I need to show:
If $[K:F]$ is finite and $u$ is algebraic over $K$, prove that $[K(u):K]\leq [F(u):F]$.
First if $u$ is algebraic over $K$, it is clear that it is also algebraic over $F$ because we can express the coefficients of polynomials over $K$ as linear combinations of elements of $F$ since $[K:F]$ is finite. What we would like to claim is that a polynomial over $K$ that is irreducible is also irreducible over $F$, because then we can use the minimal polynomial of $[K(u):K]$ and say it must have the lowest degree hence $[K(u):K]\leq [F(u):F]$. The problem though, is that a polynomial over $K$ might not exist as a polynomial over $F$ because $K$ contains elements that are not in $F$ by definition. So what I would like to do is show that the minimal polynomial $[K(u):K]$ must have a lower degree than the minimal polynomial $[F(u):F]$, which is where I am stuck on.