Show that the set {${\alpha \in F_{5^6} | F_5(\alpha)=F_{5^6} }$} contains 15480 elements

Question

Question: Show that the set {${\alpha \in F_{5^6} | F_5(\alpha)=F_{5^6} }$} contains $15480$ elements

Since this number is so large, I think there is some trick to get the answer.

Also $15480$ is quite near to $5^6$, since $5^6$-15480=145

Would these 145 elements , $\alpha$, be extended roots?

Any help would be much appreciated

Answer 1

Note that $\Bbb F_{5^6}$ contains exactly one(!) subfield isomorphic to $\Bbb F_{5^3}$ and one isomorphic to $\Bbb F_{5^2}$, and they intersect in $\Bbb F_5$. As every proper subfield must be contained in one of these subfields, we conclude that $\Bbb F_{5}(\alpha)\ne \Bbb F_{5^6}$ iff $\alpha\in\Bbb F_{5^3}\cup \Bbb F_{5^2}$. Hence there are $|\Bbb F_{5^3}|+|\Bbb F_{5^2}|-|\Bbb F_{5^3}\cap\Bbb F_{5^2}|=5^3+5^2-5=145$ "bad" $\alpha$.