathbb{Q}(\sqrt[3]{11},\zeta)$ with degree $2$ (the minimal polynomial of $\zeta$ is $p(x)=x^{2}+x+1$), then the elements $1,\zeta$ form a basis for $\mathbb{Q}(\sqrt[3]{11},\zeta)$ over $\mathbb{Q}(\sqrt[3]{11})$.

Therefore, a basis for $\mathbb{Q}(\sqrt[3]{11},\zeta)$ over $\mathbb{Q}$ has 6 elements and is obtained by multiplication of the elements of the basis for for $\mathbb{Q}(\sqrt[3]{11})$ over $\mathbb{Q}$. and the elements of the basis for $\mathbb{Q}(\sqrt[3]{11},\zeta)$ over $\mathbb{Q}(\sqrt[3]{11})$, that is, $$\{1, \sqrt[3]{11}, \zeta \sqrt[3]{11},\sqrt[3]{11}^2, \zeta \sqrt[3]{11},\zeta^2 \sqrt[3]{11}^2 \}.$$