Prove that $[E(u): F(u)] \leq [E:F]$.

Question

Given tower of fields $K\supseteq E\supseteq F $, prove that for $u\in K$ algebraic over $F$ and $[E:F]$ finite. This is a problem from W. Keith Nicholson's book.

My idea is that $[E:F]$ is finite hence algebraic. The minimal polynomial of $E(u)/F(u)$ must divide the minimal polynomial of $E/F$. Therefore the result holds?

Answer 1

If $e_1,...,e_n$ is an $F$-basis of $E$, it generates the $F(u)$-vector space $E(u)$.

An element of $E(u)$ is $\sum a_iu^i, a_i\in E$, you can write $a_i=\sum b_{ij}e_j$, this implies that $E(u)$ is generated by $e_1,...,e_n$ as a $F(u)$-vector space.