Let $\mathbb{F}=\mathbb{K}[u]$ where $u$ is transcendental over $\mathbb{K}$. Show that if $\mathbb{K} \subsetneq \mathbb{E} \subseteq \mathbb{F}$ then $u$ is algebraic over $E$.
I'm guessing that I need to use $\mathbb{K}[u] \cong \mathbb{Q}(\mathbb{K}[u])$ somehow but I don't really know how to start. Any hints or a step by step solution would much appreciated.
Sorry, I'm not going to use blackboard bold, but here goes.
Let $p\in E$, $p\not \in K$ since $K\subsetneq E$, then since $E\subseteq F$, we have $p=\frac{f(u)}{g(u)}$, for $f$ and $g$ polynomials in $K[x]$. Then $pg(x)-f(x)\in E[x]$ has $u$ as a root since $pg(u)-f(u)=f(u)-f(u)=0$, note that this polynomial is necessarily nonzero since $p\not\in K$. Thus $u$ is algebraic over $E$.