Does every extension of degree 4 over a field $F$ contain a sub-extension of degree 2 over F? If yes, prove it. If not give a counterexample.
I just want to know if my procedure is right.
There is a theorem that says that if $F \subseteq F_1 \subseteq K$ are fields then $[K:F] = [K:F_1][F_1:F]$, therefore $[F_1:F]$ divides $[K:F]$, since in our case $[K:F] = 4$ and $[F_1:F] = 2$, then $2|4$ and our assertion is right.
Thanks
A counterexample can be given by taking $F=\mathbb Q$ and $K$ to be the extension field by adjoining a root of an irreducible polynomial of degree 4 whose splitting field has the alternating group $A_4$ as its Galois group. Then $F$ is the fixed field of a subgroup $H$ of order 3 of $A_4$, and the subfields of $K$ are in one-to-one correspondence with the subgroups of $A_4$ containing $H$; in particular, a subfield of degree 2 would correspond to a subgroup of $A_4$ of order 6, but $A_4$ has no such subgroup.