How to find a minimal polynomial $\sqrt{3}$ over $\Bbb{Q}(\sqrt[6]{3})$?
$\Bbb{Q}(\sqrt{3}) \subset \Bbb{Q}(\sqrt[6]{3})$, because $(\sqrt[6]{3})^{3}=\sqrt{3}$
$[\Bbb{Q}(\sqrt[6]{3}): \Bbb{Q}]=[\Bbb{Q}(\sqrt{3}):\Bbb{Q}] * [\Bbb{Q}(\sqrt[6]{3}): \Bbb{Q}(\sqrt{3}) ]$
$6=2*3$, so minimal polynomial has degree of $3$. But I don't know what to do now. Usually i start with:
$x=\sqrt{3}$
but in this case $\Bbb{Q}(\sqrt[6]{3})$ confuses me.
$\sqrt{3} \in \mathbb{Q}(\sqrt[6]{3})$, so its minimal polynomial is $x - \sqrt{3}$.
$x^6-3=(x^3-\sqrt{3})(x^3+\sqrt{3})$, and the minimal polynomial of $\sqrt[6]{3}$ over $\mathbb{Q}(\sqrt{3})$ is $x^3-\sqrt{3}$.
If "minimal polynomial over $F$" means the minimal polynomial with coefficients in $F$ then the minimal polynomial for $\sqrt3$ over $\Bbb{Q}(\sqrt[6]{3})$ is $t-\sqrt3$.
The fact that $[\mathbb{Q}(\sqrt[6]{3}):\mathbb{Q}(\sqrt{3})]=3$, doesn't tell you that the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt[6]{3})$ is of degree three, instead it tells you that the minimal polynomial of $\sqrt[6]{3}$ over $\mathbb{Q}(\sqrt{3})$ is of degree three and indeed it's $x^3 - \sqrt{3}$.
On the other side proving that $\mathbb{Q}(\sqrt{3}) \subset\mathbb{Q}(\sqrt[6]{3}) $ is enough to conclude that the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt[6]{3})$ is $x - \sqrt{3}$.
