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

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?

• There is no such thing as "the minimal polynomial of $E/F$". Minimal polynomials refer to elements of an extension, not extensions themselves. Your idea does work in some cases: if $E/F$ is generated by a single element, then $E(u)/F(u)$ is generated by that same element, and the minimal polynomial of the element in the latter extension divides the minimal polynomial in the former. Can you extend this to the case of $E/F$ generated by multiple elements? Commented Apr 20, 2019 at 20:27

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.