Let $f = t^6-3 \in F [t] $. Construct a splitting field $K $ of $f $ over $F $ and determine $[K : F] $ for each of the cases: $F = \mathbb{Q, Z/5Z, Z/7Z} $. Do the same thing if $f $ is replaced by $g=t^6+3$.
I am just lost on how to do this one. One way to construct auch a field would be to find irreducible factors of $f $ in each of the three cases and just mod $F[t]$ out by that irreducible polynomial, and then modding the resulting field by another remaining irreducible polynomial, and so on, until it stops. However, the difficulty lies in figurimg out the irreducible factors of $f $ or $g$. It is clear (by pluggin in numbers) that $f $ and $g $ have no linear factors in any of the fields. A friend told me that I can show that$ f $ is itreducible by a hard contradiction, but I am not sure.
I realize that there are lots of parts to my question; any help, however partial, would be great and may be selected as an answer.
In each case the splitting field is $K(\alpha,\zeta)$ where $\alpha$ is a sixth root of $\pm3$ and $\zeta$ is a primitive sixth root of unity, so a zero of $X^2-X+1$.
I would work out $K(\zeta)$ first of all, then ask what the degree of $\alpha$ is over that. For instance $\Bbb Q(\zeta)=\Bbb Q(\sqrt{-3})$. If $\alpha^6=-3$ then $\alpha^3\in\Bbb Q(\sqrt{-3})$ so $|\Bbb Q(\alpha,\sqrt{-3}):\Bbb Q(\sqrt{-3})|\le3$.
On the other hand, $\Bbb F_7(\zeta)=\Bbb F_7$ and $\Bbb F_5(\zeta)=\Bbb F_{25}$ as $6\mid(7-1)$ and $6\nmid(5-1)$. In $\Bbb F_7$, $3$ is neither a square nor a cube, so $\alpha$ has degree $6$ over $\Bbb F_7$ etc.