Let $F=K(u)$ where $u$ is transcendental over $K$. Prove that it is algebraic over $E$, where $K \subset E \subseteq F$.
The method I tried for the above question was as follows: Choose $v \in E/K$ then $$v=\frac{f(u)}{g(u)} \quad , \quad f(x),g(x) \in K[x] \subset E[x]$$ Then we have that $f(u)=vg(u)$. Define $p(x) \in E[x]$ by $p(x)=f(x)-vg(x)$ so that $p(u)=0$ and hence, $u$ is algebraic over $E$. I think this is correct but I just want to make sure. Thanks!